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Torsion subgroups of elliptic curves in elementary abelian 2-extensions and monodromy of Fermat surfaces

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Torsion subgroups of elliptic curves in elementary abelian 2-extensions and monodromy of Fermat surfaces

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Content TORSION SUBGROUPS OF ELLIPTIC CURVES IN ELEMENTARY ABELIAN 2-EXTENSIONS AND MONODROMY OF FERMAT SURFACES by Ozlem Ejder 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 2017 Copyright 2017 Ozlem Ejder To Leyla ii Acknowledgments First and foremost I would like to thank my advisors Eric Friedlander and Sheldon Kamienny for their endless support. They have been a great source of inspiration and encouragement. Without them, this thesis would not be possible. I would also like to thank Burton Newman for entertaining my mathematical thoughts, sharing his knowledge and always being so enthusiastic about talking mathematics. He helped me speed up the learning process and save a lot of time. USC Math department has been a great family to me throughout these years. I would like to thank all of them. I would also like to thank Aravind Asok whom I see as a mentor and a great source of inspiration, for being always open to hear my questions and his advices. I would nally like to thank my family; my parents, my sisters Eylem and Gurbet, my brother Omer, my nephew Can, my lovely cat Leyla, and my dear Isaac for their love. iii Table of Contents Dedication ii Acknowledgments iii Abstract vi Chapter 0: Introduction 1 Chapter 1: Background on Elliptic Curves 5 1.1 Isogenies of Elliptic Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Elliptic Curves OverC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 Elliptic Curves With Complex Multiplication . . . . . . . . . . . . . . . . . 7 1.4 Reduction of Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.5 Modular Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Chapter 2: Torsion Subgroups in Elementary Abelian 2-Extensions 12 2.1 Rational Points on the Modular Curve Y 0 (n) . . . . . . . . . . . . . . . . . 12 2.1.1 The modular curve X 0 (20) . . . . . . . . . . . . . . . . . . . . . . . 13 2.1.2 The modular curve X 0 (27) . . . . . . . . . . . . . . . . . . . . . . . 14 2.2 Odd Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.3 Solutions to the Equation x 4 +y 4 = 1 . . . . . . . . . . . . . . . . . . . . . 21 2.4 Restrictions on the Torsion Subgroups . . . . . . . . . . . . . . . . . . . . . 27 2.5 E(K) tors is Non-Cyclic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.6 E(K) tors is Cyclic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2.7 More Restrictions on the Torsion Subgroups . . . . . . . . . . . . . . . . . . 45 2.8 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Chapter 3: Monodromy of Fermat Surfaces 50 3.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.1.1 Modular Interpretation of Fermat curves . . . . . . . . . . . . . . . . 50 3.1.2 Cusps of X((n)) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.1.3 Automorphisms of X((n)) . . . . . . . . . . . . . . . . . . . . . . . 52 3.1.4 Modular Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.2 Modular Symbols for Fermat Curves . . . . . . . . . . . . . . . . . . . . . . 54 3.2.1 -relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 iv 3.2.2 -relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.2.3 Homology of Fermat Curves . . . . . . . . . . . . . . . . . . . . . . . 58 3.3 Fermat Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.3.1 Denition of Monodromy . . . . . . . . . . . . . . . . . . . . . . . . 60 3.3.2 Description of the Fibration . . . . . . . . . . . . . . . . . . . . . . . 61 3.4 Calculation of the Monodromy . . . . . . . . . . . . . . . . . . . . . . . . . 63 Bibliography 68 v Abstract This thesis is focused on two problems; one is concerned mainly with arithmetic where as the other is more geometric. In Chapter 2 , we let K denote the quadratic eldQ( p d) for d =1;3 and letE be an elliptic curve dened overK. The main result of this chapter is the classication of the torsion subgroups ofE in the compositum of all quadratic extensions of K. In Chapter 3, we have a parametrized family of Fermat curves. Using the modular description of Fermat curves and the modular symbols, we rst compute a set of generators for the rst integral homology group and then compute the monodromy of this family. vi Chapter 0 Introduction Studying the set of rational solutions of a polynomial equation, which can also be seen as a geometric object, such as a curve or a surface, is one of the fundamental problems in number theory. Both the arithmetic and the geometric invariants are eciently used to understand the arithmetic of polynomials. Given a number eld K and an algebraic curve C=K, the set C(K), the set of points onC which are dened over K, has the following properties depending on the genus of the curve. 1. If C has genus 0, then C(Q) is either empty or innite. 2. If C has genus greater than 1, then C(Q) is either empty or nite. This is Faltings's celebrated theorem. Assume C has genus 1. If C(K) is not empty, then it forms a nitely generated abelian group, proven by Luis Mordell and Andr e Weil. Elliptic curves are curves of genus 1 and they are one of the main objects of this thesis. For an elliptic curve E=K, the group E(K) has the structure E(K) tors Z r : since it is nitely generated. HereE(K) tors is the nite part of this groupE(K), it is called the torsion subgroup and the integer r is called the rank of E over the number eld K. Both the rank and the torsion subgroup of elliptic curves have been studied extensively. In the rst part of this thesis, we will study the elliptic curves over the quadratic cyclotomic elds and how their torsion subgroups grow in the compositum of all quadratic extensions of the base eld. First, we give a summary of the results obtained so far on E(K) tors for a number eld K. Mazur [Maz78] in 1977 showed that the only groups that can be realized as the torsion subgroup of an elliptic curve dened over Q are the following: Z=mZ for 1m 12;m6= 11; or Z=2ZZ=2mZ for 1m 4: 1 The rank of an elliptic curve is conjectured to be arbitrarily large. In 2006, Noam Elkies found an elliptic curve overQ with rank at least 28. A similar list for the torsion groups of elliptic curves dened over a quadratic eld has been given by Sheldon Kamienny [Kam92], M.A. Kenku, and F. Momose [KM88]. Such a group has to be in the list given below. Z=mZ for 1m 18;m6= 17; Z=2ZZ=2mZ for 1m 6; Z=3ZZ=3mZ for m = 1; 2; and Z=4ZZ=4Z: If one xes the quadratic eld K, it is very likely that we will have a smaller list. In fact, the groupsZ=3ZZ=3Z andZ=3ZZ=6Z are only realized when K =Q( p 3) where as the group Z=4ZZ=4Z is only realized over the eld Q(i) since they contain the root of unity for 3 and 4 respectively (See Weil pairing). The following result is due Filip Najman [Naj10]. 1. If K = Q(i), then E(K) tors is either one of the groups from Mazur's theorem or Z=4ZZ=4Z. 2. If K =Q( p 3), then E(K) tors is either one of the groups from Mazur's theorem or Z=3ZZ=3Z orZ=3ZZ=6Z. One may also ask how the torsion subgroups of elliptic curves over a given number eld K grow over the compositum of all the quadratic extensions of K. Let F be the maximal elementary abelian two extension of K, i.e., F :=K[ p d :d2O K ] whereO K denotes the ring of integers ofK. The problem of ndingE(F ) tors whereK =Q has been studied by Michael Laska, Martin Lorenz [LL85], and Fujita [Fuj05, Fuj04]. Laska and Lorenz described a list of 31 possible groups and Fujita proved that the list of 20 dierent groups is complete. The main theorem of Chapter 2 generalizes the results of Laska, Lorenz and Fujita to the case where K is a quadratic cyclotomic eld. We nd that 1. If K =Q(i), then E(F ) tors is isomorphic to one of the following groups: Z=2ZZ=2NZ for N = 2; 3; 4; 5; 6; 8; Z=4ZZ=4NZ for N = 2; 3; 4; 2 Z=NZZ=NZ for N = 2; 3; 4; 6; 8; or f1g;Z=3Z;Z=5Z;Z=7Z;Z=9Z;Z=15Z: 2. If K =Q( p 3), then E(F ) is either isomorphic to one of the groups listed above or Z=2ZZ=32Z: The rst chapter of this thesis is reserved for the background on elliptic curves. The second chapter studies the elliptic curves over the quadratic imaginary cyclotomic elds; we rst study the points on various modular curves and we use these results in the proof of Proposition 2.1.1 and inxx 2.4. Theorem 2.4.6 gives us a large list of torsion subgroups. In xx 2.5, we analyze the growth of each non-cyclic torsion subgroup over the base eld which helps us eliminate some of the groups given in the previous list. Analyzing the case where E(K) is cyclic, we obtain more restrictions on the torsion subgroups inxx 2.7 and we prove our main result Theorem 2.8.1. The rest of the thesis i.e.,x3 is concerned with a geometric problem of computing the monodromy. The monodromy describes how certain objects behave when it runs around the singularity. As an example, consider the mapC!C takingz7!z n which is a branched covering ofA 1 C and the monodromy interchanges the sheets of the covering. Instead of looking at one curve, we may also take a family of curves parametrized by another curve, which is an example of a bration. We call the members of this family as bers, the monodromy in this case is often described in terms of homology or cohomology of a smooth ber. The rst non-trivial example is a family of elliptic curves which may also be called an elliptic surface. In [Kod64, Kod66], Kodaira classies all singularity types in elliptic brations and com- putes the local monodromy around each singularity type. Since the rst homology group of an elliptic curve is free abelian group of rank 2, the local monodromy is given as a matrix in SL 2 (Z). Although the local monodromy of elliptic surfaces are completely known, we do not have a general result on families of higher genus. For the purposes of this thesis, we will work with a surface which bers over an ane curve. Our main object inx3 is a family of Fermat curves of xed degree n, i.e., x n +y n =z n : In [Bel79], Belyi shows that for any smooth projective curve C, there exist a nite index subgroup G of SL 2 (Z) such that the quotient of the upper half plane by the action of G is isomorphic to C. For example, Fermat's cubic x 3 +y 3 = z 3 is an elliptic curve with j 3 invariant 0 and it is isomorphic to X 0 (36); hence G is 0 (36). Similarly, Fermat's quartic x 4 +y 4 =z 4 is isomorphic to the modular curve X 0 (64). This leads us to the question of whether there are subgroups of SL 2 (Z) that are known for Fermat curve of each degree n. In [Roh77], Rohrlich gives a modular description of Fermat curves in terms of certain subgroups (n) of (2). The homology of modular curves can be described by a subset of modular symbols. A modular symbol is a pair of cusps i.e.,f;g for some; inP 1 (Q) under some equivalence relation. One should see this symbol as a path from to . Manin [Man72] describes a set of generators for the group of modular symbols for a nite index subgroup G of SL 2 (Z); moreover he gives all the relations between these generators, so called the Manin symbols. Inx3, we compute the Manin symbols for the group (n) and nd a set of generators for the rst integral homology group in Proposition 3.2.3. Moreover, the group of modular symbols computed in Proposition 3.2.2 is the homology group of a Fermat curve of degree n relative to the set of cusps (or the points at innity), hence Proposition 3.2.2 shows that the relative homology group has rank n 2 + 1. Inxx 3.3, we describe a family of Fermat curves, its singularities, and the fundamental group of the parameter space. We show that the monodromy action on the bers of this family is the automorphism given by (x :y :z)7! ( n x : n y :z) and determine its action on the generators of the homology given in Proposition 3.2.3. In the rst chapter, the rank and the torsion subgroup computations are done by using the computer algebra software program Magma [BCP97]. 4 Chapter 1 Background on Elliptic Curves 1.1 Isogenies of Elliptic Curves. Let K be a number eld. Denition. LetE 1 andE 2 be elliptic curves. An isogeny betweenE 1 andE 2 is a morphism of curves :E 1 !E 2 such that (O) = O. The elliptic curves E 1 and E 2 are called isogenous if there is an isogeny :E 1 !E 2 such that (E 1 )6=fOg. The kernel of an isogeny is always nite, hence we can make the following denition. Denition. The degree of an isogeny is dened as the order of its kernel. The elliptic curves E 1 andE 2 are said to ben-isogenous if there is an isogeny of degree n between them. Also we call an isogeny f as n-isogeny if its degree is n. Let Hom(E 1 ;E 2 ) =fisogenies :E 1 !E 2 g: Let E be an elliptic curve dened over K. We call a subgroup C of E(C) K-rational if there exist an elliptic curve E 0 =K and an isogeny :E!E 0 dened over K such that C = ker. We call the endomorphism ring of an elliptic curve E as End(E) = Hom(E;E): We need the following denition before we describe endomorphism ring of an elliptic curve. Denition ([Cox89]). An orderO in a quadratic eld K is a subsetOK such that 1. O is a subring of K containing 1. 5 2. O is a nitely generatedZ-module. 3. O contains aQ-basis of K. Given an orderO, the index [O K :O] is called the conductor ofO and it is denoted by f. Theorem 1.1.1 ([Sil09]). Let K be a number eld and let E=K be an elliptic curve. Then the endomorphism ring ofE is isomorphic to eitherZ or an order in a quadratic imaginary eld. The endomorphisms of E with trivial kernel form Aut(E), the automorphism group of E. The next theorem classies the automorphism groups of elliptic curves. Theorem 1.1.2 ([Sil09]). Let K be a eld of characteristic 0 and let E=K be an elliptic curve. Then the order of Aut(E) is one of the followings: 1. 2 if j(E)6= 0; 1728. 2. 4 if j(E) = 1728. 3. 6 if j(E) = 0. Let E=K be an elliptic curve with j-invariant not equal to 0 or 1728. If E 0 is another elliptic curve isomorphic to E, then the isomorphism is dened over a quadratic extension. In that case, E and E 0 are called the quadratic twists of each other. More precisely, Denition. LetE=K be an elliptic curve. Then a quadratic twist of E is an elliptic curve over K which is isomorphic to E over a quadratic extension of K. Let E be given by the equation y 2 = f(x). If the isomorphism is dened over the eld K( p d), then the quadratic twist is denoted by E (d) and it is dened by the equation dy 2 =f(x). Also, the isomorphism between E and E (d) is given by (x;y)7! (x;y p d): If E is given by y 2 =x 3 +ax +b, then E (d) is given in the short Weierstrass form y 2 =x 3 +d 2 ax +d 3 b: Furthermore, the curves E and E (d) are isomorphic via (x;y)7! (dx;d p dy): 6 1.2 Elliptic Curves Over C There is a one to one correspondence between the elliptic curves (up to isomorphism) and the lattices (up to a homothety). The following theorem summarizes this correspondence. Theorem 1.2.1 ([Sil09]). The following categories are equivalent. (a) Objects: Elliptic curves over C. Maps: Isogenies. (b) Objects: Elliptic curves over C. Maps: Complex analytic maps taking O to O. (c) Objects: Lattices C. Maps: Map( 1 ; 2 ) =f2C : 1 2 g: Let K be a number eld and E=K be an elliptic curve given by y 2 = 4x 3 g 2 xg 3 such that =g 3 2 27g 2 3 6= 0. Then, we dene the j-invariant of E as the following: j() = 1728 g 3 2 : Theorem 1.2.2 ([Sil09]). Let E 1 =C and E 2 =C be elliptic curves. Then E 1 and E 2 are isomorphic if and only if j(E 1 ) =j(E 2 ). 1.3 Elliptic Curves With Complex Multiplication Denition. Let E=C be an elliptic curve. E is said to have complex multiplication by K if the endomorphism ring of E is an order in a quadratic imaginary eld K. Example 1. Let E=Q be the elliptic curve given by y 2 =x 3 + 1: Then, in addition toZ, End(Z) contains an element : : (x;y)7! ( 3 x;y) 7 We will use the following theorem very often. Theorem 1.3.1 ([Sil09]). Let E=K be an elliptic curve with complex multiplication. Then j(E) is integral. LetO be an order in an imaginary quadratic eldK. LetI(O) denote the set of proper fractionalO-ideals. ThenI(O) forms a group under multiplication. The principalO-ideals give a subgroup P (O)I(O) and hence we can form the quotient C(O) =P (O)=I(O) which is the ideal class group of the orderO. The order ofC(O) is called the class number ofO and denoted as h(O). See [Cox89] for further reading. Theorem 1.3.2 ([Cox89, Corollary 10.20]). LetO be an order in an imaginary quadratic eld K. Then there is a one-to-one correspondence between the ideal class group C(O) and the homothety classes of lattices withO as their full ring of complex multiplications. Theorem 1.3.3 ([Cox89, Theorem 11.1]). LetO be an order in an imaginary quadratic eld K. Then the j-invariant of the lattice (or the elliptic curve) corresponding toO is an algebraic number of degree h(O) over Q. Moreover, h(O) can be calculated by the following formula. Theorem 1.3.4 ([Cox89, Theorem 7.24]). LetO be an order of conductorf in an imaginary quadratic eld K. Then h(O) = h(O K )f [O K :O] Y p=f 1 d K p 1 p : If E is an elliptic curve with complex multiplication by an order O, then any 2 O represents an endomorphism [] of E where [] is the isomorphism betweenO and End(E;E). The following theorem describes the eld of denition of such an endomorphism. Theorem 1.3.5 ([Sil94, Theorem 2.2]). Let E be an elliptic curve dened over LC with complex multiplication by an orderO in a quadratic imaginary eld K C. Then any endomorphism of E is dened over the composite eld KL. 8 1.4 Reduction of Elliptic Curves Let K be a local eld, be a discrete valuation on K, and let E=K be an elliptic curve over a local eld K. Let y 2 +a 1 xy +a 3 y =x 3 +a 2 x 2 +a 4 x +a 6 be a Weierstrass equation for E=K such that all coecients are inO K . Then the discrimi- nant has non-negative valuation. Denition ([Sil09]). LetE=K be an elliptic curve. A Weierstrass equation is called minimal for E at v if it is minimal with respect to the condition that a 1 ;a 2 ;a 3 ;a 4 ;a 6 2O K . First choose a minimal Weierstrass form forE. We can reduce its coecients modulo to obtain a curve ~ E over a nite eld. The curve obtained may be singular. Denition ([Sil09]). Let E=K be an elliptic curve, and let ~ E be the reduced curve for a minimal Weierstrass equation. (a) E has good (or stable) reduction if ~ E is non-singular. (b) E has multiplicative (or semi-stable) reduction if ~ E has a node. (c) E has additive (or unstable) reduction if ~ E has a cusp. Theorem 1.4.1 ([Sil09]). Let E=K be an elliptic curve. Then the j-invariant j(E) is integral if and only if E has either good or additive reduction. Theorem 1.4.2 ([Sil09]). LetE=K be an elliptic curve. If E has good reduction at , then E tors (K) injects into ~ E(F q ) where F q is the residue eld ofO K at . LetE=F q be an elliptic curve over the nite eldF q . Then we have the following estimate on the size of E(F q ) which is known as Hasse bound. Theorem 1.4.3 (Hasse). Let E=F q be an elliptic curve dened over the nite eldF q with q elements. Then jE(F q )q 1j 2 p q: Moreover, there exist complex numbers and such that E(F q n) = 1 n n +q n : 9 The numbers and satisfy the following relations: jj =jj = p q and =q: 1.5 Modular Curves Let H denote the upper half plane. Then the modular group SL 2 (Z) acts on H as: " a b c d # 2 SL 2 (Z) :z7! az +b cz +d : LetG be a nite index subgroup of SL 2 (Z). Then the quotient space H=G obtained by the action ofG on H is a non-compact Riemann surface, hence it is an algebraic curve. We call this curve by Y (G). One may extend the upper half plane by adding the points P 1 (Q) to H and obtain H, so called the extended upper half plane. The quotient space turns out to be a compact Riemann surface. The number of the orbits of points inP 1 (Q) is nite, hence what we obtain is the compactication of Y (G), which we call X(G). Let G be equal to 0 (n) where 0 (n) = (" a b c d # 2 SL 2 (Z) :c 0 modn ) : In this case, we callY (G) withY 0 (n) andX(G) withX 0 (n). The modular curveY 0 (n) is a moduli space for the pairs (E;C) whereE is an elliptic curve andCE is a cyclic subgroup of order n. Moreover, if the point P on Y 0 (n) is dened over K, then the corresponding elliptic curve E is dened over K and the subgroup C is invariant under the action of the Galois group Gal( K=K) of K. The pairs (E;C) and (E 0 ;C 0 ) represent the same point on Y 0 (n) if and only if there is an isomorphism :E!E 0 such that (C) =C 0 . The following Theorem is due Ogg [Ogg73] which provides an explicit method to compute the cusps of X 0 (n) for any n. Theorem 1.5.1 ([Ogg73]). For each d dividing n, the modular curve X 0 (n) has (d;n=d) conjugate cusps which are dened over Q( (d;n=d) ) and these are all the cusps of X 0 (n). Computations To compute the rank of elliptic curves over quadratic elds, we make use of the following Lemma. 10 Lemma 1.5.2 ([LL85, Corollary 1.3]). Let d be a square-free integer. Then for an elliptic curve E=Q, the following holds: rank(E(Q( p d))) = rank(E(Q)) + rank(E (d) (Q)): Throughout the paper, we compute the rank of E(Q) and E (d) (Q) on Magma [BCP97] and use Lemma 1.5.2 to compute the rank of E(Q( p d)). The torsion subgroup of a given elliptic curve is also computed using Magma. 11 Chapter 2 Torsion Subgroups in Elementary Abelian 2-Extensions Let K be the quadratic eldQ( p d) for d =1;3 and let F be the eld K( p d :d2O K ): The eldF is called the maximal elementary abelian 2-extension ofK since its Galois group is an elementary abelian 2-group and it is maximal with respect to this property. Let E=K be an elliptic curve given by y 2 =f(x). We rst start with the following observations: 1. Iff is irreducible inK, then it remains irreducible over the eld F . Otherwise,f has a root in F and the degree of K() over K is divisible by 3 but it is not possible since K() is contained in F . 2. If E(K) does not have a point of order 2, then E(F ) cannot have a point of order 2 either. This simply follows from the fact that the points of order 2 on the elliptic curve E are given by the zeros of f i.e.,f(; 0) : f() = 0g. Therefore, if E(K) does not have a point of order 2, then f is irreducible over K and the claim follows from the rst observation. 3. If thej-invariant ofE is not 0 or 1728, then for any elliptic curveE 0 =K isomorphic to E, we have E(F ) =E 0 (F ) since E and E 0 are isomorphic over a quadratic extension of K (hence also over F ). 4. LetL be a nite elementary abelian extension of K. Then for a prime P2O L above p inO K , the residue eld ofO L is at most a quadratic extension of the residue eld ofO K =p since the Galois group of the nite eldF q n overF q is cyclic for every n. 2.1 Rational Points on the Modular Curve Y 0 (n) Proposition 2.1.1. Let E=K be an elliptic curve. Then E(C) has no K-rational cyclic subgroups of order 32; 36 or 64. 12 Proof. An ane model for the modular curve X 0 (32) is given in [Yan06, p 503] as y 2 =x 3 + 4x: Ogg's theorem [Ogg73] tells us that X 0 (32) has the following cusps. d 1 2 4 8 16 32 (d; 32=d) 1 1 2 2 1 1 We see that X 0 (32) has 8 cusps; four of them dened in Q and the other four dened in Q(i). We compute on Magma that X 0 (32) has only 8 points over Q(i) and has 4 points overQ( p 3), hence they are all cuspidal. This shows that there are no elliptic curves E=K with a K-rational cyclic subgroup of order 32. Similarly, an ane model for the modular curve X 0 (36) is also given in [Yan06, p 503] as y 2 =x 3 + 1 We again apply Ogg's method to compute the cusps of the modular curve X 0 (36). d 1 2 3 4 6 9 12 18 36 (d; 36=d) 1 1 2 1 2 1 2 1 1 The table shows that X 0 (36) has 12 cusps; six of them dened over Q and the remaining six are dened over Q( p 3). We nd that it has 6 points over Q(i) and 12 points over Q( p 3). This shows that there are no non-cuspidal K-points on X 0 (36) and so there are no elliptic curves E=K with a K-rational cyclic subgroup of order 36. Finally, E(C) has no K-rational cyclic subgroup of order 64 since otherwise it would induce a K-rational cyclic subgroup of order 32. 2.1.1 The modular curve X 0 (20) Let K =Q(i). An equation for the modular curve X 0 (20) is given in [Yan06] as y 2 = (x + 1)(x 2 + 4): 13 It is known that there are no cyclic 20-isogenies dened overQ, see Theorem 2.1 in [LL85], hence X 0 (20)(Q) has only cusps. Oggs's method tells us that there are only 6 of them. Let E be the elliptic curve given by the equation y 2 = (x + 1)(x 2 + 4). Then the Mordell-Weil group of E(Q) is isomorphic toZ=6Z by Mazur's theorem. Now the set fO; (1; 0); (0;2); (4;10); ((2i; 0); (2i 2;(2i + 4)); (2i 2;(2i 4))g is a subset of E(K). We compute on Magma that its Mordell-Weil rank over K is 0. Since E(K) contains the full 2-torsion and E(Q) has order 6, E(K) has exactly 12 elements and it is isomorphic toZ=2ZZ=6Z. See Najman's theorem on page 2. To summarize, we showed that there are 6 non-cuspidal points on X 0 (20)(K). We will study these points in more detail inxx 2.4. 2.1.2 The modular curve X 0 (27) A model for the modular curve X 0 (27) is given in [Yan06] as y 2 +y =x 3 7: Again by Ogg's method, we nd thatX 0 (27) has 6 cusps; four of them dened overQ( p 3) and the other two dened overQ. Now let K = Q( p 3). We compute that E(K) has 9 points and hence there are 3 non-cuspidal points on X 0 (27) dened over K. Similarly if K = Q(i), the group E(K) is also nite and it has 3 points which shows that there is only one non-cuspidal point on X 0 (27) dened overQ(i), in fact dened over Q. Let E 1 be the elliptic curve associated with the lattice [1; 1+ p 27 2 ]. Then [ p 27] :E 1 !E 1 and [ 9 p 27 2 ] :E 1 !E 1 dene endomorphisms of E 1 and they are cyclic of degree 27. Moreover E 1 has complex multiplication by the orderZ[ 1+ p 27 2 ] and it is given in [Cox89, p.261] that j(E 1 ) =2 15 5 3 3: 14 Therefore the endomorphisms listed above are dened overQ( p 3) by [Sil94, Theorem 2.2]. We nd a model for E 1 in the database [Col13]; the elliptic curves over Q with complex multiplication, as y 2 +y =x 3 270x 1708: Hence any elliptic curve dened over K with a K-rational cyclic subgroup of order 27 is a quadratic twist of E 1 . For our later purposes, we will state the following result of Newman, which tells us about the existence of K-rational subgroups of certain degrees in quadratic extensions of K. Theorem 2.1.2 ([New16, Theorem 8]). Let E=K be an elliptic curve. Then E(C) has no K-rational cyclic subgroups of order 24; 35 or 45 dened over a quadratic extension of K. Moreover, if E is dened over Q( p 3), E does not have a K-rational cyclic subgroup of order 20; 21 or 63 dened over a quadratic extension of Q( p 3). Let K =Q( p d) for d =1;3. Remark 2.1.3. The proof of Theorem 8 in [New16] shows that the modular curveX 0 (n)(K) has no non-cuspidal points forn = 24; 35 or 45. Moreover, ifK =Q( p 3), thenX 0 (20)(K) also does not have any non-cuspidal points. 2.2 Odd Torsion The next result gives us a method to determine the odd torsion subgroup in an elementary abelian 2-extension of a eld. The result follows from [LL85, Corollary 1.3]. Lemma 2.2.1 ([LL85]). Let E be an elliptic curve over the eld k and L be an elementary abelian 2-extension of k of degree n = 2 m , i.e., the Galois group of L=K is an elementary abelian 2-group of order n. If E(k) 2 0 =fP2E(k) : [n]P = 0 for some oddng, then E(L) 2 0 =E (d 1 ) (k) 2 0:::E (dm) (k) 2 0; for suitable d i :i = 1;:::;m inO k . Furthermore, the image of each summand E (d) (k) 2 0 is a k-rational subgroup of E(L). Using Lemma 2.2.1, we see that the odd primes dividing the order of a point in E(F ) can only be 3; 5 or 7. We will prove in Proposition 2.2.2 and Proposition 2.2.5 that E(F ) does not have a point of order 21 for K =Q( p d) for d =1;3. 15 Proposition 2.2.2. Let E be an elliptic curve dened over K =Q( p 3). Then E(F ) has no subgroup of order 21. Proof. Assume E(F ) has a subgroup of order 21. Then by Lemma 2.2.1, replacing E by a twist if necessary, we may assume that E(K) has a point of order 3 and E (d) (K) has a point of order 7 for some d inO K , hence E has a subgroup of order 21 over a quadratic extension ofK and it isK-rational by Lemma 2.2.1. This gives a contradiction to Theorem 2.1.2. Remark 2.2.3. The modular curveX 0 (21) is an elliptic curve with Mordell-Weil rank 1 over Q(i). HenceX 0 (21) can not be immediately used to determine whether an elliptic curve E can have a subgroup of order 21 dened over the eld F . We will need the following result in the proof of Proposition 2.2.5. Theorem 2.2.4 ([New16, Theorem7]). Let K be a quadratic eld and let E=K be an elliptic curve. If j(E) = 0 and p> 3 is a prime, then E(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. The proof uses the techniques from [Lem04]. Proposition 2.2.5. LetE be an elliptic curve dened overQ(i). ThenE(F ) does not have a subgroup of order 21. Proof. Let E be an elliptic curve dened over K = Q(i) and suppose that E(F ) has a subgroup of order 21. We may assume thatE(K) has a point of order 7 (by replacing with a twist if necessary) by Lemma 2.2.1. It can be found in [Kub76, Table 3, p 217] that an elliptic curve with a point of order 7 is isomorphic to E t :y 2 + (1c)xyby =x 3 bx 2 (2.1) where b = t 3 t 2 and c = t 2 t for some t6= 0; 1 in K. Therefore E is isomorphic to E t for some t2 K. Moreover, either the j invariant of E is 0 or 1728, or the isomorphism is dened over a quadratic extension of K. By [New16, Theorem 7], we know that an elliptic curve with j invariant 0 or 1728 can not have a K-point of order 7, hence E is a quadratic twist of E t and if E(F ) has a point of order 21, then so has E t ; hence we may assume E is E t . 16 We compute the third division polynomial of the elliptic curve E t as the following. (x;t) =x 4 + 1 3 t 4 2t 3 +t 2 + 2 3 t + 1 3 x 3 + t 5 2t 4 +t 2 x 2 + t 6 2t 5 +t 4 x + 1 3 t 9 +t 8 t 7 + 1 3 t 6 : Now E has a point P of order 3 dened in a quadratic extension of K and the subgroup generated by P is K-rational by Lemma 2.2.1. We claim that x(P ), the x-coordinate of the point P must be in K which forces the equation (x;t)) = 0 to have a root in K. The only points inhPi are P ,P and the point at innity. Since x(P ) = x(P ), it follows that(x(P )) =x((P )) =x(P ) and hencex(P ) is invariant under the action of the Galois group. Now, the pair (E;P ) corresponds to a point (x 0 ;s) on the curve C given by the equation C : (x;t) = 0 whereE =E s andx 0 is thexcoordinate of the pointP of order 3. Therefore it is enough to ndC(K), the set ofK-points onC. The curveC is birational (overQ) to the hyperelliptic curve ~ C :y 2 =f(u) where f(u) =u 8 6u 6 + 4u 5 + 11u 4 24u 3 + 22u 2 8u + 1 andC and ~ C are isomorphic overK outside the set of singularities which aref(0; 0); (0; 1)g. Note that we require t to be dierent than 0 or 1 in (2.1), hence it is enough to nd ~ C(K). The polynomial f(u) factors as f(u) = (u 2 u + 1)(u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1) Let g and h denote the factors of f: g(u) =u 2 u + 1 h(u) =u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1: 17 Using the Descent Theorem ([Sto11, Theorem 11]; one can also look at Example 9 and 10 in [Sto11].), it is enough to nd the points on the unramied coverings ~ C d of ~ C, which are given as the intersection of two equations in A 3 : w 2 =dg(u) =d(u 2 u + 1) and z 2 =dh(u) =d(u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1) where d is a square-free number inO K dividing the resultant of g(u) and h(u), which is 112. Therefore d belongs to the set f1;i; (1 +i); 7;i(1 +i); 7(1 +i); 7i; 7i(1 +i)g If we exclude the cases d = 1 and d = 7i, reduction of d takes valuesf2; 3; 2; 1; 1; 2g and f3; 4; 2; 2; 3; 4g with respect to the ideal (2i) and (2 +i) respectively. We will reduce the curve z 2 =d(u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1) at (2i) for the values ofd =i; (1 +i); 7; 7i(1 +i) and similarly reduce it at (2 +i) for the values d =i(1 +i); 7(1 +i). In each case described above, z 2 =dh(u) reduces to either z 2 = 2(u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1): or z 2 = 3(u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1): A quick computation on Magma shows that neither of these equations has a solution over F 5 , hence there are no K-points on ~ C d for d6= 1; 7i. Let d = 7i. Magma computes 0 as an upper bound for the Mordell-Weil rank of the Jacobian of the curve z 2 = 7i u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1 ; hence the rank of the Jacobian ofz 2 = 7ih(u) is zero. Moreover, we compute on Magma that the Jacobian of the hyperelliptic curvez 2 = (7i)h(u) has 79 and 171 points respectively over the nite eldsF 5 andF 13 (here the reduction is taken at (2i) and (2 3i)) which proves that the torsion subgroup of the Jacobian of z 2 = (7i)h(u) over K is trivial. Therefore 18 ~ C d (K) =; for d = 7i. Hence, if there is a point on the curve ~ C(K), it must arise from the covering ~ C 1 (K). Now we may assume that d = 1. We would like to nd the K-points on the curve C 2 :z 2 =h(u) =u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1: Magma computes that the Mordell-Weil rank of J(C 2 )(Q) is 2 and also that 2 is an an upper bound forJ(C 2 )(K), therefore the Mordell-Weil rank of J(C 2 ) overK is equal to its Mordell-Weil rank over Q. Also as we did above, the reduction of C 2 at the good primes (2i) and (23i) is isomorphic to the reduction ofz 2 = (7i)h(u) at (2i) and (23i) and the isomorphism is dened overF 5 andF 13 respectively, hence J(C 2 )(K) tors is also trivial. Let J denote J(C 2 ). We claim that J(K) = J(Q). Since J(K) has rank 2, we can nd x;y2 J(K) such that J(K) is generated by x and y as an abelian group. Let be the generator of the Gal(K=Q) and assume that (x) = ax +by for some a;b2 Z. Let x 0 =nx+my andy 0 =rx+sy be the generators ofJ(Q). Then sincex 0 ;y 0 are the generators (of a free abelian group), sx 0 my 0 is not zero and so a multiple of x (namely (snmr)x) is inJ(Q). Notice that any such point is xed by. Then we obtainlax +lby =lx (we use l forsnmr above to simplify the notation). SinceJ(K) has trivial torsion, it implies that (a 1)x +by = 0. We conclude that a = 1 and b = 0 i.e., x in J(Q). A similar argument shows that y is also in J(Q), consequently J(K) =J(Q). We claim that C 2 (Q) = C 2 (K). Let P be a point in C 2 (K). If P 0 denotes the point [0 : 1 : 1] on C 2 , then [PP 0 ] represents a point in J(K) which equals to J(Q). If P 0 denotes the Galois conjugate of P , then [P 0 P 0 ] must be equal to [PP 0 ], since a point in J(Q) is invariant under the action of Gal( Q=Q). Hence P =P 0 and P is in C(Q) which proves that C 2 (Q) =C 2 (K). Now we will show that ~ C(Q) = ~ C(Q(i)). Remember that the curve ~ C is given by y 2 =u 8 6u 6 + 4u 5 + 11u 4 24u 3 + 22u 2 8u + 1 and we showed by the Descent theorem that a point (u;y) in ~ C(K) corresponds to a point on the intersection of C 1 :w 2 =u 2 u + 1 and C 2 :z 2 =u 6 +u 5 6u 4 3u 3 + 14u 2 7u + 1 19 such that y =wz where (u;w)2C 1 (K) and (u;z)2C 2 (K). The rst equation w 2 =u 2 u + 1 = (u 1=2) 2 + 3=4 implies that if (u;w) is a point on C 1 (K) with u in Q, then w is also in Q. Hence, we showed that if (u;y) is a point on ~ C(K), thenu;z are both inQ sinceC 2 (K) =C 2 (Q) and by the above argument, w is also inQ. Therefore y =wz is also inQ. To summarize, we proved our claim that ~ C(K) = ~ C(Q). This implies that a pair (E;P ) onC(K) corresponds to a point (u;y) on ~ C(Q) and therefore to a point inC(Q). However, if E is dened overQ and x(P )2Q, then P is in E(Q( p d)) for some d2Q. We know by [LL85] that E does not have a subgroup of order 21 dened over a quadratic extension of Q. Therefore there is no elliptic curve E dened over K =Q(i) such thatZ=21ZE(F ). Proposition 2.2.6. Let K be a quadratic cyclotomic eld and let E be an elliptic curve dened over K. Then E(F ) 2 0 is isomorphic to one of the following groups: Z=NZ for N2f1; 3; 5; 7; 9; 15g or Z=3ZZ=3Z: Proof. By Lemma 2.2.1 and Najman's theorem, we see that the odd numbers dividing the order of E(F ) tors are products of 3; 5; 7 and 9. Since F does not contain a primitive n-th root of unity for n = 5; 7 or 9, Z=5ZZ=5Z, Z=7ZZ=7Z or Z=9ZZ=9Z can not be isomorphic to a subgroup of E(F ) by the Weil pairing. If Z=3ZZ=7Z or Z=9ZZ=7Z is a subgroup of E(F ), then it is Galois invariant by Lemma 2.2.1, hence by Proposition 2.2.2 and Proposition 2.2.5, this is not possible. (Note that if E has a Galois invariant subgroup of order 63, then it also has a Galois invariant subgroup of order 21 which is not possible for Q(i) by Proposition 2.2.5 and for Q( p 3) by Proposition 2.2.2.) Similarly, Z=5ZZ=7Z and Z=5ZZ=9Z are also not possible by Proposition 2.1.2. (See also the Remark 2.1.3 following Proposition 2.1.2). IfE(F ) contains a subgroup isomorphic toZ=3ZZ=9Z, then by Lemma 2.2.1, we may assume thatE(K) has a pointP of order 9 and has an additionalK-rational subgroupC of order 3 arising from a twist, in other words, the subgroupC is dened in a quadratic exten- sion K( p d) of K. We will show that E[3] is a subset of E(K) and obtain a contradiction since E(K) can not have a subgroup of order 27 by [Naj10, Theorem 2]. 20 Let be the generator of the Galois group of K( p d) over K. Then the image of under the map (Galois action on the set of 3-torsion points) Gal(K( p d)=K)! GL 2 (F 3 ) is " 1 0 # for some; inF 3 . Note here that is 1 modulo 3 sinceK contains a third root of unity 3 , is the determinant of the matrix of and ( 3 ) = 3 . The fact that has order 2 tells us that = 0 and E[3] is a subset of E(K) as we claimed. Hence E(F ) can not have a subgroup isomorphic toZ=3ZZ=9Z. Finally, Z=3ZZ=3ZZ=5Z can not be isomorphic to a subgroup of E(F ) either. Otherwise, E(F ) has a K-rational subgroup of order 15 and by [Naj16, Lemma 7], E is isogenous to an elliptic curve with a cyclic K-rational subgroup of order 45 contradicting Proposition 2.1.2. See [LL85, Proposition 2.2] for the case K =Q. 2.3 Solutions to the Equation x 4 +y 4 = 1 We will call a solution (x;y) (respectively (x;y;z)) of a Diophantine equation trivial if xy = 0 (respectively xyz = 0) and non-trivial otherwise. Lemma 2.3.1. Let K =Q( p d) for d =1 or3. 1. The only solutions of x 4 +y 2 = 1 dened over K are trivial. 2. The only solutions of x 4 y 4 =z 2 dened over K are trivial. 3. The only solutions of x 4 +y 4 =z 2 dened over Q(i) are trivial. Proof. 1. Consider the rational map between C :x 4 +y 2 = 1 and E :y 2 =x 3 + 4x such that f : (x;y)7! 2x 2 1y ; 4x 1y : Since theK-valued points onC maps toK-valued points onE, it is enough to look at the inverse images of K-points on E and the points where the map f is not dened. We previously computed in the proof of Proposition 2.1.1 that E(Q( p 3)) =f(0; 0); (2;4)g 21 and E(Q(i)) =f(0; 0); (2;4); (2i; 0); (2;4i)g We easily compute that the inverse image of E(Q(i)) under f is the set f(0;1); (i; 0); (1; 0):g The points where f is not dened are the points where y = 1 and so we obtain the point (0; 1). All of the solutions we found are trivial. 2. Let (a;b;c) be a solution to the given equation over K. Assume a 2 6= c, then ( 2b 2 (a 2 c) ; 4ab (a 2 c) ) is a solution to y 2 = x 3 + 4x. Let E denote the elliptic curve y 2 =x 3 + 4x. We simply compute the points (a;b;c) for each point inE(K) (which is described in the rst part) and the points (a;b;c) with a 2 =c to show that abc = 0. 3. Let f : x 4 +y 4 =z 2 ! y 2 =x 3 4x (x;y;z)7! 2x 2 y 2 z ; 4xy y 2 z A quick computation on Magma shows that the ane curve y 2 =x 3 4x has only 3 points dened overQ(i) and they are (0; 0) and (2; 0). Hence if (a;b;c) is a solution to x 4 +y 4 = z 2 over Q(i), then either a = 0 or b = 0. Notice also that if f is not dened at (a;b;c), then a = 0. Hence the only solutions dened overQ(i) are trivial. Theorem 2.3.2. Let K = Q(i). Assume that x 4 +y 4 = 1 has a nontrivial solution in a quadratic extension L of K. Then L =K( p 7). Also the only solutions are ( 1 1 + 3 p 7 2 ; 2 1 3 p 7 2 ); (i; 0); (1; 0); (0;i); (0;1) where 1;2 =i or 1; 3 =1; and i = p 1: Proof. Mordell (Chapter 14, Theorem 4 of [Mor69]) proves that ifx 4 +y 4 = 1 has a nontrivial solution in a quadratic extension L of Q, then L is Q( p 7). We will use his technique to 22 show that this result still holds if one replacesQ withQ(i). LetL be a quadratic extension of K and (a;b) be a nontrivial solution in L. Then we can nd t2L such that a 2 = 1t 2 1 +t 2 ; b 2 = 2t 1 +t 2 : We will analyze the equation in two cases: 1. t is in K 2. t is not in K. Assume t is in K. If a (respectively b) is in K, then (a;b 2 ) (respectively (a 2 ;b)) gives a solution to the equation x 4 +y 2 = 1 (respectively x 2 +y 4 = 1) which are trivial by Lemma 2.3.1. If neither a nor b is in K, then a =a 1 p w and b =b 1 p w for some a 1 ;b 1 and w in K since a 2 ;b 2 2K, hence (a 1 ;b 1 ; 1=w) is a solution to x 4 +y 4 =z 2 . Again by Lemma 2.3.1, this is not possible. Assume t = 2K. Since t is in L, K(t) is contained in L and thus L =K(t). Let F (z) be the minimal polynomial of t over K and let us dene X;Y as follows: X = (1 +t 2 )ab; Y = (1 +t 2 )b Then, it is easy to see that X 2 = 2t(1t 2 ) and Y 2 = 2t(1 +t 2 ). Since X;Y are in L, then there are c;d;e;f2K such that X =c +dt; Y =e +ft Let us dene the polynomials g(z) and h(z) as follows: g(z) = (c +dz) 2 2z(1z 2 ) h(z) = (e +fz) 2 2z(1 +z 2 ): Then g(t) = h(t) = 0 because X 2 = 2t(1t 2 ) and Y 2 = 2t(1 +t 2 ). Since F (z) is the minimal polynomial of t over K, F (z) must divide both g(z) and h(z). It follows that g and h both have exactly one root over K (not necessarily the same root) since g and h are cubic polynomials. Let us call these roots u;v respectively for g and h, i.e., g(z) = 2(zu)F (z) and h(z) =2(zv)F (z): 23 Then (2u; 2(c +du)) is a point on the ane curve E 1 : y 2 = x 3 4x. We previously computed the points on E 1 (K) in the proof of Proposition 2.3.1. There are three of them; (0; 0); (2; 0); (2; 0), hence u is either 0; 1 or1. Notice also that (2v; 2(e +fv) is a point on the ane curve E 2 : y 2 = x 3 + 4x. Thus using our previous computations of E 2 (K), we see that the only possible solutions for (2v; 2(e +fv)) are f(0; 0); (2i; 0); (2i; 0); (2; 4); (2;4); (2; 4i); (2;4i)g: This shows thatv is either 0;1 ori. We will computeF (z) for each value ofu using the identity g(z) = 2(zu)F (z) = (c +dz) 2 2z(1z 2 ) and analyze each case separately. 1. Ifu = 0, theng(0) =c 2 = 0 and hencec = 0. We obtain thatg(z) = 2z(z 2 + d 2 2 z 1) and F (z) =z 2 + d 2 2 z 1: We computeh(z) =2(zv)(z 2 + d 2 2 z 1). Notice that the constant term ofh(z) is 2v. On the other hand, h(z) = (e +fz) 2 2z(1 +z 2 ) and hence the constant term is e 2 . Therefore e 2 =2v and this is satised only when v = 0 and v =i. (a) If v = 0, then F (z) =z 2 f 2 z + 1 which contradicts with F (z) = z 2 + d 2 2 z 1 since the constant terms are not equal. (b) If v =i, then using the equality h(i) = 0, we nd that F (z) =z 2 +z( f 2 2 +i) +i f 2 2 =z 2 + d 2 2 z 1 and hence f 2 2 +i = d 2 2 andi f 2 2 =1. It follows thatf 2 = 2i andd = 0 and we arrive at a contradiction since F has to be irreducible over K. (c) If v =i, then we nd that F (z) =z 2 +z( f 2 2 i)i f 2 2 =z 2 + d 2 2 z 1 which forces d to be 0 and we arrive at a contradiction. 24 2. Ifu = 1, theng(1) = (c +d) 2 = 0 and hencec =d. We see thatg(z) =d 2 (z 1) 2 + 2z(z 2 1) and we nd that F (z) =z 2 +z( d 2 + 2 2 ) d 2 2 which tells us that the constant term ofh(z) ise 2 =d 2 v. This equation has solutions in K only if v = 0 or v = 1. (a) If v = 0, we computed F (z) in terms of the coecient f in the rst part, hence we have F (z) =z 2 f 2 2 z + 1 =z 2 +z( d 2 + 2 2 ) d 2 2 : The equality of the constant terms gives usd 2 = 2 and we obtain a contradic- tion. (b) Ifv = 1, then (e+f) 2 = 2(1+1) = 4. Now the long division ofh(z) by (z1)gives us that F (z) =z 2 ( f 2 2 1)z + e 2 2 =z 2 +z( d 2 + 2 2 ) d 2 2 and hence e 2 =d 2 =f 2 and since (e +f) 2 = 4, we get e =f = 1 F (z) =z 2 + z 2 + 1 2 : (c) Similarly if v =1, we obtain (ef) 2 =4 and h(z) = (e +fz) 2 2z(1 +z 2 ) =2(z + 1)(z 2 +z( d 2 + 2 2 ) d 2 2 ): Similar to the previous part, the long division of h(z) by (z + 1) produces F (z) =z 2 ( f 2 2 + 1)z e 2 2 =z 2 +z( d 2 + 2 2 ) d 2 2 : e 2 =d 2 and d 2 +f 2 =4 which implies thate = 0 orf = 0. Ife = 0, thend = 0 and we get a contradiction. If f = 0, then e 2 =d 2 =4 and we compute F (z) =z 2 z + 2: 25 3. If u =1, then g(1) = (cd) 2 = 0 and hence c =d. We see that F (z) =z 2 +z( d 2 2 2 ) + d 2 2 : Constant coecient of h(z) equals to 2v( d 2 2 ) = e 2 , thus v has to be a square in K. Therefore, v can be 0 or1. (a) If v = 0, then F (z) =z 2 +z( d 2 2 2 ) + d 2 2 =F (z) =z 2 f 2 2 z + 1: we nd d 2 = 2 which is not possible since d is in K. (b) If v = 1, then F (z) =z 2 +z( d 2 2 2 ) + d 2 2 =z 2 ( f 2 2 1)z + e 2 2 which implies that d 2 =e 2 and d 2 +f 2 = 4. We see that f = 0 and d 2 = 4. In this case, we compute F (z) =z 2 +z + 2: (c) If v =1, then we have F (z) =z 2 +z( d 2 2 2 ) + d 2 2 =z 2 ( f 2 2 + 1)z e 2 2 : In this case, we compute that d 2 = 1 and hence F (z) =z 2 z 2 + 1 2 : To summarize, we showed that F (z) is one of the following polynomials: z 2 + z 2 + 1 2 ; z 2 z + 2; z 2 +z + 2; z 2 z 2 + 1 2 : One can easily check that the splitting eldL=K of each polynomial listed above isK( p 7). Now we will nd all non-trivial solutions to the equation x 4 +y 4 = 1 over the eld L = K( p 7). 26 Remember that we started with a solution (a;b) in L and constructed X and Y . It is easy to see that a = X=Y . In the following, we use the notation w = p 7 for simplicity. Also j for j = 1; 2; 3 denote the integers such that 2 j = 1. 1. We foundF (z) =z 2 +z + 2 with conditions thatf = 0,d 2 =e 2 andc =d. The roots of F (z) are t = (1w)=2 and we compute a = 1 (1 + 3 w)=2: Then we nd b as 2 (1 3 w)=2. 2. Similarly, we obtained F (z) = z 2 z + 2 with the conditions c =d, e 2 = d 2 and f = 0. In this case, we nd t = (1w)=2 and a = 1 (1 + 3 w)=2: Similar to the rst case, b =i 2 (1 + 3 w)=2. 3. The case F (z) =z 2 + z 2 + 1 2 produces t = (1w)=4 and we compute a = 1 i(1 + 3 w)=2; b = 2 (1 + 3 w)=2: 4. Similarly, the polynomial F (z) =z 2 z 2 + 1 2 produces t = (1w)=4 and a = 1 i(1 + 3 w)=2; b = 2 i(1 + 3 w)=2: And the result follows. Remark 2.3.3. Unfortunately, the method of the proof of Theorem 2.3.2 does not apply to the solutions over the eld Q( p 3) since there are non-trivial solutions to the equation u 4 +v 4 =z 2 overQ( p 3). 2.4 Restrictions on the Torsion Subgroups LetL 1 andL 2 be lattices such thatL 1 L 2 for some inC. Then the mapC=L 1 !C=L 2 induced by z7!z denes an isogeny in Hom(E 1 ;E 2 ) where E 1 ;E 2 are the elliptic curves associated to the latticesL 1 ;L 2 respectively. We will denote this isogeny by [] E 1 ;E 2 . When 27 E 1 = E 2 , we will use [] E 1 and we drop the domain and the target when it is clear from the context. Proposition 2.4.1. Let E 1 and E 2 be the elliptic curves associated to the lattices [1;i] and [1; 2i]. Then any isogeny in Hom(E 1 ;E 2 ) is dened over the eld Q(i). Proof. Let :E 1 !E 2 be an isogeny. Then = [] E 1 ;E 2 for some 2C. Since and i are both in [1; 2i], it follows that = 2a+2bi for some integersa andb. Hence Hom(E 1 ;E 2 ) is isomorphic to 2(Z[i]) as an additive group and it is enough to show that the isogenies [2] E 1 ;E 2 and [2i] E 1 ;E 2 are dened overQ(i). Notice that [2] E 1 ;E 2 and [2i] E 1 ;E 2 both dene isogenies of degree 2. Since E 1 has j invariant 1728, it has a model of the form y 2 =x 3 +dx for some d. The elliptic curve E 1 has at least one 2-isogeny dened overQ since (0; 0) is a point onE 1 dened overQ (independent ofd). Note that [6 + 2i] E 1 ;E 2 is a cyclic isogeny of degree 20 in Hom(E 1 ;E 2 ). If [2] E 1 ;E 2 and [2i] E 1 ;E 2 are both dened overQ, then [2 + 6i] is also dened overQ but there are no rational cyclic 20-isogenies. See [LL85, Theorem 2.1]. Assume [2i] E 1 ;E 2 is not dened overQ. Then it is dened over a quadratic extension K (more preciselyQ( p d)). Notice that the eld of denition of the isogeny [2] E 1 ;E 2 is either Q or K. Composing the following isogenies of E 1 and E 2 , E 1 [2i] !E 2 [i] !E 1 we obtain the endomorphism [2] E 1 ofE 1 which is dened overQ. This implies that [i] E 2 ;E 1 is dened over K. Then the endomorphism E 1 [2] !E 2 [i] !E 1 is dened over K. On the other hand, the endomorphism given above is dened over Q(i) since the eld of denition of E 1 [i] !E 1 isQ(i) and E 1 [2] !E 1 is dened overQ. Thus K must equal to Q(i). (Notice here that the endomorphism [2] E 1 is the multiplication by 2 map on E 1 and hence its eld of denition isQ.) One can do a similar discussion for the case [2i] E 1 ;E 2 is dened over Q and [2] E 1 ;E 2 is not. 28 Proposition 2.4.2. Let E be an elliptic curve over K = Q(i) and F be the maximal elementary abelian 2-extension of K =Q(i). Then E(F ) cannot have a K-rational cyclic subgroup of order 20. Proof. Let E 1 and E 2 be the elliptic curves with the endomorphism rings Z[i] and Z[2i] respectively and let i be the following endomorphisms 1 = [4 + 2i] :E 2 !E 2 ; 2 = [6 + 2i] :E 1 !E 2 ; 3 = [6 2i] :E 1 !E 2 : for i = 1; 2; 3. Let i denote the dual of the isogeny i for each i = 1; 2; 3. The elliptic curves E 1 andE 2 havej-invariant inQ since their endomorphism rings are orders of class number one in the eldQ(i) and hence the isogeny 1 is dened overQ(i) by [Sil94, Theorem 2.2] and 2 ; 3 are dened over Q(i) by Proposition 2.4.1. Consequently, the 6 non-cuspidal points (seex??) on X 0 (20) over K represent the elliptic curves E 1 and E 2 up to isomorphism, that is, if an elliptic curve dened over K has a cyclic K-rational 20-isogeny, it is isomorphic to E 1 or E 2 . Let E=K be an elliptic curve such that E(F ) contains a cyclic K-rational subgroup C of order 20. Then E is isomorphic to E 1 or E 2 . Assume E is isomorphic to E 1 , then E has j-invariant 1728 and a quadratic twist E (d) (K) of E has a point of order 5 by Lemma 2.2.1, a contradiction to Theorem ?? since E (d) also has j-invariant 1728. Assume now that E is isomorphic to E 2 . The j-invariant of E 2 is given in [Cox89] as j(E 2 ) = 11 3 : We nd a model for the elliptic curve with this j-invariant in [Col13] as: E 0 :y 2 =x 3 11x 14: 29 The elliptic curve with the given equation has good reduction at 3 and it has 64 points over the nite eld F 81 . SinceE 0 (F )[5] has to inject into E 0 (F 81 ) and 5 does not divide 64, E 0 (F ) can not have a point of order 5. Now, the elliptic curve E is a quadratic twist of E 0 andE 0 (F )'E(F ), hence we arrive at a contradiction. Proposition 2.4.3. Let E be an elliptic curve dened over Q(i). Assume E has a cyclic isogeny of degree 64 dened over a quadratic extension ofQ(i), then the j-invariant of E is integral. Proof. Kenku [Ken75, Proposition 2] shows the modular curve X 0 (64) has the ane equa- tion x 4 +y 4 = 1 and that it has 12 cusps and they are the points at innity [Roh77, p.1]. Hence Theorem 2.3.2 proves that any point ofY 0 (64) dened over a quadratic extension ofQ(i) is inQ( p 7;i) and there are 32 such points. Let E i for i = 1; 2 be the elliptic curves with complex multiplication with orders Z[ 1+ p 7 2 ]; Z[ p 7] respectively and let E 3 be the elliptic curve with the lattice [1; 2 p 7]. Notice that E 3 also has complex multiplication by the order Z[2 p 7]. Now let i for i = 1; 2; 3; 4 as follows: 1 = 9 + 5 p 7 2 ; 2 = 1 + 3 p 7; 3 = 6 + 2 p 7; 4 = 10 + 2 p 7 and i the conjugate of i for each i = 1; 2; 3; 4. Then i denes an endomorphism of E i for everyi = 1; 2; 3 and 4 denes an isogeny fromE 2 toE 3 . LetF i be the kernel of i and F i be the kernel of i . The j-invariants of E 1 and E 2 are rational [Cox89] and j(E 1 ) =3 3 5 3 and j(E 2 ) = 3 3 5 3 17 3 : The isogeny [ i ] is cyclic of degree 64 for each i and Theorem 1.3.5 tells us that [ 1 ] E 1 and [ 2 ] E 2 are dened overQ( p 7). See also [Ken75, Lemma 1]. The class number of the orderO =Z[2 p 7] can be computed by the formula in Theorem 1.3.4; the conductorf ofO is 4 and the class number ofO ish = 2, henceE 3 can be dened over a quadratic extension ofQ. LetL =Q( p 7)(j(E 3 )) be the ring class eld of the order O. We know the following facts: (See also [Cox89, Proposition 9.5, p.184].) 1. The minimal polynomial f of j(E 3 ) is inZ[x] and it is of degree h(O) = 2. 30 2. All primes ofQ( p 7) that ramify in L must divide 4Z[ 1+ p 7 2 ]. First property tells us that L = Q( p 7; p d) for some d 2 Q. Consider the following diagram of eld extensions. Q Q( p 7) Q( p d) L =Q( p 7; p d) If a prime inQ ramies inQ( p d), then it is either 2 (by the second property given above) or 7 (by the fact that the only prime ramies in Q( p 7) is 7). Since L contains p 7, we may assume that d and 7 are relatively prime. Hence d is2 or1 and L =Q( p 7; p 2) or L =Q( p 7;i): We will use [Cox89, Theorem 9.2, p.180] to determineL. Letp = 29. Thenp can be written as p = 1 2 + 28:1 2 . We have the following: 1. 28 is a quadratic residue modulo 29. 2. The polynomialx 2 + 1 has a solution modulop where asx 2 2 does not. We see that L =Q( p 7;i): This shows thatE 3 is dened over the eldQ(i). Thus by Theorem 1.3.5, [ 3 ] E 2 denes an isogeny overQ( p 7;i). Now we will show that 4 is also dened overQ( p 7;i). Similar to Proposition 2.4.1, we can show that Hom(E 2 ;E 3 ) is isomorphic to 2(Z[ p 7]) and it is generated by the isogenies [2] E 2 ;E 3 and [2 p 7] E 2 ;E 3 . It follows that these iso- genies cannot be both dened over Q(i) since they generate Hom(E 2 ;E 3 ). Otherwise [10 + 2 p 7] E 2 ;E 3 would be a cyclic 64-isogeny over Q(i) but there are no such isogenies dened over Q(i) by Proposition 2.1.1. Since [2 p 7] E 2 is an endomorphism of E 2 dened over Q( p 7) (and not over Q), similar to Proposition 2.4.1, we see that either [2] E 2 ;E 3 or [2 p 7] E 2 ;E 3 is dened overQ(i; p 7). Thus (E i ;F i ) and (E i ; F i ) are rational over L and they represent 8 dierent points on Y 0 (64)(L). 31 Let (E i ;F i ) correspond to the point (X(w i );Y (w i )) on x 4 +y 4 = 1: Denote by W 64 the modular transformation corresponding to the matrix " 0 1 64 0 # : Then W 64 acts as an involution on X 0 (64). Furthermore, we know by [Ken75, Proof of Lemma 1] that W 64 sends (E i ;F i ) to (E i ; F i ). Using the transformation formula for (z), Kenku shows that W 64 maps (x;y) to (y;x). Hence if (E i ;F i ) correspond to the point (X(w i );Y (w i )) on x 4 +y 4 = 1; then (E i ; F i ) correspond to (Y (w i );X(w i )). Using the transformation formula for (z) again, we see that " 1 0 16 1 # : (x;y)7! (ix;y) " 1 0 32 1 # : (x;y)7! (x;y) " 1 0 48 1 # : (x;y)7! (ix;y): Since these matrices are in SL 2 (Z), their actions do not change the isomorphism class of the elliptic curve corresponding to the given point onY 0 (64), hence the remaining 24 points on Y 0 (64)(K( p 7)) correspond to some pair (E;F ) where E is isomorphic to E i for some i = 1; 2; 3. This proves that an elliptic curve with a cyclic isogeny of degree 64 over a quadratic extension ofQ(i) has complex multiplication and thus its j-invariant is integral. Remark 2.4.4. OverK =Q( p 3), we nd a point (2= p 5; p 3= p 5) on the curvex 4 +y 4 = 1 and a computation on Magma shows that it produces an elliptic curve with a non-integral j-invariant. 32 Proposition 2.4.5. LetK =Q( p d) ford =1 or3 and letF be the maximal elementary abelian extension of K. Assume that E is an elliptic curve dened over K. Then E(F ) does not contain a rational subgroup isomorphic to one of the following groups: Z=4ZZ=2ZZ=3ZZ=3Z; Z=2ZZ=8ZZ=3Z; Z=2ZZ=4ZZ=5Z: Moreover, if K = Q(i), then E(F ) does not contain a K-rational subgroup isomorphic to the group Z=32ZZ=2Z: Proof. For each group exceptZ=32ZZ=2Z given in the statement of the proposition, the proof follows the proof of [LL85, Proposition 2.4] by using Proposition 2.1.1, Proposition 2.4.2, Theorem 2.1.2, and Remark 2.1.3. Assume that E(F ) has a K-rational subgroup V = Z=32ZZ=2Z. We will slightly modify the argument given in the proof of Proposition 2.4 in [LL85]. Let A denote the subgroup of Aut(V ) given by the action of Gal( K=K) on V . Writing V as LS, with L =Z=32Z and S =Z=2Z, we know by Lemma 1.5 in [LL85] that there exist a subgroup A 0 of A with index 2 which A 0 stabilizes L and S. Now :E!E 0 =E=S gives us a pair (E 0 ;L 0 ) where L 0 = 0 1 (L) is of order 64 and 0 is the dual isogeny of . Thus, the pair (E 0 ;L 0 ) corresponds to a quadratic point on Y 0 (64)(K). By Lemma 2.4.3, the j-invariant of E 0 is integral and thus E 0 has either good or additive reduction [Sil09, Proposition 5.5, p.181]. By [Sil09, Corollary 7.2, p.185], E also has either good or additive reduction. Let L be the eld generated by V E(F ) and let P be a prime ofO L above p = 2i and let ~ E denote the reduction of E at prime P. IfE has good reduction at P, then E(L) injects into ~ E(F 25 ). By the Weil conjectures, we see that ~ E(F 25 ) 36. This is not possible since E(L) must have at least 64 points. Assume E has additive reduction at P. Similar to Proposition 2.4.2, by [Sil09, Proposition 3.1, p.176], E 1 (L) has no non-trivial points of order 2 k for any k. Then E 0 (L) (2) injects into ~ E ns (F 25 ). However, ~ E ns (F 25 ) is isomorphic to the additive groupF 25 and hence E 0 (L) (2) is trivial. We know by [Tat74, Addendum to Theorem 3,x6] that [E(L) :E 0 (L)] 4: 33 Therefore,E(L) (2) has at most 4 elements which contradicts our assumption thatE(L) has a subgroup isomorphic toZ=2ZZ=32Z. Theorem 2.4.6. Let K be a quadratic cyclotomic eld and F be the maximal elementary abelian extension of K. Assume that E is an elliptic curve dened over K. 1. If K =Q(i), then E(F ) tors is isomorphic to one of the following groups: Z=2 b+r ZZ=2 b Z (b = 1; 2; 3 andr = 0; 1; 2; 3) Z=2 b+r ZZ=2 b ZZ=3Z (b = 1; 2; 3 andr = 0; 1) Z=2 b ZZ=2 b ZZ=3ZZ=3Z (b = 1; 2; 3) Z=2 b ZZ=2 b ZZ=5Z (b = 1; 2; 3) or 1;Z=3Z, Z=5Z, Z=7Z, Z=9Z, Z=15Z, and Z=3ZZ=3Z . 2. If K =Q( p 3), then E(F ) is either isomorphic to one of the above groups or Z=2ZZ=32Z; Z=4ZZ=64Z Proof. The proof follows from [LL85, Proof of Theorem 2.5] using Proposition 2.4.5, Propo- sition 2.1.1, Lemma 2.4.2, Proposition 2.2.6 and Theorem 2 in [Naj10]. 2.5 E(K) tors is Non-Cyclic In this section, we study E(F ) tors when the torsion subgroup of E over K is not cyclic. Theorem 2 in [Naj10] shows that if E(K) tors is not cyclic, then it is isomorphic to one of the following groups. Z=2ZZ=2nZ for n = 1; 2; 3; 4 and (only if K =Q(i)) Z=4ZZ=4Z: We study each of these groups separately. The Proposition 2.5.5, Proposition 2.5.6, Propo- sition 2.5.7, Proposition 2.5.8 and Proposition 2.5.9 are main results of this section. 34 Let E be an elliptic curve over a number eld k with E[2]E(k). We know that the points of order 2 on E are given by the roots of the polynomial f(x) whereE is dened by y 2 =f(x). Therefore, we may assume that E is of the form y 2 = (x)(x)(x ) with ; ; in k. We will use the following results very often in this section. Remember that F denotes the maximal elementary abelian extension of K. Lemma 2.5.1 ([Kna92, Theorem 4.2]). Let k be a eld of characteristic not equal to 2 or 3, and E an elliptic curve over k given by y 2 = (x)(x)(x ) with ;; in k. For P = (x;y) in E(k), there exists a k-rational point Q on E such that [2]Q =P if and only if x;x and x are all squares in k. Furthermore, in this case if we x the sign of the square roots of x;x;x , then the x-coordinate of Q equals to p x p x p x p x p x p x +x or p x p x p x p x p x p x +x: Theorem 2.5.2 ([New17, Theorem 9]). Let K be a number eld and E=K be an elliptic curve with full 2-torsion. Then E has a model of the form y 2 = x(x +)(x +) where ;2O K . 1. E(K) has a point of order 4 if and only if ; are both squares (inO K ) or; are both squares or; are both squares in K. 2. E(K) has a point of order 8 if and only if there exist a d 2 O K , d 6= 0 and a Phytogorean triple (u;v;w) such that =d 2 u 4 ; =d 2 v 4 ; or we can replace ; by; or; as in the rst case. Proof. This is a generalization of Ono's result given in [Ono96, Main Theorem 1] and the proof can be found in Newman's article [New17, Theorem 9]. 35 By Theorem 2.5.2, we may assume that an elliptic curve E with full 2-torsion has the model y 2 =x(x +a)(x +b) with a;b2O K . We denote this curve by E(a;b). Then E(a;b) is isomorphic (over K) to E(a;ba) and E(b;ab) by the isomorphisms (x;y)7! (x +a;y); and (x;y)7! (x +b;y) respectively. The elliptic curve E (d) : y 2 = x(x +da)(x +db) is the quadratic twist of E by d2 K and if d is a square in K, then E (d) and E are isomorphic over K. Assume E has a point of order 4 in E(F ). Then using Theorem 2.5.2 together with the isomorphisms betweenE(a;b);E(a;ba), andE(b;ab), we may assume that there is a pointQ such that [2]Q = (0; 0) and thata andb are both squares. Notice that with a similar discussion, we may assume that if E(K) has a point of order 8, then M = u 4 and N = v 4 for some u;v2O K such thatu 2 +v 2 is a square inK (replacingE by a quadratic twist if necessary). The following result is on the classication of twists of elliptic curves over K. We will use this result very often in this section. Theorem 2.5.3 ([New17, Theorem 15]). Let K = Q( p D) with D =1;3, d2 K a non-square, and E=K an elliptic curve with full 2-torsion. Then, 1. If E(K) tor =Z=2ZZ=8Z, then E d (K) tor =Z=2ZZ=2Z. 2. If E(K) tor =Z=2ZZ=6Z, then E d (K) tor =Z=2ZZ=2Z. 3. If E(K) tor =Z=4ZZ=4Z, then K =Q(i) and E d (K) tor =Z=2ZZ=2Z. 4. If E(K) tor =Z=2ZZ=4Z, then E d (K) tor =Z=2ZZ=2Z unless K =Q( p 3) and d =1 in which case E d (K) tor =Z=2ZZ=4Z or Z=2ZZ=2Z. 5. If E(K) tor =Z=2ZZ=2Z, then E d (K) tor =Z=2ZZ=2Z for almost all d. For the rest ofx 2.5, E will always denote an elliptic curve given by the equation E(a;b) :y 2 =x(x +a)(x +b); a;b2K: Furthermore, we will assume that the greatest common divisor (a;b) (dened up to a unit) is square-free. Otherwise, we may replace it by the quadratic twist E (d 2 ) of E where d 2 divides both a and b. Also, [n] denotes the multiplication by n on the elliptic curve E. We will need the following lemma for the proof of Proposition 2.5.5. Lemma 2.5.4. Let be in K. If p is a square in F , then or is a square in K. If K =Q(i), then must be a square in K. 36 Proof. Letw2F be such thatw 2 = p . Thenw is a root of the polynomialf(x) =x 4 dened over K. Since f has one root in F and F is a Galois extension of K, f splits in F . Iff is reducible over K, then it has to be product of two quadratic polynomials, otherwise 3 has to divide the order of Gal(F=K). Let us write f as a product of two polynomials. x 4 = (x 2 +ex +f)(x 2 +cx +d) for e;f;c;d2K. Then ec +f +d = 0, ed +fc = 0 and c +e = 0. Therefore, replacing e byc, we obtain c(fd) = 0. So, either c = 0 or f = d. If c = 0, then f =d and so fd = implies that =f 2 in K. If f =d, then =fd, so is a square in K. Now assume thatf is irreducible overK. The Galois group off overK is an elementary abelian 2-group since it is a quotient of Gal(F=K). However, an elementary abelian 2- subgroups of S 4 is either of order 2 or the Klein four-group V 2 . Hence Gal(F=K) must be isomorphic to V 2 and in particular, it has order 4. If f remains irreducible over K(i). Then, we obtain an automorphism of order 4 in Gal(K(w)=K(i)); namely w7! iw which contradicts the fact that Galois group isV 2 . Therefore, ifK =Q(i), it can not be irreducible and we may assume that K =Q( p 3). Then f has to be reducible over K(i) and by our previous discussion, we see that or is a square in K(i). Hence if =d 2 with d2K(i), then d =bi for some b2K and =b 2 . Proposition 2.5.5. Let E : y 2 = x(x +a)(x +b) be an elliptic curve dened over K. Assume that E(K) tors =Z=2ZZ=8Z. 1. If K =Q(i), then E(F ) tors is either isomorphic to Z=4ZZ=16Z or Z=4ZZ=32Z. 2. IfK =Q( p 3), thenE(F ) is isomorphic to one of the groups above orZ=4ZZ=64Z. Proof. Since any number in K is a square in F , Z=4ZZ=16Z E(F ) by Lemma 2.5.1. By Theorem 2.5.2, we may assume that a = u 4 and b = v 4 for some u;v2O K such that u 2 +v 2 =w 2 for somew2K. We will show thatZ=8ZZ=8Z6E(F ). LetQ 2 = (x;y) be a point of order 4 such that [2]Q 2 = (a; 0). By Lemma 2.5.1, we compute that x equals to one of the followings: p u 4 + 0 p u 4 +v 4 u 4 : 37 If Z=8ZZ=8Z E(F ), then there is a point Q 3 in E(F ) such that [2]Q 3 = Q 2 and by Lemma 2.5.1, x +u 4 is a square in F i.e., x +u 4 = p u 4 p (u 4 +v 4 ) =u 2 p u 4 v 4 is a square in F . Since u 2 andu 2 = (iu) 2 are both squares in F , p u 4 v 4 is a square in F . By Lemma 2.5.4, we see that u 4 v 4 or v 4 u 4 has to be a square in K. Therefore, there exist at2K such that (u;v;t) or (v;u;t) satisfy the equationx 4 y 4 =z 2 . By part 2 of Lemma 2.3.1, uvt = 0. However, u orv can not be zero since E is non-singular. Hence t must be zero which meansa equals tob. We obtain a contradiction toE being nonsingular. Hence this shows that E(F ) does not contain a subgroup isomorphic to Z=8ZZ=8Z. Theorem 2.5.3 implies that E(F ) 2 0 = 0 and the result follows from Theorem 2.4.6. See [Fuj04, Proposition 4.1] for a similar result over Q. Proposition 2.5.6. Assume E(K) tors = Z=2ZZ=6Z. Then E(F ) tors is isomorphic to Z=4ZZ=12Z. Proof. By Theorem 2.5.3, all (non-trivial) quadratic twists of E have torsion subgroup isomorphic toZ=2ZZ=2Z. Hence, the odd part ofE(F ) tors must be isomorphic toZ=3Z. By Lemma 2.5.1, we also have that Z=4ZZ=4Z E(F ) since Z=2ZZ=2Z E(K). HenceZ=4ZZ=12ZE(F ). To prove the statement, we need to show there is no point of order 8 in E(F ). Let P 2 be in E(F ) such that [2]P 2 = P 1 = (0; 0). We compute the x-coordinate of P 2 by Lemma 2.5.1 as x(P 2 ) = p ab: Then, P 2 is either ( p ab; p ab( p a + p b)) or ( p ab; p ab( p a + p b)). Suppose that P 2 is in [2]E(F ). Then again by Lemma 2.5.1, p ab has to be a square in F and by Lemma 2.5.4, ab orab is a square in K. Suppose that ab is a square in K. Then a = da 02 and b =db 02 where (a 0 ;b 0 ) is a unit. Then P 2 = ( 1 da 0 b 0 ;da 0 b 0 ( 2 a 0 p d +b 0 p d)) where 2 1 = 2 2 = 1 and the point ( 1 da 0 b 0 ;da 0 b 0 ( 2 a 0 +b 0 )) 38 denes a point of order 4 in E (d) (K). However by Theorem 2.5.3, we know that any quadratic twist ofE has torsion subgroup isomorphic toZ=2ZZ=2Z overK. Henceab is not a square. Now assumeab is a square. Leta =da 02 andb =db 02 . IfP 2 is in [2]E(F ), then x(P 2 ) +a is also a square in F . x(P 2 ) +a = p ab +a =(da 0 b 0 )i +da 02 =da 0 (a 0 b 0 i) Hence a 0 +b 0 i =u 2 s for some u in K(i) and s2K. We also see that a 0 b 0 i = u 2 s where u denotes the Galois conjugate of u. Then (ab)=d =a 02 +b 02 = (u u) 2 s 2 is a square in K. Consider the curve E 0 =E(ab;b) :y 2 =x(x +ab)(xb): Taking the quadratic twist of E 0 =E(ab;b) by d, we obtain E 0(d) :y 2 =x(x +d(ab))(xdb): Notice that d(ab) = d 2 (a 02 +b 02 ) anddb = d 2 b 02 are squares in K. Hence E 0(d) has a point of order 4 by Lemma 2.5.1. However, this is not possible by Theorem 2.5.3 since E and E 0 are isomorphic over K. Hence P 2 is not in [2]E(F ). Using the isomorphism between E(a;ba);E :y 2 =x(x +a)(x +b), and E(a;b) we described earlier, one can show that there is no point P of order 8 in E(F ). Therefore, there is no point of order 8 in E(F ). This proves that E(F ) =Z=4ZZ=4ZZ=3Z. See [Fuj04, Proposition 4.3] for a similar result over Q. Proposition 2.5.7. Let K =Q(i). If E(K) tors =Z=4ZZ=4Z, then E(F ) tors is isomor- phic to Z=8ZZ=8Z. Proof. Suppose thatZ=4ZZ=4ZE(K). By Theorem 2.5.2, we may assume thata =s 2 and b =t 2 for some s;t2K. Let P 1 = (0; 0) and Q 1 = (s 2 ; 0) as before and x(P 2 ) =st and x(Q 2 ) =s p s 2 t 2 s 2 39 such that [2]P 2 = (0; 0) and [2]Q 2 =Q 1 . Since Q 2 has order 4, it must be in E(K) which forces s 2 t 2 to be a square in K. Let r be in K such that s 2 t 2 =r 2 : Hence, we compute thatQ 2 = (srs 2 ;isr(rs)). By lemma 9, we know thatZ=8ZZ=8Z E(F ). We want to show thatZ=16Z6E(F ). Using Lemma 2.5.1, we nd a point P 3 such that [2]P 3 =P 2 and x(P 3 ) = 1=2 p st( p s + p t + p s +t) 2 : Assume P 3 is in [2]E(F ). Then st is a square in K by Lemma 2.5.1 and Lemma 2.5.4. Since s and t are relatively prime, either s and t are both squares or they are both i times a square. (Since1 is a square, we do not need to consideriu 2 .) Now, let u;v be in K such that s = iu 2 ;t = iv 2 . Then the equation s 2 t 2 = r 2 gives usu 4 +v 4 = r 2 which has no nontrivial solutions over K by Lemma 2. Similarly, if s;t are both squares, then u 4 v 4 =r 2 . This proves that P 3 62 [2]E(F ). Similarly using Lemma 2.5.1, we can nd another point Q 3 where [2]Q 3 =Q 2 such that x(Q 3 ) +s 2 = p (srs 2 )sr + p (srs 2 )(srr 2 ) + p (srr 2 )sr +sr =s p r p rs + p sr p (rs) 2 +r p s p sr +sr =(s) p r p rs + p r p s(rs) +r p s p rs (s)r = 1=2 p sr( p r + p rs p s) 2 If Q 3 is in [2]E(F ), then x(Q 3 ) +s 2 is a square in F . By lemma 11,sr is a square in K. Hence as we discussed earlier, s andr are either both squares or both i times a square. In both cases, we obtain a non-trivial solution for the equation x 4 y 4 = z 2 which is not possible. Therefore, Q 3 is not in [2]E(F ). We will show next that there are no points of order 16 in E(F ). Let R 3 be in E(F ) such that [2]R 3 =P 2 +Q 2 =R 2 where R 2 is a point of order 4 such that [2]R 2 = (t 2 ; 0). We nd that x(R 3 ) +t 2 = (1=2) p tri( p ti + p r + p r +ti) 2 : Hence if R 3 is in [2]E(F ), then tri has to be a square. This leads to a contradiction to Lemma 2.3.1 as earlier. Hence R 3 is not in [2]E(F ). Notice that R 3 = [a]P 3 + [b]Q 3 for some a;b odd. We will assume for simplicity that a =b = 1. (The following can easily be modied for general a;b. 40 Assume that there is a pointP of order 16 inE(F ). Then 2[P ] = [k]P 3 + [l]R 3 for some k;l2Z. Dene Q = 8 > > > < > > > : [(k 1)=2]P 3 + [l=2]R 3 if k is odd, l is even [k=2]P 3 + [(l 1)=2]R 3 if l is odd, k is even [(k + 1)=2]P 3 + [(l 1)=2]R 3 if k,l are both odd Then, [2](PQ) is either P 3 ;R 3 or Q 3 which is not possible as we showed earlier. Hence, Z=16Z6E(F ) and E(F ) =Z=8ZZ=8Z. Proposition 2.5.8. Assume that E(K) tors =Z=2ZZ=4Z. 1. Let K =Q(i). Then E(F ) does not contain a subgroup isomorphic to Z=8ZZ=8Z. 2. Let K =Q( p 3). Then E(F ) contains a subgroup isomorphic to Z=8ZZ=8Z only if E 1 (K) has a point of order 4 in which case E(F ) =Z=8ZZ=8Z. Proof. By Lemma 2.5.1, we know that Z=4ZZ=8Z E(F ). We rst determine when E(F ) has a subgroup isomorphic toZ=8ZZ=8Z. By Theorem 2.5.2, we may assume that a =s 2 andb =t 2 for somes;t2O K relatively prime. Suppose thatZ=8ZZ=8ZE(F ). Let Q 1 = (s 2 ; 0) and Q 2 be in E(F ) so that [2]Q 2 =Q 1 . We compute x(Q 2 ) =s 2 +s p s 2 t 2 : Then there is a point Q 3 2E(F ) such that [2]Q 3 =Q 2 . Hence by Lemma 2.5.1, x(Q 2 ) +s 2 =s p s 2 t 2 is a square inF which implies that p s 2 t 2 is a square inF . Hence eithers 2 t 2 ort 2 s 2 is a square in K by Lemma 2.5.4. Let s 2 t 2 =r 2 for some r2K. Then we compute that Q 2 = (srs 2 ;isr(rs)) If K =Q(i), the points P 2 and Q 2 generate E[4] and they are both in E(K). This contra- dicts the fact that E(K) =Z=2ZZ=4Z. Note that there is no need to consider the case t 2 s 2 is a square separately since1 is a square in K. 41 Now we may assume thatK =Q( p 3). ThenQ 2 = (s 2 +sr;isr(rs)) is inE(K(i)). Moreover, we see that the quadratic twist E (1) of E has a point (s 2 +sr;sr(rs)) of order 4 over K. Therefore if the elliptic curve E (1) does not have a point of order 4, then Z=8ZZ=8Z6E(F ). Similarly, ifs 2 t 2 =r 2 , thenR 2 = (t 2 +rt;irt(rt)) gives rise to a K-point of order 4 on E (1) . Hence we conclude that E(F ) does not contain the full 8-torsion if E (1) (K) does not have a point of order 4. AssumeE (1) (K) has a point of order 4. Then we can show E(F ) contains a subgroup isomorphic to Z=8ZZ=8Z by computing points of order 8 as we did in the proof of Proposition 2.5.7. Next we will show that there is no point of order 16 in E(F ). We know that there is a point P 3 in E(F ) such that [2]P 3 =P 2 . We will show that P 3 62 [2]E(F ). We compute x(P 3 ) = (1=2) p st( p s + p t + p s +t) 2 : Assume P 3 is in [2]E(F ). Then p st is a square in F and so st orst is a square in K. Since s and t are relatively prime, s =du 2 and t =dv 2 for some unit d inO K . The only units inO K aref1g, hence u =1. In each case we obtain a non-trivial solution (over K) to the equation x 4 y 4 =z 2 sinces 2 t 2 ort 2 s 2 is a square. This is not possible by Lemma 2. This shows that P 3 is not in [2]E(F ). Let s 2 t 2 = r 2 . Then using the computations of points Q 3 and R 3 in the proof of Proposition 2.5.7, we see that Q 3 and R 3 can not be in [2]E(F ). This can be shown with a similar argument we used to show P 3 is not in [2]E(F ). With a similar discussion to the proof of Proposition 2.5.7 we construct a point Q in E(F ) and show that E(F ) does not contain a point of order 16. If s 2 t 2 =r 2 , for a similar argument, use x(Q 3 ) and x(R 3 ) which we computed in Proposition 2.5.7. See [Fuj04, Proposition 4.6] for a similar result over Q. Theorem 2.5.9. Let K be the quadratic eld Q(i) or Q( p 3). Assume E(K) tors = Z=2ZZ=2Z. Then E(F ) tors is isomorphic to one of the groups listed in Proposition 2.5.5, Proposition 2.5.6, Proposition 2.5.8, Proposition 2.5.7, or the group Z=4ZZ=4Z. Proof. A quadratic twist of E can have torsion subgroup isomorphic to Z=2ZZ=8Z, Z=2ZZ=6Z, Z=2ZZ=4Z, Z=2ZZ=2Z or Z=4ZZ=4Z. Note that since E and E (d) are isomorphic over a quadratic extension of K, E(F ) tors =E (d) (F ) tors : 42 Hence if E (d) (K) tors 6 = Z=2ZZ=2Z for some d2 K, then E(F ) tors will be one of the groups listed in Proposition 2.5.5, Proposition 2.5.6, Proposition 2.5.8, Proposition 2.5.7. Therefore we may assume thatE (d) (K) =Z=2ZZ=2Z for alld2K. The rest of the proof is same as the proof of Proposition 2.5.6. See [Fuj04, Proposition 4.5] for a similar result overQ. 2.6 E(K) tors is Cyclic LetE :y 2 =f(x) be an elliptic curve withE(K) tors =Z=NZ. IfN is odd, then there is no point of order 2 inE(K). Since the 2-torsion points onE are ( i ; 0) where i are the roots of f, E(K) tors being odd implies f is irreducible over K. Therefore, f is irreducible over F . ThenE(F ) tors is also odd and we analyzed this case inx 2.2. Hence, we assume that N is even. We will need the following lemma to show that Z=8ZZ=8Z is not contained in E(F ) when E(K) is cyclic. Lemma 2.6.1. Let K be the eldQ( p 3) and let F be the maximal elementary abelian 2- extension of K. Suppose a2K. Then p ai can not be a square in F . Proof. Suppose that p ai is a square inF . Then the proof of Lemma 2.5.4 shows that ai is a square in K(i). Then ai =b 2 (1 +i) 2 or ai =b 2 (1i) 2 with b2K: Then a = 2b 2 or a =2b 2 . Hence p ai is equal to b(1i) and we obtain that (1i) is a square in F . Now let = p 1 +i. Hence ( 2 1) 2 + 1 = 0 and we see that is a root of the polynomial f(x) =x 4 2x 2 + 2: We observe that the degree of the splitting eld of f is 8 and that F has to contain the splitting eld of f since it contains , hence the Galois group of f over K has to be an elementary abelian 2-group. Notice that the Galois group off is a subgroup ofS 4 . SinceS 4 does not have any elementary abelian 2-subgroup of order 8, we get a contradiction. Hence neither p 1 +i nor p 1i is a square inF and this proves that p ai can not be a square in F for any a2K. 43 Proposition 2.6.2. Let E be an elliptic curve over K and suppose that E(K) tors =Z=2NZ for some integer N. 1. If K =Q(i), then Z=4ZZ=4Z6E(F ). 2. If K =Q( p 3), then Z=8ZZ=8Z6E(F ). Proof. Suppose E be given by the equation y 2 =f(x). Then f has exactly one root in K. Without loss of generality, we may assume that E is given by y 2 =f(x) =x(x)(x ) where denoted the complex conjugate of. We may write and asa+b p c andab p c for some square-free c since they are dened over a quadratic eld. Then = 2b p c: If the point Q 1 = (; 0) is in [2]E(F ), then by Lemma 2.5.1, is a square in F and hence either c orc is a square by Lemma 2.5.4. If K =Q(i), then Q 1 62 [2]E(F ) which proves thatZ=4ZZ=4Z6E(F ) when K =Q(i). Let K = Q( p 3). If Z=4ZZ=4Z E(F ), then c =1 and we may assume that = a +bi. Let P 2 be a point such that [2]P 2 = (0; 0). Suppose that E[8] E(F ). Then P 2 is in [2]E(F ) and Theorem 2.5.2 implies that x(P 2 ) = p ; x(P 2 ) and x(P 2 ) are all squares in F . Hence by Lemma 2.5.4, =a 2 +b 2 =d 2 or =a 2 +b 2 =d 2 for some d in K. We obtain either (x(P 2 ))(x(P 2 ) ) = 2d(da) =e 2 (2.2) or (x(P 2 ))(x(P 2 ) ) = 2d(d +ai) =f 2 (2.3) for some e;f2K. We can parametrize a;b;d as a =k(m 2 n 2 ); b = 2kmn; and k(m 2 +n 2 ) 44 for some k;m;n2O K . We set d =ki(m 2 +n 2 ) if a 2 +b 2 =d 2 . (Note that switching the order of parametrization of a and b does not change the result.) Equation 2.2 and 2.3 gives us either m 2 +n 2 or 2(m 2 +n 2 ) is a square in K. Suppose 2(m 2 +n 2 ) is a square in K, then m 2 +n 2 is divisible by 2 and so is m 2 n 2 . This means thata;b;d are all divisible by 2. (Notice that 2 remains as a prime inO K .) In this case, we can replace E by the quadratic twist E (2) of E since E(F ) =E (2) (F ) and they both have cyclic torsion subgroup over K. Therefore it is enough to consider the case wherea;b are not both divisible by 2. We will assume that m 2 +n 2 is a square in K, then we compute x(Q 2 ) where Q 1 = (; 0) = [2]Q 2 . We know that x(Q 2 ) = p ( ) = 2(mni) p mni is a square inF , hence p mni has to be a square inF but this is not possible by Lemma 2.6.1. Proposition 2.6.3 ([Fuj04, Lemma 13]). Let E(K) be cyclic. Then E(F ) contains a point of order 4 if and only if there exist a d2O K such that E (d) (K) has a point of order 4. Proof. The proof follows the proof of the statement when K =Q and it is given in [Fuj04, Lemma 13]. 2.7 More Restrictions on the Torsion Subgroups Proposition 2.7.1. LetE be an elliptic curve overK. ThenE(F ) cannot have a subgroup isomorphic to Z=4ZZ=4ZZ=5Z. Proof. Suppose thatZ=4ZZ=4ZZ=5ZE(F ). Assume that E(K) is cyclic. Then by Proposition 2.6.3, E (d) (K) has a point of order 4 for some d2 K. Since E and E (d) are quadratic twists, E(F ) = E (d) (F ) and hence E (d) (F ) has a Galois invariant subgroup of order 20 which is not possible by Proposition 2.1.1. Similarly, the result holds in the case where E(K) contains the full 2-torsion by our results inxx 2.5. Proposition 2.7.2. LetE be an elliptic curve overK. ThenE(F ) cannot have a subgroup isomorphic to Z=12ZZ=12Z. 45 Proof. If E(K) contains Z=2ZZ=2Z, then E(F ) cannot have a subgroup isomorphic to Z=3ZZ=3Z by the results ofxx 2.5. Hence we may assume that E(K) is cyclic. If K =Q(i), then E[4] is not contained in E(F ) by Lemma 2.6.2. Let K = Q( p 3). We may assume that E has a rational subgroup C of order 3 and a point P of order 4 by Lemma 2.6.3 (replacing E by a twist if necessary). Let : E! E 0 := E=C, then E 0 has a cyclic isogeny of order 9 dened over K by [Naj16, Lemma 7] since E has an additional K-rational 3-cycle. The image of P , (P ) is in E 0 (K) and it is of order 4 since the order of the ker() is relatively prime to 4. Then E 0 has a K-rational subgroup of order 36 which is not possible by Proposition 2.1.1. Hence, there is no such curve over the eldsQ(i) orQ( p 3). Proposition 2.7.3. LetE be an elliptic curve overK. ThenE(F ) cannot have a subgroup isomorphic to Z=4ZZ=32Z. Proof. Suppose that E is an elliptic curve dened over K with Z=4ZZ=32Z E(F ). Then Proposition 2.5.6, Proposition 2.5.9, and Lemma 2.6.3 implies that we may assume that E(K) has a point of order 4. Let P 2 denote such a point in E(K). We pick generators x;y for the 2-adic Tate module T 2 (E) of E such that xP 2 (mod 4): Notice that E[4] E(F ). Then the 2-adic representation of the group Gal( F=F ) is given as follows: Gal( F=F )! Aut(T 2 (E)) 2 :7! 1 + 4a 4c 4b 1 + 4d ! (2.4) for a ;b ;c and d inZ 2 . Note that F contains a primitive 8th root of unity. Hence det( 2 ()) 1 (mod 8) since det( 2 ()) 8 = ( 8 ) = 8 . Computing the determinant, we obtain that a +d 0 (mod 2). Let E 0 = E=hP 1 i where P 1 = [2]P 2 and be the morphism E! E 0 . We will 46 choose the generators of T 2 (E 0 ) as x 0 and y 0 where 2x 0 = (x) and y 0 = (y). Hence, we nd the 2-adic representation of Gal( F=F ) on T 2 (E 0 ) as: 0 2 :7! 1 + 4a 8c 2b 1 + 4d ! (2.5) Since E(K) has a point of order 4, E 0 (K) contains full 2-torsion. See [Sil09, Example 4.5]. Hence by Lemma 2.5.1, the full 4-torsion E 0 [4] is contained in E 0 (F ). Hence the representation in (2.5) tells us that b is divisible by 2 for every 2 Gal( F=F ). Let b = 2b 0 . Also notice that E 0 (F ) must have a point of order 16 since E(F ) has a point of order 32. Let kx 0 +ly 0 (mod 16) be such a point for some k;l2Z and at least one of k;l is not divisible by 2. Since this point is in E(F ), it is xed under the action of Gal( F=F ). Then we obtain from the representation in (2.5) that (1 + 4a )k + 8c lk (mod 16) (2.6) (4b 0 )k + (1 + 4d )ll (mod 16) (2.7) Assumea is a unit inZ 2 for some, then so isd sincea +d 0 (mod 2). An easy computations show that k and l are both 0 (mod 2) which is a contradiction. Hence a 0 (mod 2) for all 2 Gal( F=F ) and so is d . Proposition 2.6.2 together with the results ofxx 2.5 imply that eitherE(K) =Z=4ZZ=4Z orE(F ) does not contain E[8]. If E(K) =Z=4ZZ=4Z, then we showed in Proposition 2.5.7 that it can not have a point of order 16. Hence, we may assume that E(F ) does not contain E[8] and it implies that b 0 is not divisible by 2 for some 1 . See the representation in 2.5. Once again using 2.6 and 2.7, we compute thatc 0 (mod 2) for all. Then the repre- sentation ofT 2 (E) in 2.4 implies thatE[8] is contained inE(F ) and we get a contradiction. See [Fuj04] for the case K =Q. Corollary 2.7.4. LetE :y 2 =x(x +a)(x +b) be an elliptic curve dened over K. Assume that E(K) tors =Z=2ZZ=8Z. Then E(F ) tors =Z=4ZZ=16Z: Proof. The statement follows from the Proposition 2.5.5 and Proposition 2.7.3. 47 Proposition 2.7.5. Let E=K be an elliptic curve. Then E(F ) cannot be isomorphic to Z=4ZZ=8ZZ=3Z. Proof. We will proceed the same way as in Proposition 2.7.3. The representation of Gal( F=F ) on T 2 (E) and T 2 (E 0 ) is same as given in the proof of Proposition 2.7.3. Since E 0 (F ) contains full 2 torsion and also a point of order 3, E 0 (F ) =Z=4ZZ=12Z by Propo- sition 2.5.6. Hence b 0 (mod 2) for all . If a 0 (mod 2) for all , then so is d and y 0 (mod 8) is stabilized under the action of Gal( F=F ). However, E 0 (F ) does not have a point of order 8. Hence a is not divisible by 2 for some 1 . Then similar to Proposition 2.7.3, rewriting 2.6 and 2.7 modulo 8, we get a contradiction. 2.8 Main Result Theorem 2.8.1. Let K be a quadratic cyclotomic eld, let E be an elliptic curve over K, and let F be the maximal elementary abelian 2-extensions of K. 1. If K =Q(i), then E(F ) tors is isomorphic to one of the following groups: Z=2ZZ=2NZ (N = 2; 3; 4; 5; 6; 8) Z=4ZZ=4NZ (N = 2; 3; 4) Z=NZZ=NZ (N = 2; 3; 4; 6; 8) orf1g;Z=3Z, Z=5Z, Z=7Z, Z=9Z, Z=15Z. 2. If K =Q( p 3), then E(F ) is either isomorphic to one of the groups listed above or Z=2ZZ=32Z: Proof. We begin with the list given in Theorem 2.4.6. Suppose that E(F ) (2) 6= 1. If E(F ) 2 0 = 1, then either Z=8ZZ=8Z is not contained in E(F ) or E(F ) =Z=8ZZ=8Z by our results inxx 2.5 and Proposition 2.6.2. Then with the notation of Theorem 2.4.6, if b = 3, then r = 0. By Proposition 2.7.3, if b = 2, then r 2. We obtain the groups: Z=2ZZ=2NZ for N = 1; 2; 4; 8; Z=4ZZ=4NZ for N = 1; 2; 4; Z=8ZZ=8Z 48 andZ=2ZZ=32Z if K =Q( p 3). Suppose E(F ) 2 0 =Z=3Z. IfZ=2ZZ=2Z is contained in E(K), then E(K) =Z=2Z Z=6Z and we showed that E(F ) =Z=4ZZ=12Z. Otherwise, we know that E(F ) cannot contain Z=4ZZ=4Z if K = Q(i). Similarly E(F ) cannot contain Z=8ZZ=8Z if K = Q( p 3). Along with Proposition 2.7.5, we are left with three possible groups: Z=2ZZ=6Z; Z=2ZZ=12Z; Z=4ZZ=12Z The case E(F ) 2 0 = Z=3ZZ=3Z follows from Proposition 2.7.2, Proposition 2.6.2 and Theorem 2.4.6. The only possible group is Z=6ZZ=6Z: Similarly whenE(F ) 2 0 =Z=5Z, it follows from Proposition 2.6.2, Lemma 2.7.1 and Theorem 2.4.6 that the only option is Z=2ZZ=10Z: The case where E(F ) (2) = 1 was studied in the rst section and we found the groups f1g;Z=3Z;Z=5Z;Z=7Z;Z=9Z;Z=15Z andZ=3ZZ=3Z: Remark 2.8.2. Every group we listed in Theorem 2.8.1 except Z=2ZZ=32Z also appears as the torsion subgroup of some elliptic curve dened over Q in its maximal elementary abelian 2 extensions. We were able to prove neither the nonexistence of an elliptic curve dened overQ( p 3) with such a subgroup in E(F ) nor give an example of such a curve. 49 Chapter 3 Monodromy of Fermat Surfaces The background on the modular description of Fermat curves and the modular symbols are given inxx 3.1. Inxx 3.2, we compute the modular symbols for Fermat curves and the generators of the rst integral homology group. After describing the bration and the induced action on a generic ber, we compute the monodromy on the rst homology group H 1 (C(n);Z) inxx 3.4. 3.1 Background 3.1.1 Modular Interpretation of Fermat curves Let H denote the extended upper half plane H[P 1 (Q). The group SL 2 (Z) acts on H by fractional linear transformations. Let M = " a b c d # be in SL 2 (Z). Then M:z := az +b cz +d The quotient H=G of H by the action of a nite index subgroup G of SL 2 (Z) is a projective curve and it is denoted as X(G). In the following, we describe (n); subgroups of SL 2 (Z) such that the Fermat curve x n +y n =z n is isomorphic to X((n)) for every n2Z 1 . We begin with dening a principle congruence subgroup of SL 2 (Z). (2) = (" a b c d # 2 SL 2 (Z) :a;d 1 mod 2 andb;c 0 mod 2: ) 50 It is known that (2) =hA;Bi with A := " 1 2 0 1 # and B := " 1 0 2 1 # : Let (n) = hA n ;B n ; (2) 0 i be the subgroup of (2) generated by A n ;B n , and the commutator subgroup (2) 0 of (2). Then H!X((2)) induces a map X((n))!X((2)): The modular curve X((n)) is isomorphic to the Fermat curve C(n) :x n +y n =z n and the modular curve X((2)) is a genus 0 curve (See [Roh77],[Lon08]). Moreover, the map X((n))!X((2)) corresponds to the projection map g :C(n)!P 1 (x :y :z)7! (x n :z n ) The morphism g has degree n 2 with ramication at the points 0; 1 and1 which are cusps ofX((2)). At each of these points, the ramication degree isn and hence we can conclude that the modular curve X((n)) has 3n cusps and there are exactly n of them lying above each of the points 0; 1, and1. Thus the cuspidal points on C(n) are the points [x :y :z] with xyz = 0 which are known as the points at innity. 3.1.2 Cusps of X((n)) Dene a i ;b i , and c i for 1in as: a i = (0 : i : 1); b i = ( i : 0 : 1); c i = ( i : 1 : 0) where =e =n and is a primitiven'th root of unity. Using the mapg described above, we see that the pointsa i ;b i andc i correspond to the cusps above 0; 1 and1 respectively. The point c i corresponds to the equivalence class (B j :1) for some j since the Galois group of the morphism g is (2)=(n) . 51 Lemma 3.1.1. Let denote the quotient map H! H=(n). Then (A k :1) =(B k :1): Proof. We will do induction on k. If k = 1, then (A:1) =(B:1) since BA 1 :1 = 1 and A:1 = (ABA 1 B 1 )B:1. Let 1 ; 2 be in (2). Then we have ( 1 2 :z) =( 2 1 :z) since (n) contains the commutator subgroup of (2). Assume (A k1 :1) =(B k1 :1): Remember that (z) =(z 0 ) if and only if there exist a 2 (n) such that z 0 = z. Then A k :1 =A:(A k1 :1) =A( 1 B k1 ):1 = 2 1 :B k1 A:1 = B k :1: where 1 ; 2 ; in (n). 3.1.3 Automorphisms of X((n)) It is proven in [Tze95] that the automorphism group of the Fermat curve x n +y n = z n is generated by two automorphisms of order n which are given by (x :y :z)7! ( n x :y :z); (x :y :z)7! (x : n y :z) and the permutation group S 3 on three letters. On the other side, the normalizer of (n) in SL 2 (Z) acts on X((n)). Notice that A is in the normalizer of (n) and it stabilizes B k :1 (or the points c i ) since (AB k :1) =(B k A:1) =(B k 1). The only automorphism of order n of X((n)) xing c k for all k is given by : (x :y :z)7! (x :y :z): Hence the automorphism corresponds to some power of A. 52 3.1.4 Modular Symbols Let M 2 be the free abelian group on the set of symbolsf;g with ;2P 1 (Q) modulo the 3-term relations f;g +f; g +f ;g = 0 and modulo any torsion. Dene a left action of SL 2 (Z) onM 2 by lettingg2SL 2 (Z) act by gf;g =fg;gg where the action of g = " a b c d # on and is as it is dened inxx 3.1.1. Let G be a nite index subgroup of SL 2 (Z). Then we can dene M 2 (G), the group of modular symbols for G as the quotient ofM 2 by the submodule generated by the set fxg:x :x2M 2 ;g2Gg and modulo any torsion. Suppose g2 SL 2 (Z). Then the Manin symbol associated to g is gf0;1g. Notice here that ifGg =Gg 0 , thengf0;1g =g 0 f0;1g inM 2 (G). Hence, there is a well-dened Manin symbol associated to each right coset of G in SL 2 (Z). We will use the notation [x] for the Manin symbol of Gx. (For more details, one can read [Ste07].) Dene a right action of G on the set of Manin symbols as follows: f0;1g:(hg) =h (gf0;1g): In other words, the group SL 2 (Z) acts on Manin symbols as h:[g] := [gh] for anyg;h2 SL 2 (Z): Manin [Man72] showed that every modular symbol can be written as a Z-linear combi- nation of Manin symbols and determined the relations between these generators. Let = " 0 1 1 0 # ; = " 0 1 1 1 # ; J = " 1 0 0 1 # : 53 Theorem 3.1.2 ([Man72]). If x is a Manin symbol, then x +x = 0 (3.1) x +x +x 2 = 0 (3.2) xxJ = 0: (3.3) Moreover, these are all the only relations between Manin symbols. LetG be a nite index subgroup of SL 2 (Z). Then the following result of Manin describes the integral homology of a modular curve using modular symbols. Theorem 3.1.3 ([Man72]). The representation of any class h2 H 1 (X(G);Z) as a sum P n k f k ; k g of Manin symbols can be chosen so that P k n k (( k )( k )) = 0 as a zero dimensional cycle on X(G). 3.2 Modular Symbols for Fermat Curves We would like to computeM 2 ((n)). Hence, we will begin with nding coset representatives of (n) in SL 2 (Z). We already mentioned that the cosets of (n) in (2) are given by A i B j : 0i;j (n 1): Lemma 3.2.1. A complete set of right coset representatives of (2) in SL 2 (Z) consists of the following matrices: " 1 1 1 0 # ; " 0 1 1 1 # ; " 0 1 1 0 # ; " 1 1 0 1 # ; " 1 0 1 1 # ; " 1 0 0 1 # : Proof. We rst label the matrices with the given order above as i 's for i = 1::: 6. Then one can show that i 1 j 62 (2) for any i6=j. Since the index of (2) in SL 2 (Z) is 6, we are done. Notice that = 2 and = 3 . By Lemma 3.2.1, the set of right cosets of (n) in SL 2 (Z) is given by fA i B j k : 0i;j (n 1) and 1k 6:g In the following computations, the powers of A and B are considered inZ n . We will begin with computing the relation 3.1 in Theorem 3.1.2. 54 3.2.1 -relations We rst compute that 1 =A 1 J 4 ; 2 = 5 ; 3 = 6 J; 4 =A 1 ; 5 =J 2 ; 6 =: Now there is a unique Manin symbol for each right coset of (n) in SL 2 (Z). Thus we have: [A i B j 1 ] = [A i1 B j 4 ]; [A i B j 2 ] = [A i B j 5 ]; [A i B j 3 ] = [A i B j ]; [A i B j 4 ] = [A i+1 B j 1 ]; [A i B j 5 ] = [A i B j 2 ]; [A i B j 6 ] = [A i B j 3 ]: Notice that the right cosets of (n) in (2) commute since (n) contains the commutator subgroup of (2). Now using Equation 3.1, we have the following equalities: [A i B j 1 ] + [A i1 B j 4 ] = 0 (3.4) [A i B j 2 ] + [A i B j 5 ] = 0 (3.5) [A i B j 3 ] + [A i B j ] = 0: (3.6) ThereforeM 2 ((n)) is generated by f[A i B j k ] : 0i;j (n 1) andk = 1; 2; and 6:g 3.2.2 -relations Similarly we compute k for k = 1::: 6 as the folowing: 1 =A 1 ; 2 =A 1 ; 3 =A 1 4 ; 4 =AB 1 5 J; 5 =B 3 J; 6 =: And the Manin symbol of k is: [A i B j 1 ] = [A i1 B j ]; [A i B j 2 ] = [A i+1 B j 1 ]; [A i B j 3 ] = [A i1 B j 4 ]; [A i B j 4 ] = [A i+1 B j1 5 ]; [A i B j 5 ] = [A i B j+1 3 ]; [A i B j 6 ] = [A i B j 2 ]: Hence by Equation 3.2 we obtain: 55 [A i B j ] + [A i B j 2 ] + [A i+1 B j 1 ] = 0 (3.7) [A i B j 3 ] + [A i1 B j 4 ] + [A i B j1 5 ] = 0: (3.8) Now using -relations, Equation 3.8 becomes [A i B j ] + [A i B j 1 ] + [A i B j1 2 ] = 0 (3.9) Hence by Equation 3.7, we conclude that the set of Manin symbols f[A i B j ] and [A i B j 2 ] : 0i;j (n 1):g generateM 2 ((n)). Therefore using Equation 3.9 and Equation 3.7, we obtain the following relation: [A i+1 B j ] + [A i+1 B j1 ] = [A i B j ] + [A i B j ] (3.10) for 0i (n 1) and 0jn 1. Remember that 2 =. Proposition 3.2.2. The group of modular symbols for the Fermat group (n) is free of rank n 2 + 1 and it is generated by f[A i B j ] : 1i (n 1); 0j (n 1)g f[A n1 B j ] : 0j (n 1)g and [B n1 ]: Proof. We will use x i;j := [A i B j ] and y i;j := [A i B j ] to ease the notation. We put Lexico- graphical order on the setfx i;j ;y i;j :i = 1;:::;n; j = 0;:::n 1g as follows: For any i;j;k;l, x i;j y k;l and x i;j x k;l or y i;j y k;l if and only if i<k or i =k and jl: Also note that the index set for i;j is Z n . Hence we see that if i =n 1 and j = 0, then x i+1;j1 =x 0;n1 . 56 We already showed that x i;j and y i;j generate the group of modular symbols for the Fermat group (n). Furthermore we know from Equation 3.10 that they satisfy the relation x i;j +y i;j =x i+1;j +y i+1;j1 : (3.11) Equation 3.11 gives us a system of linear equations. LetR be the (n 2 2n 2 ) matrix obtained by this system of equations with respect to the ordered basis fx i;j ;y i;j :i = 1;:::;n 1; j = 0;:::n 1g: Let R i;j denote the ij'th row of R, i.e., the row corresponding to the equation: x i;j +y i;j x i+1;j y i+1;j1 = 0: Notice that for i n 2 the rst non-zero entry of each R i;j is 1 and it is the coecient of x i;j . Assume i =n 1. For j xed, we add P i6=(n1) R i;j to R n1;j and denote the new row we obtain by R n1;j 0 . Notice that the rst n 2 entry of R n1;j 0 is 0 since n1 X i=0 (x i;j x i+1;j ) = 0: The equation corresponding to R n1;j 0 gives us that n1 X i=0 x i;j +y i;j x i+1;j y i+1;j1 = n1 X i=0 (y i;j y i+1;j1 ) = 0 Ifj 1, then the rst non-zero term ofR n1;j 0 is the coecient ofy 0;j1 . Similarly ifj = 0, then it is the coecient of y 0;0 . Now let R n1;j 1 = X kj R n1;k 0 and replace R n1;j 0 by R n1;j 0 . Assume j6=n 1, then the rst non-zero entry of R n1;j 1 is 1 and it is the coecient of y 0;j . If j =n 1, then R n1;n1 1 is the zero row since X k X i y i;k y i+1;k1 ! = 0: 57 To summarize, we obtain a matrix of the form: 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 1 . . . 0 . . . 1 0 1 . . . 1 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 Hence fy i;j : 1i (n 1); 0j (n 1)g fx n1;j : 0j (n 1)g and [y 0;n1 ] are the free variables and they generate the group of modular symbols for (n). The proof of the proposition gives us that if j6=n 1, then X kj n1 X i=0 (y i;k y i+1;k1 ) ! = n1 X i=0 y i;n1 y i;j = 0: (3.12) We will need Equation 3.12 in Sectionx 3.4. 3.2.3 Homology of Fermat Curves In this section, we will describe the rst integral homology group of Fermat curves in terms of modular symbols. Proposition 3.2.3. The homology group H 1 (X((n));Z) of X((n)) is generated by i;j := [A i ] [A i+1 B n1 ] + [A i+1 B j ] [A i B j+1 ] for 1i (n 2) and 0j (n 2). 58 Proof. We rst compute that f0;1g =f1; 0g: Then A i f1; 0gA i B j f1; 0g =fA i :1;A i B j :1g and A i+1 B j1 f1; 0gA i+1 B n1 f1; 0g =fA i+1 B j1 :1;A i+1 B n1 :1g: We proved inxx 3.1.2 that (A i (1)) =(B i (1)). Hence, (A i+1 B n1 :1) =(A i+1 gA n1 :1) for someg2 (n): =(g 0 A i+1 A n1 :1) for someg 0 2 (n): =(A i :1): One can similarly show that (A i+1 B j1 :1) =(A i B j :1): Therefore we obtain that (A i :1)(A i B j :1) +(A i+1 B j1 :1)(A i+1 B n1 :1) = 0: Hence by Theorem 3.1.3, i;j = [A i ] [A i+1 B n1 ] + [A i+1 B j1 ] [A i B j ] is inH 1 (X((n));Z) for everyi;j such that 1in 2 and 0jn 2. We will show that i;j 's areZ-linearly independent. To ease the notation, we will use y i;j for [A i B j ] as in the proof of Proposition 3.2.2. Hence i;j denotes the cycle y i;0 y i+1;n1 +y i+1;j y i;j+1 and suppose that X i;j n i;j i;j = 0: (3.13) 59 We will prove by induction that n i;j = 0 for alli = 1;:::;n 2 and for allj = 0;:::;n 2. We know by Proposition 3.2.2 that y i;j 's areZ-linearly independent. If j is xed, then the coecient of y 1;j+1 in the sum 3.13 isn 1;j . Hence n 1;j = 0 for all 0jn 3: Similarly the coecient of y 1;0 is P j n 1;j and thus n 1;n2 = 0: This proves the base step of the induction. Assume n i1;j = 0 for all j = 0;:::;n 2. If j 1, then the coecient of y i;j isn i;j1 +n i1;j and using the induction hypothesis, we obtain n i;j1 = 0 for all j = 1;:::;n 2: For j = 0, we look at the coecient of y i ; 0 which equals to n i1;0 + P j n i;j . Since we showed n i;j = 0 for every jn 3 we see that n i;n2 must also equal to 0. Hence the set f i;j : 1in 2; 0jn 2g is linearly independent and they form a basis for the homology group since the genus of the Fermat curve is 1=2(n 1)(n 2) which implies that the rank of the rst homology group is (n 1)(n 2). 3.3 Fermat Surfaces 3.3.1 Denition of Monodromy Let X and Y be algebraic varieties over C. Suppose that f :X!Y is a bration. Let y 0 be inY and letX 0 be the ber off aty 0 . For in 1 (Y;y 0 ), deneH :X 0 [0; 1]!Y as H(x;t) := (t) for all x2X 0 . Then by the homotopy lifting property of f, there exists a lift ~ H :X 0 [0; 1]!X as given in the following diagram (See [DK01] for more details.). 60 X 0 f0g X X 0 [0; 1] Y f ~ H H The commutativity of the above diagram induces a morphism X 0 ! X 0 (at t = 1) which induces a homomorphism on the homology groups. Hence we obtain an action of the fundamental group 1 (Y;y 0 ) of Y on H 1 (X 0 ;Z): : 1 (Y;y 0 )! Aut(H 1 (X 0 ;Z)) which we call the monodromy action. It is worth noting here that the induced mapX 0 !X 0 is a self homotopy equivalence and it is unique up to a homotopy. 3.3.2 Description of the Fibration Let X denote the Fermat surface given by the equation x n +y n +z n =w n : inP 3 C . Consider the rational map f :X!P 1 given by f : (x :y :z :w)7! (z :w): We rst blow up the variety X successively at the points of the set B =f(x :y : 0 : 0)2P 3 : x n +y n = 0g: We obtain a variety X, a proper morphism f : X!P 1 and a birational map ' : X!X. A generic ber of f is given by the equation x n +y n = 1. The bers of f are projective curves since f is proper. Then a generic ber of f is isomorphic to the curve x n +y n =z n : The singular bers are given by x n +y n = 0 when z n = w n . Let U denote the variety f 1 (P 1 S) with S =f[ k n : 1]2P 1 :k = 0;:::;n 1:g: 61 We obtain a bration U! V where V =P 1 S by Ehremann's theorem [Ehr95]. Let y 0 be the point [0 : 1]2P 1 . To describe the monodromy action, we rst formulate the generators of the fundamental group 1 (V;y 0 ) in the following way: Ifn is even, then take leaves of the polar roser = 2cos( n 2 ) and otherwise take leaves of r = 2cos(n) (in the complex plane given by w = 1). When n = 3, one can visualize these generators as in the following gure. 0 30 60 90 120 150 180 210 240 270 300 330 0 0:5 1 1:5 2 2 3 1 Figure 3.1: Generators of 1 (V;y 0 ) in w = 1 plane for n = 3. Let k be the loop around the singular point [ k n : 1] and let denote the monodromy representation. First we notice that k e 2ki n is 1 and hence k n = n 1 . Consider (t) = 1 k (t) n = 1 1 (t) n . The function has the same start and the end point, i.e., it is a loop. Since the lines at ==2n (respectively ==n) are tangent to 1 when n is odd (respectively when n is even), (t) will be tangent to the lines ==2 or. Hence we can nd continuous functions r(t) and g(t) such that 1 k (t) n =r(t)e 2ig(t) : and g(0) = 0 and g(1) = 1. Let (u :v :w) be a point on the ber x n +y n =z n . Note that k (t) lies in the complex plane w = 1. We would like to nd a lift k for k such that f( ) = . Since (t) lies in the plane w = 1, (t) also should be in w = 1. We will assume that w6= 0. Since X and X are isomorphic on w6= 0, we may assume that U is given by the equation x n +y n =w n (1u n ) for u n 6= 1. Remember that we pick the base point y 0 as u = 0. Let (x :y :w : 0) be a point in the ber of u = 0. We observe the followings: 1. The functionr(t) only takes positive real values and there exist a function whosen'th power is r(t) such that r 1=n (t) is also positive real for every t. 62 2. Dene ~ :I!U ~ (t) = ((t)x;(t)y :w :(t)) where (t) =r(t) 1=n e 2ig(t) n . 3. The path (t) takes values in U, i.e., (t)w n (1(t) n ) = (1(t) n )(x n +y n ): 4. The path begins at ~ (0) = (x :y :w) and ends at ~ (1) = ( n x : n y :w). Hence ~ induces an action on the ber x n +y n =w n as (x :y :w)7! ( n x; n y :w): In particular, this action only depends on the choice of the n'th root of unity. 3.4 Calculation of the Monodromy Inxx 3.3, we showed that the action of each generator of the fundamental group on a generic ber is identical. We will compute the monodromy by describing the action of A on the modular symbols i;j using our results fromxx 3.1.3. Remember that C(n) denotes the Fermat curve x n +y n =z n . Theorem 3.4.1. Let k be the generators of 1 (V;y 0 ) as given in Sectionx 3.3. Then for any k the action of k on the homology group H 1 (C(n);Z) is given by the following: i;j 7! i+1;j for 1in 3; 0jn 2: and if i =n 2, then n2;0 7! n2 X k=1 k;n1k and n2;j 7! j1 X l=0 0 @ nl2 X s=jl s;l s;n2s+jl 1 A n2 X i=j+1 i;ni+j1 for 1jn 2. 63 Proof. We recall from equation (3.12) that n1 X i=0 (y i;n1 y i;j ) = 0 for any 0jn 2. If i<n 2, then it is clear that A i;j = i+1;j . Assume i =n 2, then A n2;j =A(y n2;0 y n1;n1 +y n1;j y n2;j+1 ) =y n1;0 y 0;n1 +y 0;j y n1;j+1 =y n1;0 y 0;n1 +y 0;j y n1;j+1 + n1 X i=0 (y i;n1 y i;j ): =y n1;0 y n1;j+1 + n1 X i=1 (y i;n1 y i;j ) We will separate the cases j = 0 and j 1. Assume j = 0, then A n2;0 =y n1;0 y n1;1 + n1 X i=1 (y i;n1 y i;0 ) =y n1;1 +y 1;n1 + n1 X i=2 y i;n1 n2 X i=1 y i;0 = n2 X i=1 (y i;ni y i+1;ni1 ) + n2 X i=1 (y i+1;n1 y i;0 ) = n2 X i=1 (y i+1;n1 y i;0 +y i;ni y i+1;ni1 ) = n2 X i=1 i;ni1 : 64 Assume j 1. The proof is similar to the case j = 0 but it requires more work. A n2;j =y n1;0 y n1;j+1 + n1 X i=1 (y i;n1 y i;j ) =y n1;0 y n1;j+1 n1 X i=1 y i;n1 + n1 X i=1 y ni;j =y n1;0 y n1;j+1 +y j+1;n1 y n1j;j + n1 X i=1 i6=j+1 y i;n1 n1 X i=1 i6=j+1 y ni;j = (y j+1;n1 y n1;j+1 ) + 0 @ n1 X i=j+2 y i;n1 y i1j;j 1 A + j X i=1 y i;n1 y n+i1j;j ! +y n1;0 y n1j;j : since n1 X i=j+2 y ni;j = n1 X i=j+2 y i1j;j and j X i=1 y ni;j = j X i=1 y n+i1j;j : We rst compute that n1 X i=j+2 (y i;n1 y i1j;j ) = n1 X i=j+2 (y i;n1 y i1;0 ) + n1 X i=j+2 (y i1;0 y i1j;j ): We can easily calculate that 0 @ n1 X i=j+2 y i;n1 y i1;0 1 A + (y j+1;n1 y n1;j+1 ) = n1 X i=j+2 (y i;n1 y i1;0 ) + n2 X i=j+1 (y i;n+ji y i+1;n+ji1 ) = n2 X i=j+1 (y i+1;n1 y i;0 +y i;n+ji y i+1;n+ji1 ) = n2 X i=j+1 i;ni+j1 : 65 Now we will show that 0 @ n1 X i=j+2 y i1;0 y i1j;j 1 A +y n1;0 y n1j;j + j X i=1 y i;n1 y n+i1j;j (3.14) = j1 X l=0 0 @ nl2 X s=jl s;l s;n2s+jl 1 A : (3.15) Left hand side of equation (3.14) becomes n1 X i=j+1 y i;0 y ij;j + j X i=1 y i;n1 y n+i1j;j : We can rearrange the rst sum as the following: n1 X i=j+1 y i;0 y ij;j = n1 X i=j+1 0 @ i X l=ij+1 y l;il y l1;il+1 1 A = j1 X l=0 n2j X k=0 y 1+j+kl;l y j+kl;l+1 ! = j1 X l=0 0 @ nl2 X s=jl y 1+s;l y s;l+1 1 A : We may also rearrange the second sum as: j X i=1 y i;n1 y n+i1j;j = j X i=1 n2+ij X k=i y k;n1+ik y k+1;n2+ik ! = j1 X l=0 0 @ nl2 X s=jl y s;n1s+jl y s+1;n2s+jl 1 A : 66 Combining these two sums, = j1 X l=0 0 @ nl2 X s=jl y 1+s;l y s;l+1 1 A + j1 X l=0 0 @ nl2 X s=jl y s;n1s+jl y s+1;n2s+jl 1 A = j1 X l=0 0 @ nl2 X s=jl y 1+s;l y s;l+1 +y s;n1s+jl y s+1;n2s+jl 1 A = j1 X l=0 0 @ nl2 X s=jl s;l s;n2s+jl 1 A : Corollary 3.4.2. 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Abstract This thesis is focused on two problems

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Creator Ejder, Özlem (author)

Core Title Torsion subgroups of elliptic curves in elementary abelian 2-extensions and monodromy of Fermat surfaces

School College of Letters, Arts and Sciences

Degree Doctor of Philosophy

Degree Program Mathematics

Publication Date 04/21/2017

Defense Date 03/22/2017

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag elliptic curves,Fermat curves,Fermat surfaces,growth of torsion,monodromy,OAI-PMH Harvest

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Friedlander, Eric (committee chair), Guralnick, Robert (committee member), Kamienny, Sheldon (committee member), Kocer, Yilmaz (committee member)

Creator Email ejder@usc.edu,ozheidi@gmail.com

Permanent Link (DOI) https://doi.org/10.25549/usctheses-c40-365666

Unique identifier UC11255925

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Legacy Identifier etd-EjderOzlem-5265.pdf

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Rights Ejder, Özlem; Ejder, Ozlem

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