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Growth of torsion of elliptic curves over noncyclotomic imaginary quadratic fields of class number 1
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Growth of torsion of elliptic curves over noncyclotomic imaginary quadratic fields of class number 1
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Content
GROWTH OF TORSION OF ELLIPTIC CURVES OVER NONCYCLOTOMIC
IMAGINARY QUADRATIC FIELDS OF CLASS NUMBER 1
by
Irmak Balc ¸ık
A Dissertation Presented to the
FACULTY OF THE USC GRADUATE SCHOOL
UNIVERSITY OF SOUTHERN CALIFORNIA
In Partial Fulfillment of the
Requirements for the Degree
DOCTOR OF PHILOSOPHY
(MATHEMATICS)
August 2022
Copyright 2022 Irmak Balc ¸ık
To my teachers
ii
Acknowledgements
The author acknowledges the fellowship support awarded by the Department of the Mathematics at
USC for Fall 2020. That semester was the most fruitful times for this project.
I’m indebted to my advisor Sheldon Kamienny for continuous guidance, encouragement and
support. I feel fortunate enough for his suggestion of the thesis topic. I greatly benefited from
his lectures, ideas and beautiful perception of the mathematics world. For the time to come,
conversations with my advisor will continue to be a source of inspiration.
I’m grateful to my committee member Eric Friedlander for his outstanding interest and perspec-
tive in the research which enrich the content of this thesis. I would like to thank Susan Montgomery
whom I see as a mentor and very thankful to her throughout my years at USC for being always
open to hear my struggles and help me go through them.
I would also like to thank
¨
Ozlem Ejder for sharing her knowledge and always being patient with
my mathematical thoughts which help me a lot with speeding up the learning process.
I have been fortunate enough to have excellent friends over the years - Maria Allayioti, Lernik
Asserian, Zhanerke Temirgaliveya, Jiyeon Park, Ezgi Kantarci O˘ guz, Mengsha Yao, Bet¨ ul Mut-
lug¨ un. I would like to write separately for each of them. They teach me a lot and offer great
perspectives.
Finally, I would like to thank my family; my parents, my brother Fırat, my niece S ¸afak for their
infinite support and love.
iii
Table of Contents
Dedication ii
Acknowledgements iii
Abstract vi
List of Tables vii
Chapter 1: Introduction 1
Chapter 2: Background on Elliptic Curves 8
2.1 Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 Weil Pairing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.5 Modular curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Chapter 3: Torsion over Quadratic Fields 20
Chapter 4: K-rational points on X
0
(N) 24
4.1 N=21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 N=27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.3 N=28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.4 N=30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.5 N=33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.6 N=35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.7 N=40 & 48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.8 N=45 & 55 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.9 N=77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Chapter 5: Growth in Case E(K)[2]6=Z
2
Z
2
48
5.1 Certain Restrictions on Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5.2 Classification of Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
5.3 Examples of Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
iv
Chapter 6: Growth in Case E(K)[2] =Z
2
Z
2
55
6.1 Classification of Twists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
6.2 Growth of Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
6.3 Examples of Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
References 83
v
Abstract
LetK be a noncyclotomic imaginary quadratic field of class number one andE an elliptic curve
defined overK: We study the relation ofE(L)
tor
withE(K)
tor
whereL is any quadratic extension
ofK: IfE(K)
tor
( E(L)
tor
; we say that torsion grows. IfE satisfies one of the following cases:
E(K)[2]'Z
1
orE(K)[2]'Z
2
Z
2
; we determine the complete classification of possible torsion
structures appearing asE(L)
tor
for any quadratic extensionL ofK:
vi
List of Tables
4.1 Representatives (E;C) of isomorphism classes corresponding to noncuspidal ra-
tional points ofX
0
(21) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 Representatives (E;C) of isomorphism classes corresponding to exceptional K-
rational points ofX
0
(28) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.3 Representatives (E;C) of isomorphism classes corresponding to exceptional K-
rational points ofX
0
(33) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.4 The rational points onX
0
(77)
+
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4.5 Noncuspidal quadratic points ofX
0
(77) . . . . . . . . . . . . . . . . . . . . . . . 47
5.1 Growth of Torsion overK =Q(
p
2) . . . . . . . . . . . . . . . . . . . . . . . . 54
6.1 K-rational points onE
C
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
6.2 Determination of
1
(E
C
(K)
tor
) . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
6.3 Example of GrowthE(K)
tor
'Z
2
Z
2
overK =Q(
p
2) . . . . . . . . . . . . 81
6.4 Example of GrowthE(K)
tor
'Z
2
Z
2
overK =Q(
p
7) . . . . . . . . . . . . 82
6.5 Example of GrowthE(K)
tor
'Z
2
Z
4
overK =Q(
p
7) . . . . . . . . . . . . 82
6.6 Examples of Growth ofE(K)
tor
'Z
2
Z
6
overK =Q(
p
D) . . . . . . . . . . . 82
6.7 Example of GrowthE(K)
tor
'Z
2
Z
6
over anyK2S . . . . . . . . . . . . . . 82
6.8 Examples of Growth ofE(K)
tor
'Z
2
Z
6
overK =Q(
p
19) . . . . . . . . . 82
6.9 Examples of Growth ofE(K)
tor
'Z
2
Z
8
overK =Q(
p
7) . . . . . . . . . . 82
vii
Chapter 1
Introduction
The theory of algebraic curves is a masterful culmination of algebra and geometry. Arithmetic
algebraic geometry is mainly concerned with the study of set of points lying on those curves that can
be represented in a special form. In this thesis, we are interested in elliptic curves, a central branch
of mathematics having connections with various other subfields, ranging from number theory to
complex analysis, and from cryptography to mathematical pyhsics.
Given a number fieldK, an algebraic curveC=K, a pointP onC is called K-rational if the
coordinates ofP are defined overK. C(K) denotes the set of K-rational points onC: There is a
trichotomy forC(K) determined by the “genus”g2N of the curve. To give a very rough idea of
what genus is, algebraic curves for geometers are analogous to Riemann surfaces. These surfaces
have a certain number of handles called the genus.
1. g = 0, thenC(K) is either empty or infinite.
2. g> 1, thenC(K) is either empty or finite (by Faltings).
There remains the caseg = 1 which sits in between these two extremes. It is harder to determine
whether a curve of genus 1 has a rational point. The Hasse principle fails in this case. IfC of genus
1 has aK-rational point, thenC is an elliptic curve and in this caseC(K) can be finite or infinite,
and deciding which can be challenging.
More precisely, an elliptic curveE over a fieldK is a smooth projective curve of genus 1 with a
distinguishedK-rational point. As a consequence of Riemann-Roch theorem, if char(K)6= 2; 3;E
has an affine model of the formy
2
=x
3
+Ax+B for someA;B2K with =16(4A
3
+27B
2
)6=
1
0: There is just one more projective point “at infinity” where all vertical lines meet, namely
O = [0 : 1 : 0]: One of the aspects that makes the theory of elliptic curves so rich is that the set
E(K) can be equipped with a group structure which is geometric in nature. In fact, around 1900
Poincare conjectured thatE(K) is finitely generated abelian group whenK is the field of rational
numbers. In 1922, Mordell proved the landmark result describingE(Q) and in 1928, Andre Weil
generalized this result not only for elliptic curves over algebraic number fields but also for abelian
varieties. Their work culminated in the following theorem:
Theorem 1.1 (Mordell 1922 - Weil 1928). LetK be a number field andE an elliptic curve defined
overK: ThenE(K) is a finitely-generated abelian group and hence
E(K)'E(K)
tor
Z
r
whereE(K)
tor
denotes the torsion part ofE(K) andr is the rank ofE(K).
The rank of an elliptic curve is mysterious. It may happen (and often does throughout) that
r = 0. There is no known effective algorithm to compute the rank. Note that the Mordell-Weil
theorem implies thatE(K)
tor
is always finite. This fact prompts a natural question: what abelian
groups can appear in this context? In this work, we are interested in the torsion part of the group
E(K). The most significant result regarding the torsion part ofE(Q) is the following theorem due
to Mazur.
Theorem 1.2 (Mazur [27]). Let E be any elliptic curve overQ. Then,E(Q)
tor
must be one of the
following 15 groups:
Z=nZ; 1n 12; n6= 11
Z=2ZZ=2nZ; 1n 4
(1.1)
Subsequently, the work of Mazur was generalized by Kenku, Momose and Kamienny to the set
of all quadratic fields.
2
Theorem 1.3 (Kamienny [17], Kenku and Momose [20]). Let K be a quadratic number field and
E an elliptic curve over K. ThenE(K)
tor
is isomorphic to one of the following 26 groups:
Z=nZ; 1n 18; n6= 17
Z=2ZZ=2nZ; 1n 6
Z=3ZZ=3nZ; 1n 2
Z=4ZZ=4Z :
(1.2)
Kamienny’s result provided a strong evidence for the Strong Uniform Boundedness Conjecture
which was eventually proven by Merel.
Theorem 1.4 (Merel [28]). For every integer d there existsB
d
0 such that for every number
field K with [K :Q] =d and every elliptic curveE=K we have
jE(K)
tor
jB
d
:
With the above results in hand, researchers began to study the deeper question of classifying
all the possible torsion groups that can occur for each fixedd> 2: Over cubic fields, the analogue
of Mazur’s theorem has been established last year [8]. Although we still do not have a complete
classification analogous to (1.1) over quartic fields, Jeon, Kim and Park [16] determined which
groups can appear for infinitely many non-isomorphic elliptic curves. Moreover, Derickx, Kami-
enny, Stein and Stoll [9] showed that 17 is the largest prime dividingjE(K)
tor
j asE varies over
any quartic fieldK: Our work is a step in the direction of classifying growth of torsion in quadratic
extensions of base fields larger thanQ: More precisely, we ask “for a given elliptic curve defined
over a quadratic number fieldK; how doesE(L)
tor
relate withE(K)
tor
where [L : K] = 2?”. If
E(K)
tor
(E(L)
tor
; we say that torsion grows. In this thesis, we answer this question for the setS
which consists of all the noncyclotomic imaginary quadratic fields of class number one, that is
S =fQ(
p
2);Q(
p
7);Q(
p
11);Q(
p
19);Q(
p
43);Q(
p
67);Q(
p
163)g:
3
We choose to work over the base fields inS because the complete classification of possible
torsion structures occurring over these fields is known by Theorem 1.5 (stated below) and Theorem
3.1. The method used in their paper is outlined by Kamienny and Najman [18].
Theorem 1.5 (Sarma and Saika [36]). Let E be an elliptic curve over an imaginary quadratic
number field K of class number one. Then, as K varies,E(K)
tor
is isomorphic to one of the groups
appearing in Theorem 3, excluding the groupsZ=13Z;Z=16Z andZ=18Z.
Newman [31] studied the case of quadratic cyclotomic fields which areQ(
p
1) andQ(
p
3):
Initially, our goal was to complement Newman’s work with the same methods relying on explicit
computations. However, it turned that the computations became quite infeasible. As a result, we
developed a more abstract approach, which has the benefit of not in fact requiring the fields to be of
class number one. So, over the course of this work, our methods are more general than they might
seem.
Our first result for elliptic curves with trivial 2-torsion will complete the classification of torsion
growth for all imaginary quadratic fields of class number one. Our second result for elliptic curves
with full 2-torsion will generalize both Kwon’s work [25] in whichK =Q and Newman’s work
[31] in whichK =Q(
p
1) orQ(
p
3): Kwon’s result has been extended by Tornero et. al. [12]
to the cases in whichE(Q)
tor
is cyclic thus finishing the classification overQ: Our methods didn’t
explicitly rely upon the facts that the ring of integersO
K
is UFD and has a finite unit group whereas
Kwon’s methods do. Instead, we generalized Newman’s methods more efficiently without using
Grobner basis, (notably Lemma 7) as well as using new techniques ranging from the symmetric
squares of modular curves to the relative Chabauty, from the Galois action to the parametrization of
elliptic curves with specified torsion [35], which can be easily extended to arbitrary number fields.
We now give a more detailed overview of the paper. The second chapter of this thesis is
reserved for the background on elliptic curves. The third chapter studies the torsion structures of
elliptic curves over quadratic number fields, with a particular focus on the noncyclotomic imaginary
quadratic fields of class number one.
4
Inx4, as a major step towards the main results, we classify the isomorphic elliptic curves with
certainN-isogenies. Generally speaking, understandingK-rational cyclicN-isogenies for given
number fieldK is equivalent to finding noncusidalK-rational points on the modular curvesX
0
(N).
More precisely, for givenN the set of noncuspidalK-rational points on the modular curveX
0
(N)
parametrizes theK-isomorphism classes of pairs (E;C) whereE is an elliptic curve defined over
K andCE(K) is a cyclic Gal(K=K)-invariant subgroup of orderN: A major difficulty arises
when the model of X
0
(N) is of high genus. This is the case for N = 77 in which X
0
(77) has
genus 7 and is the hardest case that we address. Results of Bars [2] and Harris-Silverman [14]
assert that ifX
0
(N) is of genus greater than 2, then it has finitely many quadratic points, except
28 values ofN which are 22,23,26,28,29,30,31,33,35,37,39,40,41,43,46,47,48,50, 53,59,61,65,71,
79,83,89,101,and 131: In particular, the set of all quadratic points onX
0
(77) is finite. By using the
relative symmetric Chabauty along with the Mordell-Weil sieve, we classify all the quadratic points
ofX
0
(77): This approach is presented in more detail by earlier work of [3] in which the highest
genus handled is 5:
In the cases N = 28; 30; 33; 35; 40 the curve X
0
(N) is hyperelliptic of genus 2 or 3 whose
Jacobian J
0
(N)(Q) is finite. As the set of K-rational points on X
0
(N) is finite by a theorem
of Faltings, and does not form a group structure, the main trick for studying quadratic points on
X
0
(N) is to instead study the rational points on its symmetric squareX
0
(N)
(2)
: By the existence
of a degree 2 map fromX
0
(N)!P
1
(overQ) the set of quadratic points are classified into two
subsets: the ones which arise as a preimage of rational point onP
1
, callednonexceptional and the
ones which do not arise in this way, calledexceptional: This comes down to finding the rational
points on the Jacobian J
0
(N)(Q) of X
0
(N): Since J
0
(N)(Q) <1; the exceptional points are
finite. Every nonzero rational point ofJ
0
(N) has a unique Mumford representation, see Theorem
4.145 of [1]. Magma lists points ofJ
0
(N)(Q) in Mumford representation. From these Mumford
representations, one can extract (exceptional) points on X
0
(N): In fact, Bruin and Najman [5]
classified all exceptional quadratic points on these modular curvesX
0
(N) with this method. We
complement their lists forN = 30; 35 and show that there are no nonexceptionalK-rational points
5
lying on these two curves for anyK inS: For this purpose, we benefit from the classification of
points on quadratic twists ofX
0
(N) determined in [33].
In the casesN = 45; 55 the curvesX
0
(N) are nonhyperelliptic curves of genus 3; 5 respectively.
The quadratic points on these curves are determined by Siksek and Ozman [34]. We compute
whether each point lying on the curves gives rise to aK-rational cyclicN-isogeny pointwise defined
in a quadratic extension ofK: In the casesN = 21; 27;X
0
(N) is an elliptic curve with positive rank
over someK2S which makes the enumeration of all quadratic points computationally infeasible.
We then shift our perspective and resolve these cases using the division polynomial method.
Inx5, we present a number of known results as well as prove various restrictions on growth which
help us narrow the focus in the rest of this chapter. An important ingredient for the classification of
torsion growth is the description ofT
K
(G), the set of finite abelian groupsH (up to isomorphism)
such that there is an elliptic curve E=K with E(K)
tor
' G and E
d
(K)' H where E
d
is the
quadratic twist ofE byd2K: With the classification of torsion of quadratic twists, we prove the
main result of this chapter for elliptic curves overK with trivial 2-torsion.
Inx6, we study elliptic curves with full 2-torsion. More precisely, an elliptic curve E over
K with full 2-torsion admits a Weierstrass model of the formy
2
= x(x +)(x +) where;
inO
K
: Moreover, the existence of points of order 3, 4 or 8 is equivalent to finding solutions to
certain equations involving and . (See Theorem 6.1 below). These characterizations can be
used to eliminate points of order that are multiples of 3; 4 or 8. We start with proving Theorem 6.4
which classifies the torsion structures on quadratic twists. IfL =K(
p
d); andE=K is an elliptic
curve withE(K)
tor
given, thenE
d
(K)
tor
sets a bound forE(L)
tor
(see Proposition 3.1). Hence if
one wants to study the relation ofE(L)
tor
withE(K)
tor
; the first task is to classify which torsion
groups occur asE
d
(K)
tor
: To do so, we make use of the parametrization of elliptic curves with the
described torsionZ=2ZZ=2nZ (1 n 6): Twisting and equating parametrized models of
curves give rise to systems of Diophantine equations. Finally, we prove our main result Theorem
6.3. With the classification of twists in hand, we obtain a finite list of possible torsion structures
ofE(L)
tor
for a given torsion groupE(K)
tor
: Furthermore, the growth occurs inL if and only if
6
there exists P 2 E(K)
tor
with P 2 2E(L)
tor
n 2E(K)
tor
: The criteria of the existence of such a
point is given in Lemma 6. Applying this criteria yields systems of Diophantine equations as in
the previous section. A difficulty arises when we deal with torsion points of high order such as 10
or 12. In order to work around this problem, we shift our perspective and use the parametrization
of elliptic curves with torsionZ=2ZZ=10Z orZ=2ZZ=12Z described in [35] together with
the duplication formula of a point on an elliptic curve.
Notation and Computations
Unless it is stated, the number field K is inS and L denotes a quadratic extension over K: We
interchangebly use the following representations of the sets
S =fQ(
p
D) : D2f2;7;11;19;43;67;163gg
T =fQ(
p
D) :D2f1;2;3;7;11;19;43;67;163gg
Let [n] denote the multiplication by n map on an elliptic curve E and E[n] the kernel of [n].
Throughout, we useZ
n
for the cyclic group of ordern: Given an elliptic curveE=K withE(K)[2]'
Z
2
Z
2
; we work with the Weierstrass modelE(;) :y
2
=x(x +)(x +) where;2O
K
that is the the ring of integers of K: The rank and torsion subgroup computations are done by
using the computer algebra software programs Magma and Sage. The code that verifiesX
0
(77)
computations can be found here:
https :==github:com=irmakbalcik=X077
7
Chapter 2
Background on Elliptic Curves
LetK be a number field andK denote the algebraic closure ofK. We start with introducing the
theory of isogenies between elliptic curves.
Definition 1. An elliptic curveE defined overK is a smooth projective curve of genus 1 with a
distinguishedK-rational point O:
Elliptic curve over a number fieldK can be represented by a Weierstrass equation of the form
y
2
=x
3
+Ax +B; A;B2K (2.1)
=16(4A
3
+ 27B
2
)6= 0 (2.2)
where is the discriminant. Elliptic curve can be embedded in the projective plane by homoge-
nizing its equation
y
2
z =x
3
+Axz
2
+Bz
3
and the specified point is taken to be the point at infinity O = [0 : 1 : 0]: The relationships between
curves are described by morphisms. For projective curves, a morphism : E! E
0
is defined
by a polynomial mapping
(x :y :z) = (
0
(x;y;z) :
1
(x;y;z) :
2
(x;y;z))
8
where
i
’s are homogeneous polynomials of equal degree. Note that in affine coordinates, will
induce a rational map
(x;y) =
0
(x;y; 1)
2
(x;y; 1)
;
1
(x;y; 1)
2
(x;y; 1)
:
A homomorphism is a morphism of elliptic curves that respects the group structure of the curves.
Since every morphism is a (unique) composition of a homomorphismE!E
0
and a translation on
E
0
, any morphismE!E
0
mapping O to O must be a homomorphism.
2.1 Isogenies
Definition 2. LetE andE
0
be elliptic curves overK: An isogeny fromE toE
0
is a nonconstant
morphism :E!E
0
satisfying(O) = O:
Two elliptic curvesE andE
0
overK areK-isogenous if there is a nontrivial isogeny fromE
toE
0
defined overK: The kernel of a non-zero isogeny is always finite and the degree of an isogeny
is defined as the order of its kernel. It is simply saidN-isogeny for a degreeN-isogeny. Given an
elliptic curveE=K, we call a subgroupC ofE(K) of orderN an N-isogeny if there exists an
elliptic curveE
0
and an isogeny :E!E
0
withC =ker(): Due to V´ elu [40], given an elliptic
curve E=K and a finite subgroup C of E(K); there exists up-to-isomorphism a unique elliptic
curveE
0
and an isogeny :E!E
0
withker() =C: The field over whichE
0
and are defined
is the minimal extensionL=K for whichC is invariant under the action of Gal(K=L) (each field
automorphism in Gal(K=K) acts on pointsP2E(K) via its action on the coordinates ofP ). We
then say thatC isL-rational.
A morphism from an elliptic curve E over K to itself that fixes the distinguished point O is
called an endomorphism: Except the zero morphism, every endomorphism is an isogeny. Let
End(E) denote the endomorphism ring ofE. For every ellliptic curveE and for every integerm,
the multiplication-by-m map [m] is a homomorhism fromE to itself. So End(E) always contains
the copy ofZ, in the form of multiplication-by-m maps. We need the following notion before we
describe further the endomorphism ring of an elliptic curve.
9
Definition 3. An orderO in a quadratic fieldK is a subsetOK such that
•O is a subring ofK containing 1.
•O is a finitely generatedZmodule.
•O contains aQbasis ofK:
LetO be an order in an imaginary quadratic fieldK: LetI(O) be the set of proper fractional
O-ideals. ThenI(O) is a group under multiplication. The principal idealsP (O) ofI(O) forms a
subgroup and we obtain
C(O) =I(O)=P (O)
which is the ideal class group of the orderO: The order ofC(O) is callled theclassnumber of
O: Then,K has class number one if and only if its ring of integersO
K
is a principal ideal domain.
Theorem 2.1. LetK be a number field and letE=K be an elliptic curve. Then the endomorphism
ring ofE is isomorphic to eitherZ or an order in a quadratic imaginary field. In the later, we say
E has complex multiplication and writeE with CM.
The endomorphisms of E with trivial kernel form the subgroup Aut(E); the automorphism
group ofE. In order to classify the automorphism groups of elliptic curves, we need the following
notion.
Two elliptic curvesE andE
0
defined overK areisomorphic if there exists isogenies
1
:E!
E
0
and
2
: E
0
! E whose composition is the identity. For two elliptic curves over K to be
isomorphic overK, it is necessary and sufficient that their renownedj-invariants, defined in terms
of the coefficients of equation 2.1 by the formula
j =
1728(4A
3
)
to be equal. See Corollary 1 below.
10
Proposition 2.1. SupposeE :y
2
=x
3
+Ax +B,E
0
:y
2
=x
3
+A
0
x +B
0
are elliptic curves over
K: IfE andE
0
are isomorphic overK then there exists ad2K
such that
A =d
4
A
0
and B =d
6
B
0
and in this case =d
12
0
:
Corollary 1. Let K be a field withchar(K)6= 2; 3 andE=K,E
0
=K elliptic curves.ThenE andE
0
are isomorphic overK if and only ifj =j
0
:
In particular, this enables us to compute the automorphism group (overK) of an elliptic curve.
Corollary 2. Let K be a field with char(K)6= 2; 3; E=K an elliptic curve. and Aut(E) the
automoprhism group ofE:
• j6= 0; 1728 then Aut(E)'Z=2Z
• j = 1728 then Aut(E)'Z=4Z
• j = 0 then Aut(E)'Z=6Z:
Note that we can have elliptic curvesE andE
0
defined overK such that there is an isomorphism
E!E
0
defined overK but not overK: In this case we sayE andE
0
are twists: If the isomorphism
defined over a quadratic extension of K; then E and E
0
are called quadratic twist of each other.
More precisely,
Definition 4. Let E=K be an elliptic curve. Then a quadratic twist of E is an elliptic curve
E
0
=K which is isomorphic toE over a quadratic extension ofK.
If E;E
0
are isomorphic with j = j
0
6= 0; 1728, they are in fact quadratic twists. Then the
isomorphism is defined over some quadratic extensionK(
p
d). IfE is given by the form
E :y
2
=x
3
+ax +b
11
then its d-quadratic twist E
d
is given by the equation: y
2
= x
3
+d
2
ax +d
3
b which is also
isomorphic to the model
E
d
:dy
2
=x
3
+ax +b:
Furthermore, the curvesE andE
d
are isomorphic via
E
d
!E
(x;y)7! (x;y
p
d)
Hence ifd is a square inK thenE andE
d
are isomorphic over all extensions ofK and ifd is
not a square inK then we haveE(K(
p
d))'E
d
(K(
p
d)):
2.2 Torsion
In this section, we concentrate on torsion subgroup of an elliptic curve.
Definition 5. LetE be an elliptic curve andn2 Z withn 1: The n-torsion subgroup ofE;
denoted byE[n], is the set of points of order dividing n, so
E[n] =fP2E(K) : [n]P = Og
The torsion subgroup of E is the set of points of finite order, E
tor
=
S
1
n=1
E[n]: If E is an
elliptic curve over K, then E(K)[n] denotes the n-torsion points of E(K): Furthermore, for a
Galois extensionF ofK; every2 Gal(F=K) induces a permutation on then-torsion points. This
induces
Gal(K=K)! GL
2
(F
2
)
the modp Galois representation attached toE: Furthermore, ifK(E[n]) denotes the field generated
over K by the coordinates of the points in E[n]; then K(E[n]) is Galois over K: In fact, for an
elliptic curveE=K the structure ofn-torsion is well-understood overK.
12
Theorem 2.2 ([38], Cor.6.4). If n is a positive integer then we have
E(K)[n] =Z=nZZ=nZ:
Corollary 3. IfE is an elliptic curve overK; thenE(K)
tor
'Z=mZZ=nZ wherem dividesn:
Proof. AssumejE(K)
tor
j =n and we haven =p
k
1
1
p
k
2
2
:::p
kr
r
for some primesp
i
withk
i
2N: Since
E(K)[p
k
i
i
] is a subgroup ofE(C)[p
k
i
i
] andZ=pZZ=qZ'Z=pqZ whenp andq are coprime, the
claim immediately follows from Theorem 2.2.
There are several algorithms for computing E(K)
tor
: We will first study how to find a good
bound on the size of E(K)
tor
: The most efficient way is to look at the reductions of the curve.
LetO
K
be the ring of integers of K and m its maximal ideal. If E is an elliptic curve over K,
it has a Weierstrass equation with coefficients inO
K
: y
2
+a
1
xy +a
3
y =x
3
+a
2
x
2
+a
4
x +a
6
:
This equation is not unique, but we can take it with minimal discriminant. Then we can look at
the reduction of this elliptic curve at the maximal ideal m, where we take the same Weierstrass
equation with coefficients in the fieldO
K
=m. Now the question is whether this reduced formula
gives us again an elliptic curve. This may not be the case since the non-singularity condition might
not hold in this situation (i.e = 0 (mod m)): If we get an elliptic curve, then the next question
is the structure of the group defined over the new reduced curve. If the reduced curve is still an
elliptic curve, we say that E has good reduction at m. If this is not the case, we say that E has
bad reduction atm. The important fact is that E has bad reduction atm if and only ifm divides
the discriminant of E. If a curve E has bad reduction, we have two cases to consider. Either it
has a cusp, and we have a cuspidal (or additive) reduction, or it has a node, and we have a nodal
(or multiplicative) reduction. The next result shows us how efficient this point of view is in our
understanding ofE(K)
tor
:
Theorem 2.3 ([38]). LetE=K be an elliptic curve and letn 1 be an integer that is relatively
prime tochar(F
q
) whereF
q
is the residue field ofO
K
at m. IfE has good reduction at m then
E(K)[n] injects into
~
E(F
q
):
13
As an application of Theorem 2.3, we can consider reducing the elliptic curves overK =Q.
LetE=Q be an elliptic curve defined byy
2
=x
3
+Ax +B whereA;B2Q, and letp be a prime
that doesn’t divide the discriminant (E): The elliptic curveE then has good reduction atp.
If we reduce A and B modulo p, we obtain an elliptic curve
~
E = E (mod p) defined over the
finite fieldF
p
. Thus, from a single curveE=K we get an infinite family of elliptic curves, one for
each primep whereE has good reduction. Now repeated use of Theorem 2.3 provides an efficient
method for finding the torsion subgroup of E(Q): Now one may naturally ask how does
~
E(F
p
)
vary withp? LetE=F
q
be an elliptic curve over the finite fieldF
q
. In addition to the known result
that
~
E(F
p
) is either cyclic or product of two cyclic groups, we have the following estimate on the
size ofE(F
q
) which is known as Hasse bound.
Theorem 2.4 (Hasse). LetE=F
q
be an elliptic curve defined over the finite fieldF
q
: Then,
jE(F
q
)q 1j 2
p
q:
While these results enable us to find a list of possible torsion structures for E(K)
tor
; the
determination ofE(K)
tor
will be strongly relying on reducibility of division polynomials.
Division polynomials
Let E be an elliptic curve over a field K defined y
2
= x
3
+Ax +B: We define the division
polynomials [41]
n
2Z[x;y;A;B] by
0
= 0;
1
= 1;
2
= 2y;
3
= 3x
4
+ 6Ax
2
+ 12BxA
2
2n
= (2y)
1
n
[
n+2
]
2
n1
n
2
n+1
]; n 3
2n+1
=
n+2
3
n
n1
3
n+1
; n 2:
14
We also define the polynomials
n
=x
2
n
n+1
n1
!
n
= (4y)
1
(
n+2
2
n1
n2
2
n+1
):
which are closely related to torsion points by the following theorem.
Theorem 2.5. Consider the elliptic curvey
2
= x
3
+Ax +B defined over a number fieldK: Let
P = (x;y) be a point on the curve, and letn be a positive integer. Then,
nP =
n
(x)
2
n
(x)
;
!
n
(x;y)
3
n
(x;y)
:
By the theory of isogenies, thex-coordinate of eachn-torsion point is precisely determined by
the roots ofn-th division polynomial
n
(x):
2.3 Weil Pairing
For given elliptic curveE=K; there is a relation betweenK andE(K)
tor
via the Weil pairing. We
make use of the Weil pairing to deduce restrictions on the growth of torsion.
Proposition 2.2 ([38]). LetE=K be an elliptic curve, n a positive integer andU
n
the set ofnth
roots of unity inK. There exists a map
E[n]E[n]!U
n
such that
1. (bilinear)ha;b +ci =ha;bi:ha;ci andhb +c;ai =hb;ai:hc;ai for alla;b;c2E[n]:
2. (alternating)ha;ai = 0 for alla2E[n] and henceha;bi =hb;ai
1
for alla;b2E[n]:
3. (nondegenerate) Ifa2E[n] andha;bi = 1 for allb2E[n] thena = 0:
15
4. (Galois-invariant)(ha;bi) =h(a);(b)i for alla;b2E[n] and2Gal(
K=K):
Letfe
1
;e
2
g be basis ofE[n] then we can obtain a more concrete description of this pairing.
Using (1.) above, for anya
1
;a
2
;b
1
;b
2
2Z=nZ we have
ha
1
e
1
+a
2
e
2
;b
1
e
1
+b
2
e
2
i =he
1
;e
2
i
det(M)
where
M =
2
6
4
a
1
b
1
a
2
b
2
3
7
5
Furthermorehe
1
;e
2
i must be a primitiventh root of unity
n
, as otherwise using (1.) above we
can contradict (3). With respect to the basis above, we have a Galois representation where
: Gal(
K=K)! GL
2
(Z=nZ)
7!M
and hence(
n
) =h(e
1
);(e
2
)i =
det(M
)
n
:
2.4 Jacobian
This section presents the basic properties of divisors and introduces the jacobian of a curve. Given
a smooth projective curveC defined overK, we can attach an abelian group to the curve as follows.
Definition 6. A divisor D is a formal sum of points inC
D =
X
P2C
m
P
P; m
P
2Z
where only a finite number of the m
P
are non-zero. The deg(D) of D is the sum
P
P2C
m
P
:
The order of D at P is the integer m
P
, we write ord
P
(D) = m
P
: The support of D is the set
supp(D) =fP2C(K) :ord
P
(D)6= 0g of points occurring inD with a nonzero coefficient.
16
Definition 7. A divisorD =
P
P2C
m
P
P is eective ifm
p
0 for allP:
The set of all divisors, denotedDiv(C), forms an additive group under the addition rule
X
P2C
m
P
P +
X
P2C
n
P
P =
X
P2C
(m
P
+n
P
)P
The divisors of degree 0 forms a subgroup ofDiv(C) which we denote byDiv
0
(C). In particular,
iff2K(C)
then the associated divisordiv(f) =
P
P2C
ord
P
(f)P2Div
0
(C) ([38], Prop.3.1).
Sinceord
P
is a valuation, the map
div :K(C)
!Div
0
(C)
is a homomorphism of abelian groups. This yields the following definitions.
Definition 8. A divisor D 2 Div
0
(C) is principal if it has the form D = div(f) for some
f2K(C)
: The set of all principal divisors ofC forms a subgroup ofDiv
0
(C), denoted byP (C):
Then one can obtain the quotient:
J(C) :=Div
0
(C)=P (C)
which is called the Jacobian of C. (in [38],Pic
0
(C)). Two degree-0 divisorsD;D
0
are said to
be linearly equivalent ifDD
0
is principal; we writeDD
0
: We usually write [D] for the linear
equivalence class of a divisorD2 Div
0
(C): Although J(C) could in principal be presented as
a projective abelian variety, for many purposes it seems easier not to work with explicit defining
equations forJ(C). For example, sinceJ(C) is a projective variety, it can be embedded in some
projective spaceP
N
overK: Already forg = 2,N = 4
g
1 is large to deal with. One advantage of
the identification ofJ(C) as the group of linear equivalence classes of degree-0 divisors onC(K)
is that we can represent points onJ(C) by divisors onC; which makes computations rather easy.
IfC is a curve of genus zero, thenJ(C) is trivial (just a point).
17
IfC is a curve of genus one, thenJ(C) is a one dimensional abelian variety, hence an elliptic
curve.
If C has a K-rational point (so is an elliptic curve itself), then J(C) and C are isomorphic.
Otherwise,C andJ(C) cannot be isomorphic, sinceJ(C) always has aK-rational point, namely
the identity of the group law. In fact,J(C) will be of our interest for the curves of genusg 2: If
C(K) is non-empty, then we choose aK-rational pointQ onC, there is an embedding overK;
i :C!J(C)
P7! [PQ]:
The absolute Galois group Gal(K=K) acts on C via the natural action of Gal(K=K) on the
coordinates. This action can be extended linearly toDiv
0
(C) which induces a well-defined action
onJ(C) sincediv(f)
= (div(f))
.
Definition 9. A divisorD =
P
m
P
P2Div(C) is defined overK if for all2Gal(K=K); we
haveD
=D; where
D
=
X
m
P
P
=
X
m
P
P
:
The subset ofDiv(C) defined overK forms the subgroup ofK-rationaldivisors:
Note thatD is defined overK does not imply that the points in the support ofD are defined
overK. But ifD = D
for all2 Gal(K=K) then it must be the casem
P
= m
P
for all2
Gal(K=K): LetJ(C)(K) denote the set of all points inJ(C) defined over K. By the Mordell-Weil
theorem,J(C)(K) has a structure of finitely generated abelian group. The problem of finding the
setC(K) ofK-rational points onC can now be stated equivalently in the form of finding the set
J(C)(K)\i(C): The advantage of this point of view is that we can make use of the additional group
structure onJ(C)(K) to obtain information onC(K): A very trivial case whereJ(C)(K) =f0g
impliesC(K) is a singleton.
18
2.5 Modular curves
LetH =fz2C : Im(z) > 0g be the upper half plane. There is an action of the modular group
= SL
2
(Z) onH as
2
6
4
a b
c d
3
7
5
:z =
az +b
cz +d
LetG be a finite index subgroup of and consider the action ofG onH: Then, the quotient space
Y () =H=G is a non-compact Riemann surface. By adjoining the (finitely many) cusps toY (G);
one can obtain a compact Riemann surface denoted byX(G). In fact, this curve has a model over
Q:X(G) can also be obtained as a quotient space. This is done by extendingH toH
=H[P
1
(Q).
Any subgroupG of is still acting onH
. We can obtainX(G) asH
=G which turns out to be
a compact Riemann surface. The number of the orbits of points inP
1
(Q) is finite, hence what we
obtain is the compactification ofY (G), which isX(G):
LetN 1 be an integer. The most important subgroup for our purpose is the following:
0
(N) =
8
>
<
>
:
2
6
4
a b
c d
3
7
5
2 j
2
6
4
a b
c d
3
7
5
2
6
4
0
3
7
5
modN
9
>
=
>
;
Then, the modular curveX
0
(N) =X(
0
(N)) has a model overQ: The main reason for our interest
in these curves is the (non-cuspidal) points of these curves can be used to classify elliptic curves
together with certain torsion data. Their moduli interpretation is summarized as follows. The set of
the non-cuspidalK-rational points onX
0
(N) parametrizesK-isomorphism classes of pairs (E;C)
where E=K is an elliptic curve and C is a cyclic subgroup of E(K) of order N such that C is
Gal(K=K)-invariant. Two pairs (E;C), (E
0
;C
0
) are equivalent iff there exists aK-isomorphism
:E!E
0
such that(C) =C
0
:
19
Chapter 3
Torsion over Quadratic Fields
The results in this chapter provide the context and motivation for the main results of this thesis.
From now on, we let K denote a quadratic number field. Let E be an elliptic curve given by
y
2
=f(x) withf is a monic cubic polynomial overK: There are a couple of observations we can
make:
Remark 1. (i) Iff is irreducible overK, then it remains irreducible overL. Otherwise,f has a
root inL and the degree ofK() overK is divisible by 3 which is not possible sinceK() is
contained inL. HenceE(K)[2]'Z
1
thenE(L)[2]'Z
1
:
(ii) If E(K) has a point of order 2, then E
d
(K) has a point of order 2; too. This simply
follows by if f(x) = x
3
+ ax
2
+ bx + c for some a;b;c 2 K and for d 2 K, d 6= 0 let
f
d
(x) = x
3
+dax
2
+d
2
bx +d
3
c. Ifr2 K andf(r) = 0, thenf
d
(dr) = 0 and this induces a
bijection between theK-roots off andf
d
. HenceE(K)[2]'E
d
(K)[2]:
In the next proposition, we generalize Kwon’s result [25] to quadratic number fields which
allows us to bound the growth ofE(K)
tor
in a quadratic extension ofK:
Proposition 3.1. LetK be a quadratic number field,L =K(
p
d) a quadratic extension ofK and
let denote the generator of Gal(L=K). The map
E(L)
tor
h
!E
d
(K)
tor
P7!P(P )
20
is a homomorphism with kernelE(K)
tor
and hence induces an injection
E(L)
tor
=E(K)
tor
,!E
d
(K)
tor
Proof. In order to see the image of h lies in E
d
(K)
tor
, we look at the Gal(L=K) action over
h(E(L)
tor
).
(P(P )) =(P )((P )) =(P )P =(P(P ))
So, ifh(P ) = (x;y) then(h(P )) = (x;y) which forces the coordinates to be (x;z
p
d) where
x and z are in K. Hence, we have h(E(L)
tor
) E
d
(K)
tor
: Further, it fixes the identity so is a
homomorphism. To find the kernel ofh, we observe that
P(P ) =Q(P ),(PQ) =PQ,PQ2K
This shows thatker(h) =E(K)
tor
: Therefore it induces an injection as desired.
It follows that the growth of torsion over a quadratic extension can be controlled by the torsion
over the base field. Due to the classification of torsion of elliptic curves over quadratic fields, we
can determine an upper bound for the growth. This motivates us to find the sharpest upper bound on
the growth of torsion for the smaller family of elliptic curves defined over an imaginary quadratic
field of class number one. Furthermore, there are at most finitely many extensions in which the
torsion subgroup grows, i.e. the torsion subgroups are stable under almost all quadratic extensions
of the base field.
Corollary 4. LetE be an elliptic curve overK: Then, for all but finitely many square-free integers
d2K; we have
E(K(
p
d))
tor
=E(K)
tor
Proof. This proof is a variant of [25, Corollary 2]. By Proposition 3.1,E(K(
p
d))
tor
=E(K)
tor
is
isomorphic to a subgroup ofE
d
(K)
tor
. Since any point inE
d
(K)
tor
has a possible order 1,2,...16 or
21
18 by Theorem 1.2,E(K)
tor
(E(K(
p
d)) if and only if there existsP2E(K)
tor
;Q2E(K)
tor
,
andm2f2; 3;:::16; 18g such that
mQ =P; K(Q) =K(x;y) =K(
p
d) whereQ = (x;y):
Note that there are only finitely many P 2 E(K)
tor
and m2f2; 3;:::; 16; 18g. Thus, there are
only finitely manyQ2 E(K)
tor
such thatmQ = P: Among them only finitely manyQ such that
K(Q) =K(
p
d) for somed2K: Excluding thosed’s which appear asK(Q) =K(
p
d)), we get
E(K(
p
d))
tor
=E(K)
tor
for all otherd’s.
The following result will be especially of great importance in determining odd torsion part upon
quadratic base change.
Lemma 1 ([12],Cor 4). Given an elliptic curveE overK and an odd positive integern, we have
E(K(
p
d))[n]'E(K)[n]E
d
(K)[n]:
LetG(K) denote the set of groups that appear as torsion groups of elliptic curves defined over
K. Theorem 1.3 describesG(K) asK varies over all quadratic fields. On the other hand, Corollary
4 shows that ifK is a quadratic field andL a quadratic extension ofK thenG(K) is contained in
G(L). Thus a large number of elements ofG(L) are already known. If we fixK as a noncyclotomic
imaginary quadratic field of class number one, we will more likely to get a smaller list than (1.2),
which is followed by the next result.
Theorem 3.1 (Saika and Sarma,[36]). LetE be an elliptic curve over the specified number field.
1. E(Q(
p
2))
tor
is isomorphic to either one of the Mazur’s groups,Z
11
orZ
2
Z
10
:
2. E(Q(
p
7))
tor
is isomorphic to either one of the Mazur’s groups,Z
11
,Z
14
orZ
15
:
3. E(Q(
p
11))
tor
is isomorphic to either one of the Mazur’s groups,Z
14
,Z
15
orZ
2
Z
10
:
4. E(Q(
p
19))
tor
is isomorphic to either one of the Mazur’s groups,Z
11
,Z
2
Z
10
, orZ
2
Z
12
:
22
5. E(Q(
p
43))
tor
is isomorphic to either one of the Mazur’s groups,Z
11
,Z
14
orZ
2
Z
12
:
6. E(Q(
p
67))
tor
is isomorphic to either one of the Mazur’s groups,Z
14
,Z
15
orZ
2
Z
12
:
7. E(Q(
p
163))
tor
is isomorphic to either one of the Mazur’s groups,Z
14
, orZ
2
Z
12
:
23
Chapter 4
K-rational points on X
0
(N)
Given anyK inS; we study theK-rational cyclicN-isogenies of elliptic curves defined overK.
The values ofN will vary according to the prime factors of the possible torsion subgroups realized
overK. For an elliptic curveE over any number fieldK, recall that a subgroupC ofE(K) is a
K-rational N-isogeny if there exists an elliptic curveE
0
=K and an isogeny :E!E
0
defined
over K such that C = ker(): Equivalently, a subgroup C of E(K) of order N is a K-rational
N-isogeny if it is invariant under the action of Gal(K=K): Note thatC might beK-rational even
though it contains points that are not. For a givenN-isogenyC, Magma describes a polynomial
f
C
whose roots are precisely thex-coordinates of the points inC. IfC is pointwise defined in a
quadratic extension L of K then f
C
must split completely over L i.e. f
C
must have irreducible
factors of degree at most 2 over K: We say then C is pointwise defined over L. One can
immediately observe the following.
Proposition 4.1. Let K be a number field and E=K an elliptic curve withjE(K)
tor
j = m: Let
d2 K be a non-square. IfH is a subgroup ofE
d
(K) of odd ordern; thenE has aK-rational
mn-isogeny pointwise defined overK(
p
d)):
Proof. With respect to short Weierstrass form fixed forE; there is an isomorphism :E
d
!E via
((x;y)) = (x;
p
dy): SinceH is a subgroup pointwise defined overK; then the image(H) ofH
forms a Gal(K=K)-invariant subgroup ofE(K(
p
d)) of odd ordern: On the other hand, by Lemma
1,(H)\E(K)
tor
is trivial, henceG :=(H)E(K)
tor
is isomorphic to a Gal(K=K)-invariant
24
subgroup of E(K(
p
d)): Therefore, G is a K-rational mn-isogeny of E pointwise defined over
K(
p
d) as desired.
With the analysis of certain isogenies throughout this chapter, our goal is to prove the following.
Theorem 4.1. Given any K inS and an elliptic curve E over K, E has no K-rational cyclic
N-isogeny pointwise defined over a quadratic extension ofK forN = 21; 28; 30; 33; 35; 40;
45; 48; 55; 77. In particular, there exists noK-rational cyclic 27-isogeny pointwise defined over a
quadratic extension of anyK inS except forQ(
p
2) andQ(
p
11):
Finding quadratic points. While classifying the quadratic points on a modular curveX
0
(N);
the difficulty arises whenX
0
(N) has, for several of our applications, a genus that is too high. To
circumvent this difficulty we study instead rational (rather than quadratic) points on the symmetric
square ofX
0
(N).
Definition 10. The symmetric squareX
(2)
of an algebraic curveX is the quotient space of the
cartesian productXX :
X
(2)
:= (XX)=S
2
by the action of the symmetric groupS
2
:
Let X
0
(N)
(2)
denote the symmetric square of X
0
(N): Note that since X
0
(N) is a compact
Riemann surface,X
0
(N)
(2)
is therefore a complex manifold. ForX
0
(N) withg 1; the symmeric
squareX
0
(N)
(2)
is closely related to the JacobianJ
0
(N) ofX
0
(N): To be more precise, letK be
any quadratic number field and let denote the non-trivial element in Gal(K=Q): Suppose thatE
is an elliptic curve overK withC aK-rational cyclicN-isogeny. The pair (E;C) corresponds to a
K-rational pointP onX
0
(N): LetP
denote the image ofP under: The pairfP;P
g gives rise
to a well-defined rational pointD :=P +P
ofX
0
(N)
(2)
: Note that the setX
0
(N)
(2)
consists of
pairsfP;P
g of a genuinely quadratic pointP onX
0
(N) and its Galois conjugate, as well as pairs
fP;Qg of rational pointsP;Q onX
0
(N): Therefore,X
0
(N)(K) can be put insideX
0
(N)
(2)
(Q):
25
As a result, classifying all the quadratic points on the curveX
0
(N) amounts to studying the set of
rational points ofX
0
(N)
(2)
: LetJ
0
(N) denote the Jacobian ofX
0
(N): There is a map
h :X
0
(N)
(2)
!J
0
(N)
byh(D) = [D 2P
0
] whereP
0
is a cusp inX
0
(N)(Q): This map is injective as long asX
0
(N)
has genus greater than 2; and is not hyperelliptic. This is easily seen as follows: Ifh(D) = 0 then
D 2P
0
is the divisor of a degree two function onX
0
(N). Such a function exists only ifX
0
(N)
has genus 2; or is hyperelliptic.
IfX
0
(N) is a hyperelliptic curve of genus 2 such that its Jacobian has rank 0 overQ; then
Bruin and Najman [5] showed that for the hyperelliptic involution and a cuspC2X
0
(N)(Q) the
map
:X
0
(N)
(2)
!J
0
(N)
D7! [DC(C)]
is an isomorphism onto its image, except above 0, and
1
(f0g) is isomorphic toP
1
: Hence, we
have
X
0
(N)
(2)
(Q) =
1
(f0g)[
1
(J
0
(N)(Q)f0g):
We call the points on X
0
(N) coming from J
0
(N)(Q)f0g under as exceptional: Oth-
erwise, they are called nonexceptional: SinceJ
0
(N)(Q) is finite by assumption, the preimage
1
(J
0
(N)(Q)f0g) is finite. The points ofJ
0
(N)(Q) is easy to compute. We will adapt this
strategy to do explicit computations for some values of N of our interest.
IfX
0
(N) is a nonhyperelliptic curve of genus between 3g 5 whoseJ
0
(N)(Q) is finite, then
h is particularly injective. By pulling back finitely many points inJ
0
(Q) underh, it is theoretically
possible to determine X
0
(N)
(2)
(Q); hence the quadratic points of X
0
(N) follows. Siksek and
Ozman [34] described all the quadratic points on these curvesX
0
(N): Their approach is extended
26
by Box [3] to the non-hyperelliptic curves X
0
(N) of 3 g 5 whose J
0
(N)(Q) is infinite,
which requires the (relative) symmetric Chabauty developed by Siksek [37] in combinatation with
a Mordell-Weil sieve which will be covered in more detail as part of the analysis ofX
0
(77):
4.1 N=21
X
0
(21) is an elliptic curve with the modely
2
+xy =x
3
4x 1: By Magma,X
0
(21)(Q) has rank
0, torsionZ
2
Z
4
and 4 cusps. The torsion does not change when the base fieldQ is extended
to any K inS: As shown in Table 4.1, each of four noncuspidal points corresponds to a cyclic
21-isogeny defined over an extension ofK of degree at least 3. So none of these rational points
produce a cyclic 21-isogeny over a quadratic extension ofK:
Point j(E) E f
C
(-1/4,1/8) 3375/2 [20/441, -16/27783] (1,3,6)
(2,-1) -189613868625/128 [-1915/36, -48383/324] (1,3,6)
(-1,2) -1159088625/2097152 [-505/192, -23053/6912] (1,3,6)
(5,-13) -140625/8 [-1600/147, -134144/9261] (1,3,3,3)
Table 4.1: Representatives (E;C) of isomorphism classes corresponding to noncuspidal rational
points ofX
0
(21)
The rank ofX
0
(21) only goes up inK =Q(
p
43) andQ(
p
67): Hence a search of points
onX
0
(21)(K) is computationally infeasible to determine whetherE(K) has a cyclic 21-isogeny.
Instead, we will use the division polynomial method to prove the following.
Proposition 4.2. For fixedK2fQ(
p
43);Q(
p
67)g; letE be an elliptic curve overK andE
d
run through the quadratic twists ofE. IfE(K)'Z
7
thenE
d
(K)6'Z
3
for all nonsquared2K.
Proof. Assume E(K)' Z
7
and E
d
(K)' Z
3
for some non-square d2 K. Then E(K) has a
point of order 7 and it has an additionalK-rational subgroupC of order 3 (arising from the twist).
By Table 3 in [24], we take the families of elliptic curves overK with 7-torsion as
E
t
:y
2
+ (1c)xyby =x
3
bx
2
(4.1)
27
whereb =t
3
t
2
andc =t
2
t for somet6= 0; 1 inK: We may assumeE isE
t
0
for somet
0
.
Let P be a point on E generating the K-rational subgroup C. Then, P has order 3 which
is defined over K(
p
d). Write x(P ); for the x-coordinate of P: We claim that x(P ) is in K:
SincehPi is aK-rational subgroup consisting onlyfP;Pg withx(P ) =x(P ); it follows that
x(P )
= x(P
) = x(P ) hence x(P ) is invariant under the action of Gal(K=K). Now the pair
(E;P ) corresponds to a point (x(P );t
0
) on the curveC given by the equation(x;t) = 0 where
(x;t) is the third division polynomial ofE
t
:
(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
)
It comes down to find the set C(K) of K-rational points on C. By Magma, C has singular
points atf(0; 0); (0; 1)g andC is birational (overQ) to the hyperelliptic curve
~
C :v
2
=f(u) =u
8
6u
6
4u
5
+ 11u
4
+ 24u
3
+ 22u
2
+ 8u + 1
outside the singular points. By the choice oft6= 0; 1 in (4.1), it is enough to find
~
C(K): Now
~
C is
a hyperelliptic curve of genus 3 and its defining polynomial factors as:
f(u) = (u
2
+u + 1)(u
6
u
5
6u
4
+ 3u
3
+ 14u
2
+ 7u + 1)
Let
~
C
(2)
be the symmetric square of
~
C. The set
~
C
(2)
(Q) consists of the equivalence classes of
pairsfP;Pg of a quadratic point and its Galois conjugate, as well as pairsfP;Qg of rational points
in
~
C: To study all quadratic points on a curve of genus greater than 2, one often studies the rational
points on the symmetric square of the curve. Write
~
J for the Jacobian of
~
C: The map
:
~
C
(2)
!
~
J
fP;Qg7! [P +Q1
+
1
]
28
is an injection outside of 0. The copy ofP
1
inside
~
C gets contracted to 0, under: So
1
(0) is
isomorphic toP
1
: We have
~
C
(2)
(Q) =
1
(0)(Q)[
1
(
~
J(Q)nf0g):
We first compute
1
(
~
J(Q)nf0g): By 2-descent computation in Magma,
~
J(Q) has Mordel-Weil
rank 0. As
~
J has a good reduction at 5, the map
~
J(Q)!
~
J(F
5
) is injective. Since the group
generated by the 2-torsion point corresponding to the factorization off(u) on the right-hand side
of the curve equation and the difference of the two points at infinity on
~
C, which are1
+
= [1; 1; 0]
and1
= [1;1; 0], surjects onto
~
J(F
5
)'Z
2
Z
52
, it is therefore equal to
~
J(Q):
Now consider a pointP
0
2
~
C(K) and writeP
0
for its image under the nontrivial automorphism
of K: Then P
0
+P
0
is an effective divisor of degree 2 on
~
C; defined overQ: For the effective
rational divisor1
+
+1
of degree 2, the linear equivalence class ofP
0
+P
0
1
+
1
is a
rational point on
~
J: It follows thatP
0
+P
0
is linearly equivalent toD +1
+
+1
for some divisor
D of degree 0 with [D]2
~
J(Q): We may now assume thatx(P
0
) = 2Q. It remains to enumerate all
104 elements of
~
J(Q) for [D]6= 0 in Magma and check if these divisors are of degree 2 with the
points in the support defined overK:
1
p
3
2
; 0
!
;
p
2;4
p
2 + 5
;
p
2;4
p
2 5
;
2
p
2
2
;
5 4
p
2
4
!
;
2
p
2
2
;
5 4
p
2
4
!
;
1
p
2; 11 8
p
2
;
1
p
2;11 8
p
2
:
None of these points is defined over the base fields of interest. This implies that there are no
exceptionalK-rational points on
~
C: i.e. noK-points withx-coordinate not inQ:
This leaves the case whenx(P
0
)2Q. There are rational points which are two points at infinity
and four points withx(P
0
) = 0 or 1. For all other such points, P
0
+P
0
is linearly equivalent to
1
+
+1
, and so [D] = 0. Thus P
0
= (u;
p
f(u)) with u2 Q and so the nonexceptional
29
K-rational points would have rationalu withf(u) =43v
0
orf(u) =67v
0
for some rational
v
0
: However, there are no such points, sincef(u) is always positive. Therefore, we have shown that
~
C(K) =
~
C(Q) =f1
+
;1
; (0; 1); (0;1); (1; 1); (1;1)g:
Now the pair (E;P ) associated with (x(P );t
0
) onC(K) corresponds to a point on
~
C(K), so to
a rational point (u;v) on the curve
~
C and therefore to a rational point onC: However, ifE is defined
overQ andx(P )2Q; thenP2E(Q(
p
d)) for somed2Q: ThenE(Q(
p
d)) has a subgroup of
order 21, but this is impossible by [26].
Remark 2. Proposition 4.2 in fact shows that there is no cyclic 21-isogeny pointwise defined in a
quadratic extension ofK for anyK2S: Moreover, forK =Q(i), one can find a similar result in
[10] using a different technique.
4.2 N=27
The modular curveX
0
(27) is an elliptic curve with the modely
2
+y =x
3
7: OverQ,X
0
(27) has
rank 0 and torsionZ
3
with 2 cusps. By Magma, the torsion does not change over anyK inS; and the
rank only goes up overQ(
p
2) andQ(
p
11): In case of positive rank, it is not computationally
efficient to go through infinitely manyK-rational points ofX
0
(27): To circumvent this difficulty,
we applied the divison polynomial method as used in the previous caseN = 21 and obtained the
following elliptic curves over K with a subgroup of order 27 pointwise defined over a quadratic
extension ofK:
Remark 3. (1) LetK =Q(
p
2) andE=K be an elliptic curve given by
y
2
+
1
27
(950w 619)xy +
1
243
(16720w 210862)y =x
3
+
1
243
(16720w 210862)x
2
30
wherew =
p
2: Magma computes thatE(K)
tor
'Z
9
andE(K(
p
3))
tor
'Z
3
Z
9
:
(2) LetK =Q(
p
11) andE=K be an elliptic curve given by
y
2
+
1
27
(2072w 4265)xy +
1
243
(949568w + 377548)y =x
3
+
1
243
(949568w + 377548)x
2
wherew =
p
11. By Magma,E(K)
tor
'Z
9
andE(K(
p
3))
tor
'Z
3
Z
9
:
4.3 N=28
X
0
(28) is a hyperelliptic curve of genus 2 with the model
y
2
+ (2x
3
+ 3x
2
3x)y =x
4
3x
3
+ 4x
2
3x + 1: (4.2)
Note thatX
0
(28) has 6 cusps which are all defined overQ. As known,X
0
(28)(K) can be embedded
in its symmetric productX
0
(28)
(2)
(Q); we instead study rational points ofX
0
(28)
(2)
: SinceX
0
(28)
has the hyperelliptic involution i = w
7
, we have : X
0
(28)
(2)
(Q)! J
0
(28)(Q) mapping the
pairfP;Qg to the equivalence class [P +Q1
+
i(1
+
)] where1
+
= [1; 2; 0]. Now, is
an injection except above 0; so it boils down to pull back the rational points of J
0
(28): One can
compute in Magma
#J
0
(28)(Q) = 36
C
0
(28)'Z
6
Z
6
whereC
0
(28) is the cuspidal subgroup ofJ
0
(28)(Q) generated by the classes of the divisorsP
1
P
2
whereP
1
;P
2
are rational cusps ofX
0
(28): It then follows thatJ
0
(28)(Q) = C
0
(28): To classify
the exceptional points ofX
0
(28) which are the ones coming fromJ
0
(28)(Q)nf0g; we enumerate
31
all the non-zero 35-elements in Magma. The divisors of degree 2 with the support defined in an
imaginary quadratic field with class number one are the rational ones except
p
3 + 1
2
; 1
;
p
3 + 1
2
;1
;
p
7 1; 0
;
p
7 + 3
2
;1
;
p
7 + 1
4
; 0
:
These points are all the exceptional quadratic points onX
0
(28): Each point is associated with
an elliptic curve as in Table 4.2. As a result, none of the exceptional ones give rise to a cyclic
28-isogeny pointwise defined in a quadratic extension ofK:
Point Elliptic Curve f
C
p
3+1
2
; 1
1=98(135
p
3585)+1=343(660
p
31782)
(1,1,3,3,6)
p
3+1
2
; 0
1=98(135
p
3585)+1=343(660
p
31782)
(1,1,3,3,6)
p
7+1
4
;
p
7+5
8
[54=7;54=7] (1,1,3,3,6)
p
7+3
4
;
p
7+3
8
[85=63;38=63] (1,1,3,3,6)
p
7+1
2
;
p
7+1
2
[45=7;54=7] (1,1,3,3,6)
Table 4.2: Representatives (E;C) of isomorphism classes corresponding to exceptionalK-rational
points ofX
0
(28)
It remains to classify the nonexceptionalK-rational points which are not coming fromJ
0
(28)(Q)nf0g:
LetP be aK-rational nonexceptional point ofX
0
(28): ThenP is in the form of (x;
p
f
28
(x))
wheref
28
(x) is the discriminant of 4.2
f
28
(x) =x
6
3x
5
+ 25x
4
30x
3
+ 25x
2
12x + 4:
One can easily observe that f
28
(x) happens to be positive for all real numbers and so P must
be defined in a real quadratic field. Therefore, there are no nonexceptionalK-rational points on
X
0
(28):
32
4.4 N=30
The modular curveX
0
(30) is a hyperelliptic curve of genus 3 with the model
y
2
+ (x
4
x
3
x
2
)y = 3x
7
+ 19x
6
+ 60x
5
+ 110x
4
+ 121x
3
+ 79x
2
+ 28x + 4 (4.3)
which can be simplified as
y
2
=f
30
(x) =x
8
+ 14x
7
+ 79x
6
+ 242x
5
+ 441x
4
+ 484x
3
+ 316x
2
+ 112x + 16
= (x
2
+ 3x + 1)(x
2
+ 6x + 4)(x
4
+ 5x
3
+ 11x
2
+ 10x + 4)
X
0
(30) has 8 cusps which are all rationals. WriteJ
0
(30) for the Jacobian ofX
0
(30): By [5,
Proposition 3], J
0
(30)(Q)' Z
2
Z
4
Z
24
: Note that there are only two points at infinity on
X
0
(30) which are1
+
= [1; 1; 0] and1
= [1; 0; 0]: SinceX
0
(30)
(2)
(Q) is embedded inJ
0
(30)(Q)
outside of 0, we first study the finitely many points inJ
0
(30)(Q)f0g: LetP2X
0
(30)(K) with
x(P ) = 2Q and letP denote its Galois conjugate. ThenP +P1
+
1
is a rational divisor of
degree 0 and hence defines an element inJ
0
(30)(Q): It follows thatP +P is linearly equivalent to
D +1
+
+1
whereD is a divisor of degree 0 representing one of the 192 elements inJ
0
(30)(Q):
AsX
0
(30) is hyperelliptic, for [D]6= 0; the divisorD +1
+
+1
is linearly equivalent to at most
one effective divisor of degree 2. Using the Mumford representations in Magma for the elements
ofJ
0
(30)(Q)nf0g; we obtain the only such divisors are sums of two rational points, except
3
p
5
2
; 0
!
;
3
p
5; 0
;
3
p
7
2
;
9 3
p
7
2
;
3
p
7
2
;
9 3
p
7
2
;
3
p
7
4
;
27 15
p
7
32
;
3
p
7
4
;
27 15
p
7
32
:
The ones defined overQ(
p
7) correspond to the following points on (4.3) withw =
p
7
1
2
w 3
;w 3
;
1
2
w 3
;
1
2
w 15
;
1
4
w 3
;
1
16
5w 9
;
33
1
4
w 3
;
1
32
5w + 9
One can easily check in Magma that the minimum extension ofQ(
p
7) in which these points
produce a cyclic 30-isogeny has degree 4.
It leaves the case x(P ) 2 Q: Then (x(P );
p
f
30
(x)); lies on the d-twist X
d
0
(30) where
K =Q(
p
d): By the choice ofK2S,d =p wherep belongs to the setf2; 7; 11; 19; 43; 67; 163g:
By Corollary 3.5 in [33], this twist does not haveQ
p
-rational points, therefore this leaves alone the
rational cusps.
4.5 N=33
The modular curveX
0
(33) is a hyperelliptic curve of genus 3 with the model
y
2
+ (x
4
x
2
1)y = 2x
6
2x
5
+ 11x
4
10x
3
+ 20x
2
11x + 8
which can be considered as a quadratic equation iny as: y
2
g(x)yh(x) = 0 whereg(x) =
x
4
+x
2
+ 1 and h(x) = 2x
6
2x
5
+ 11x
4
10x
3
+ 20x
2
11x + 8. So its discriminant is
defined byf
33
(x) =x
8
+ 10x
6
8x
5
+ 47x
4
40x
3
+ 82x
2
44x + 33: Galbraith [11, Table 4]
derivesy
2
= f
33
(x) as a model forX
0
(33): Note that there are only 4 rational cusps. LetJ
0
(33)
denote the Jacobian ofX
0
(33): andi =w
11
for the hyperelliptic involution ofX
0
(33): We compute
thatJ
0
(33)(Q) has rank 0 by 2-descent computation in Magma with good reduction atp = 5; 7:
Moreover, asX
0
(33)(Q) consists entirely of cusps, we compute the group structure of the cuspidal
subgroup C
0
(33) to beZ
10
Z
10
which is generated by the difference of two points at infinity,
1
+
= [1; 1; 0], i(1
+
) = [1; 0; 0] and the difference of a rational point and one of the points at
infinity. Also, we compute
J
0
(33)(F
5
)'Z
10
Z
20
J
0
(33)(F
7
)'Z
2
Z
2
Z
10
Z
10
:
34
As J
0
(33)(Q) injects into both of these groups J
0
(33)(F
5
) and J
0
(33)(F
7
); we show that
J
0
(33)(Q) = C
0
(33): By enumerating all the 99 elements of J
0
(33)(Q)nf0g in Magma, we
compute all the exceptional quadratic points as in Table 4.3. Each point corresponds to the
isomorphism class of a pair (E=K;C) whereC is a cyclic 33-isogeny defined in an extension of
K of at least degree 10.
This leaves to determine the nonexceptionalK-rational points onX
0
(33): GivenK =Q(
p
d)
inS withd2f2;7;11;19;43;67;163g; the nonexceptionalK-rational points would
have rational x with f
33
(x) =dy
2
0
for some rational y
0
: Since f
33
(x) is decomposed into the
conjugate factors overQ(
p
11) as (x
2
x + 3)(q(x) +r(x)
p
11)(q(x)r(x)
p
11) for some
polynomialsq(x);r(x) with rational coefficients. One can observe thatx
2
x + 3 is positive for
allx2R. Moreoverf
33
(x) = 0 if and only ifq(x)
2
+ 11r(x)
2
= 0 i.e. bothq(x) andr(x) are zero
which is impossible. Thusf
33
happens to be positive for allx2R so it cannot be equal tody
2
0
for any rationaly
0
: Therefore, there are no nonexceptionalK-rational points onX
0
(33):
Point Elliptic Curve f
C
p
2;
p
2 + 2
360
p
2 + 420;1008
p
2 5600
(1,5,10)
p
2;
p
2 + 1
360
p
2 + 420;1008
p
2 5600
(1,5,10)
p
2 1;7
p
2 2
1
8
(1404
p
2 2157);
1
16
(44171
p
2 + 6156)
(1,5,10)
p
2 1;5
p
2 5
1
8
(4356
p
2 + 363);
1
16
(64009
p
2 + 100188)
(1,5,10)
p
2
2
;
4
p
25
4
[
8349
2
;
1
2
(847
p
2 + 209088)] (1,5,10)
p
2
2
;
p
2 + 2
1
2
(5040
p
2 + 28011);
1
2
(880397
p
2 488304)
(1,5,10)
p
7+1
2
;1
[
1
128
(296769
p
7 + 323829);
1
512
(7944695
p
7 +
104422307)
(1,5,10)
p
7+1
2
;
p
7 + 1
1
128
(27591
p
7 140211);
1
512
(2704139
p
7 +
4638407)
(1,5,10)
3
p
7+1
4
;
9
p
7+33
4
1
8
(2553
p
7 + 4365);
1
8
(2645
p
7 + 96359)
(1,5,10)
3
p
7+1
4
;
9
p
7+93
32
1
8
(33
p
7 2475);
1
8
(11033
p
7 + 6611)
(1,5,5,5)
1
p
11
2
;
p
11 + 1
[1056;13552] (1,5,10)
Table 4.3: Representatives (E;C) of isomorphism classes corresponding to exceptionalK-rational
points ofX
0
(33)
35
4.6 N=35
The modular curveX
0
(35) is a hyperelliptic of genus 3. Galbraith [11, Table 4] gives the following
model
X
0
(35) :y
2
=x
8
4x
7
6x
6
4x
5
9x
4
+ 4x
3
6x
2
+ 4x + 1
= (x
2
+x 1)(x
6
5x
5
9x
3
5x 1)
Note that the set of cusps is entirely defined overQ: Let J
0
(35) denote the Jacobian of X
0
(35).
J
0
(35)(Q) has rank 0 by 2-descent computation in Magma. AsJ
0
(35) has good reduction at 3, the
reduction mapJ
0
(35)(Q)!J
0
(35)(F
3
)'Z
2
Z
24
is injective. We find
Z
2
Z
24
=h[3(0;1) 31
+
]; [1
1
+
]i
which therefore impliesJ
0
(35)(Q) =Z
2
Z
24
: Since finding quadratic points onX
0
(35) amounts
to finding the rational points on its symmetric square X
0
(35)
(2)
, we consider the following map
:X
0
(35)
(2)
!J
0
(35) mappingfP;Qg7! [P +Q1
+
1
]: LetP be a quadratic point in
X
0
(35)(K) with x-coordinate not inQ and letP denote its Galois conjugate. The linear equivalence
class of the degree 0 divisorP +P1
+
1
gives rise to a nonzero rational point onJ
0
(35):
Magma represents all 47 elements of J
0
(35)(Q)nf0g as effective divisors minus a multiple of
one of the points at infinity. We check that none of the points is of this form except (
1
p
5
2
; 0).
Therefore, there is no exceptionalK-rational points onX
0
(35):
If x(P )2 Q; there are rational points which are two points at infinity and two points with
x(P ) = 0; for all other such points, the Galois conjugate P must be the image of P under the
hyperelliptic involutionw
35
ofX
0
(35) given byy7!y. Thusy(P ) must be
p
d times a rational
number whereK =Q(
p
d) withd a non-square. Thereby,P gives rise to aQ-rational point on
the quadratic twistX
d
0
(35) ofX
0
(35) byd. The quadratic fields of interest are in fact of the form
Q(
p
d) where d =p with p prime in the setf2; 7; 11; 19; 43; 67; 163g: However, by Corollary
36
3.5 in [33] the quadratic twistX
d
0
(35) ofX
0
(35) does not even haveQ
p
-rational points, let alone
rational points which shows that
X
0
(35)(K) =X
0
(35)(Q) =f1
+
;1
; (0; 1); (0;1)g:
4.7 N=40 & 48
The modular curvesX
0
(40) andX
0
(48) are hyperelliptic curves of genus 3 and admit the following
models by [11] or [13]:
y
2
=f
40
(x) =x
8
+ 8x
6
2x
4
+ 8x
2
+ 1
y
2
=f
48
(x) =x
8
+ 14x
4
+ 1:
X
0
(40) has only rational cusps whereas X
0
(48) has quadratic cusps which are defined over
Q(
p
1): Let P be a nonexceptional quadratic point on X
0
(N); where N = 40 or 48. Then P
is of the form (x;
p
f
N
(x)) withx2Q andf
N
(x) forN = 40; 48 as defined above. Since the
polynomialf
N
(x)> 0 for allx2R; thenP must be defined over a real quadratic field. Therefore,
there are no non-exceptional quadratic points onX
0
(N) defined over anyK2S: On the other hand,
the quadratic exceptional points are completely classified in [Table 11,15 [5]]. All the exceptional
quadratic points of X
0
(40) are defined over Q(
p
1). By Magma each gives rise to a cyclic
40-isogeny pointwise defined in an extension ofQ(
p
1) of degree at least 8. In conclusion, we
have
X
0
(40)(K) =X
0
(40)(Q) =f1
+
;1
; (1;4); (1;4); (0;1)g
for anyK2S: In the caseX
0
(48), the exceptional quadratic points are defined overQ(
p
1) but
they are entirely cuspidal points as pointed above. As a result, for anyK2S we have
X
0
(48)(K) =X
0
(48)(Q) =f1
+
;1
; (1;4); (1;4); (0;1)g
37
where1
+
= [1; 1; 0] and1
= [1;1; 0]: Therefore, there are no K-rational cyclic 40 and
48-isogeny for anyK inS:
4.8 N=45 & 55
X
0
(45) is a nonhyperelliptic curve of genus 3 with a projective model
x
2
y
2
+x
3
zy
3
zxyz
2
+ 5z
4
:
The quadratic points onX
0
(45) are determined in [34] by using a slightly different model. There
are only two noncuspidal quadratic points defined over an imaginary quadratic field of class number
one. They areP = (
1w
2
;1; 1) andQ = (1;
w1
2
; 1) wherew =
p
11: By Magma, these points
do not produce a cyclic 45-isogeny pointwise defined in a quadratic extension ofQ(
p
11):
X
0
(55) is a nonhyperelliptic curve of genus 5. The quadratic points ofX
0
(55) are determined
in [34]. By Magma, we observe that the ones defined over an imaginary quadratic fieldK of class
number one do not yield a cyclic 55-isogeny upon a quadratic extension ofK:
4.9 N=77
The modular curve X
0
(77) is a nonhypelliptic curve of genus 7. However, X
0
(77) is not in the
database of the Small Modular Curves package in Magma. Instead, we compute the model by using
the code written by Ozman and Siksek [34].
We start with computing the torsion subgroup of the Jacobian. WriteJ
0
(77) for the Jacobian
ofX
0
(77): Note that all the cusps ofX
0
(77) are rationals. LetC
0
(77) be the cuspidal subgroup of
J
0
(77)(Q) generated by the classes of differences of the cusps. In the next lemma, we show that
the cuspidal subgroup ofX
0
(77) in fact satisfies the equalityC
0
(N) =J
0
(N)(Q)
tor
:
Lemma 2. The torsion subgroup ofJ
0
(77)(Q) is equal to its cuspidal subgroupC
0
(77) which is
isomorphic toZ
10
Z
60
:
38
Proof. In Magma, one computes the group structure ofC
0
(77) to beZ
10
Z
60
: As 3 and 13 are
primes of good reduction forX
0
(77), we compute that
J
0
(77)(F
3
)'Z
10
Z
420
J
0
(77)(F
13
)' (Z
2
)
3
(Z
30
)
2
Z
7380
Since J
0
(77)(Q)
tor
injects into both of these groups, then [J
0
(77)(Q)
tor
: C
0
(77)] divides 7 and
2
3
:3
3
:5:41 respectively. But these are coprime. Hence we showJ
0
(77)(Q)
tor
=C
0
(77):
Generally speaking, given a divisor N
1
of level N with (N;N=N
1
) = 1, the Atkin-Lehner
involutionw
N
1
acts as a permutation on the set of cusps ofX
0
(N): LetB
0
(77) denote the group
generated by the Atkin-Lehner involutions ofX
0
(77): As shown in [21], the automorphism group
Aut(X
0
(77)) ofX
0
(77) isB
0
(77) and we haveB
0
(77) =hw
7
ihw
11
i: One can easily check that
B
0
(77) acts transitively on the set of cusps. Let =hw
77
i be the subgroup of Aut(X
0
(77)) and
consider the quotient curve X
0
(77)
+
= X
0
(77)= with the projection : X
0
(77)! X
0
(77)
+
:
This map has degreedeg = #: We now denote the Jacobian ofX
0
(77)
+
byJ
+
0
(77) to avoid the
confusion with the JacobianJ
0
(77) ofX
0
(77):
Lemma 3. rk(J
0
(77)(Q)) = rk(J
+
0
(77)(Q)).
Proof. We use howJ
0
(77) is isogenous to a product of abelian varieties associated to newforms,
following [19, p. 234]. More presicely, we have
J
0
(77)A
f
1
A
f
2
A
f
3
A
f
4
A
2
f
5
(4.4)
where denotes isogeny overQ and A
f
i
’s are simple modular abelian varieties corresponding
to Galois orbits f
i
of weight 2 and level 77. It follows by (5) that rk(J
0
(X)) =
P
5
i=1
m
i
A
f
i
where m
i
denotes multiplicity. We also found that J
+
0
(77) A
f
1
A
f
5
: With the modular
abelian varieties package in Magma, we checked that the L-functionL(A
f
i
; 1) is non-zero for the
39
eigenforms f
i
in (5) not conjugate to f
1
. By the theorem of Kolyvagin and Logachev [23], this
implies that the correspondingA
f
i
has analytic rank 0 so is the algebraic rank overQ. Therefore,
rk(J
0
(77)(Q)) =A
f
1
=rk(J
+
0
(77)(Q)):
We now need information about the Mordell-Weil group ofJ
+
0
(77)(Q): Note thatX
0
(77)
+
is
a hyperelliptic curve of genus 2 by [15]. Using the algorithm in [39] implemented in Magma as
2-descent on Jacobians of hyperelliptic curves, we compute thatJ
+
0
(77)(Q) has Mordell-Weil rank
1 as well as its generators.
We would like to describe all quadratic points onX
0
(77): One way of finding quadratic points on
X
0
(77) is by pulling back of rational points onX
0
(77)
+
under: We will call points onX
0
(77) not
arising as pullback of a rational point onX
0
(77)
+
exceptional: Our aim is to prove the following:
Theorem 4.2. There are no exceptional quadratic points onX
0
(77):
For this purpose, we use the Chabauty method for symmetric products of curves developed by
Siksek in [37] combining with a Mordell-Weil sieve. In order to apply this, we slightly change how
we view quadratic points similarly as in the previous cases ofN: For a given smooth curveX over
Q; we view a pair of quadratic pointP onX and its Galois conjugatefP;Pg as a rational point
on the symmetric productX
(2)
: The pairfP;Pg is represented as an equivalence class of degree
2 effective divisorD := P +P inX
(2)
(Q): In the particular caseX = X
0
(77), if we pull back
P 2 X
0
(77)
+
(Q),
(P ) is a pair of quadratic points on X
0
(77): We call such quadratic points
non-exceptional. Hence we can rephrase Theorem 4.2 asX
0
(77)
(2)
(Q) =
(X
0
(77)
+
(Q)):
Morevoer,X
0
(77)
+
(Q) can be completely determined by explicit Chabauty command built in
Magma. Once we know the full list of points inX
0
(77)
+
(Q), combining with Theorem 4.2, this
gives us a complete list of all quadratic points on X
0
(77): In light of these facts, we found that
X
0
(77)
+
has precisely six rational points and so by the theorem above, X
0
(77) has precisely six
pairs of quadratic points. We are able to list these quadratic points explicitly by using the equations
for the degree 2 map fromX
0
(77)! X
0
(77)
+
: These equations come for free while computing
the model ofX
0
(77):
40
Point Coordinates
Q
1
[1:-1:0]
Q
2
[1:1:0]
Q
3
[-1:-2:1]
Q
4
[-1:2:1]
Q
5
[0:-1:1]
Q
6
[0:1:1]
Table 4.4: The rational points onX
0
(77)
+
Chabauty Step : Let X be a smooth projective curve overQ of genus 2 with Jacobian
J(X) of rankr
X
: We suppose that there exists an involutionw ofX defined overQ such that the
quotient mapX!X=hwi has degree 2. DefineC :=X=hwi and denote byg
C
for the genus of
C andr
C
for the rank of the Jacobian ofC. Forp a prime of good reduction for bothX andC;
Coleman gives the following bilinear pairing
X
=Qp
J(X)(Q
p
)!Q
p
;
!;
"
X
i
(P
i
Q
i
)
#!
7!
X
i
Z
P
i
Q
i
! (4.5)
where
X
=Qp
is the space of regular differentials on the curveX
=Qp
: The kernel on the left is 0 and
on the right is J(X)(Q
p
)
tor
: Define V to be the annihilator of J(X)(Q) with respect to 4.5. As
being an algebraic curve overQ,X has a minimal regular model overZ
p
which we denote byX:
TheZ
p
-schemeX is smooth whenX
Fp
is non-singular. We considerV :=V\
X
=Zp
and
e
V denote
its reduction modulop. The symmetric Chabauty differs from the usual Chabauty-Coleman method
in the way that the annihilating differentials must also satisfy Tr(!) = 0 where Tr:
X=Qp
!
C=Qp
is the trace operator. More precisely, we defineV
0
=V\ ker(Tr) and let
e
V
0
be its image under
the reduction modulop: Note thatmodpresiduedisc of a pointQ2X
(2)
(Q) consists of points
P2X
(2)
(Q) such that
e
P
e
Q where denotes the reduction modp:
Now let!
1
;:::;!
r
be a basis for
e
V and letQ be an element ofX
(2)
(Q): IfQ =Q
1
+Q
1
as an
effective degree 2 divisor, we sayQ
1
has multiplicityd
1
= 2: Otherwise we haveQ = Q
1
+Q
2
whereQ
1
;Q
2
are distinct with multiplictyd
j
= 1: Fix an extensionv toL =Q(Q
1
;Q
2
) abovep:
41
Lett
f
Q
j
be a uniformiser atQ
j
for eachj2f1; 2g: We can expand each!
i
as a formal power series
atQ
j
as!
i
=
P
1
k=0
a
k
(!
i
;t
f
Q
j
)t
k
f
Q
j
dt
f
Q
j
We then define for a positive integerm
v(!
i
;t
f
Q
j
;m) =
a
0
(!
i
;t
f
Q
j
);
a
1
(!
i
;t
f
Q
j
)
2
;:::;
a
m1
(!
i
;t
f
Q
j
)
m
!
Let
e
A be the followingr 2 matrix:
e
A =
0
B
B
B
B
B
B
B
@
v(!
1
;t
f
Q
1
;d
1
) v(!
1
;t
f
Q
2
;d
2
)
v(!
2
;t
f
Q
1
;d
1
) v(!
2
;t
f
Q
2
;d
2
)
.
.
.
.
.
.
v(!
r
;t
f
Q
1
;d
1
) v(!
r
;t
f
Q
2
;d
2
)
1
C
C
C
C
C
C
C
A
Note that whenQ
1
=Q
2
, we haved
1
= 2 with the assumption thatp> 2:
Theorem 4.3 (Symmetric Chabauty, Siksek). LetQ =fQ
1
;Q
2
g2X
(2)
(Q): Forp a prime of good
reduction forX; pick!
1
;:::;!
r
as a basis for
e
V as above. Supposep> 3 when [F
p
(
f
Q
1
);F
p
] = 1:
If rank(
e
A) = 2 thenQ is alone in its modp residue class belonging toX
(2)
(Q):
Proof. This is the same statement as in [3, Theorem 2.1] which is a variant of the special case of
Theorem 1 in [37].
We particularly make use of Theorem 4.3 for the points in X
0
(77)
(2)
(Q)n
(X
0
(77)
+
(Q)):
To deal with the points in
(X
0
(77)
+
(Q)) we need to argue in a slightly different way using
the relative version of Chabauty. The relative Chabauty differs from the symmetric Chabauty in
the way that the annihilating differentials must satisfy Tr(!) = 0 where Tr is the trace map: Tr:
X=Qp
!
C=Qp
: More precisely, we defineV
0
=V\ ker(Tr) and let
e
V
0
be its image reduction
modulop:
Theorem 4.4 (Relative Symmetric Chabauty, Siksek). LetQ =fQ
1
;Q
2
g2
(C(Q))X
(2)
(Q)
be a non-exceptional point. Letp be a prime of good reduction for bothX andC, morever assume
42
thatp> 3 when [F
p
(
e
Q
1
);F
p
] = 1: Lett
f
Q
1
be a uniformiser atQ
1
and!
1
;:::;!
s
be a basis for
e
V
0
:
If for somei2f1;:::sg;
!
i
dt
f
Q
1
j
t
g
Q
1
=0
6= 0
Then every element ofX
(2)
(Q) in the residue class ofQ is in fact belonging to
(C(Q)):
Proof. This is a modification of special case of [37, Theorem 2].
In order to apply Theorem 4.4 to the caseX =X
0
(77) andC =X
0
(77)
+
; we need to compute
the image
e
V
0
ofV
0
=V\ ker(Tr:
X=Qp
!
C=Zp
) under the reduction modp where bothX and
C have a good reduction atp: For this purpose, we need the following lemma from [3].
Lemma 4. Let X be a smooth projective curve with minimal proper regular modelX=Z
p
and
denote
e
X =X
Fp
: Then the reduction map
X=Zp
!
e
X=Fp
is surjective.
Proposition 4.3. LetV be the image of 1w
77
:
X=Q
!
X=Q
: Assumer
X
=r
C
: Forp a good
reduction for bothX andC; we have
e
V
0
= (1e w
77
)(
e
X
) where denotes the reductionmodp:
Proof. Since w
77
is an involution, (1w
77
)(
X
) =ker(1 +w
77
) which is by the definition the
kernel of the trace map. Next we show that all the differentials inV indeed annihilateJ(X)(Q) with
respect to (4.5) i.e.
R
D
0
! = 0forall!2V andD2J(X)(Q): Sincer
X
=r
C
by Lemma 3, then
the indexN = [J(X)(Q) :
(J(C)(Q))] is finite. LetD2J(X)(Q): AsN:D2
(J(C)(Q));
there exists 2J(C)(Q) such thatN:D =
(): By the properties of Coleman integration, we
have
Z
D
0
! =
1
N
Z
N:D
0
! =
1
N
Z
()
0
! =
1
N
Z
D
0
Tr(!) = 0
for every!2 (1w
77
)(
X
): Let
f
W be the image of 1e w
77
: We want to show that
f
W =
e
V: Since
!2 ker(1 e w
77
) thene !2 ker(1 e w
77
); we have
e
V
f
W: Now considere !2
f
W ande v such that
e ! = (1w
77
)(e v): By Lemma 4,e v lifts to somev in
X=Zp
: Then (1w
77
)(v)2V is reduced to
e !; as claimed.
The Mordell Weil sieving step : We adapted here the sieve used in [37] to apply it to
the case X = X
0
(77): The goal is to sieve for potential unknown quadratic points and obtain a
43
contradiction. To start with, we need to choose primesp
1
;:::;p
k
: A detailed guideline on how to
choose primes and in which order can be found in [4]. In order to make the sieve work, we further
need the following info:
• A finite index subgroupGJ(X)(Q) with generatorsD
1
;:::D
n
.
• A positive integerI withI:J(X)(Q)G:
• A non-empty setLX
(2)
(Q) of known nonexceptional quadratic points.
For the moment, we letp :=p
i
for somei: Define :Z
n
!GJ(X)(Q) by (a
1
;:::;a
n
) =
P
n
i
a
i
D
i
: Write :X
(2)
(Q)!G byD7!I:[D1] where1 is a rational degree 2 divisor and
let{
p
be the Abel-Jacobi map onX
(2)
(F
p
) with the basepointf 1: Then we define
p
to make the
following diagram commute
L X
(2)
(Q) G Z
n
e
X
(2)
(F
p
) J(F
p
)
i
p
p
LetP 2 X
(2)
(Q)nL be a (hypothetical) unknown point. We will argue that suchP cannot
exist. Since the diagram commutes, note that
g
(P ) =
p
(
e
P )2 Im(
p
): Define the set
H
p
:=fS2
e
X
(2)
(F
p
) :
p
(S)2 Im(
p
)g
andM
p
H
p
as the set of pointsS satisfying one of the following conditions
• S = 2
e
L
• S =
e
R for someR2L not satisfying the conditions of Theorem 4.3.
• S =
e
R for someR2L\
(C(Q)) not satisfying the conditions of Theorem 4.4 in the case
of degree 2 mapX!C:
44
Notice that by the choice of our unknown pointP ,
e
P is inM
p
: We use the Chabauty step to
eliminate as many of the points inM
p
as possible. The more points that pass the Chabauty step,
the better since this enables us to limit more possibilites for
e
P: Since(M
p
)
p
(Z
n
); we form
the following ker(
p
)-cosets
W
p
:=
1
p
((M
p
)):
If (P ) = a
1
D
1
+ ::: + a
n
D
n
then (a
1
;:::;a
n
) 2 w + ker(
p
) for some w + ker(
p
) in
W
p
: Hence the set of cosets W
p
will carry the information of the list of possibilities for
e
P:
The beauty of the sieve lies in the observation that W
p
can be compared for different values
of p: This means that if we choose a different prime of good reduction, namely p
0
; then the
preimage
1
p
0
((P )) = (a
1
;:::;a
n
)2 w
0
+ ker(
p
0) for somew
0
+ ker(
p
0) inW
p
0 as above. So
(a
1
;:::;a
n
)2 W
p
\W
p
0: Then we repeat this process for eachp in the listp
1
;:::;p
k
of primes of
good reduction. We conclude the following.
Proposition 4.4 (Sieving Principle). If
T
k
i=1
W
p
=;; thenX
(2)
(Q) =L:
Magma Output
The model obtained for the modular curveX
0
(77) as a curve inP
6
:
x
1
x
3
x
2
2
+ 21x
2
x
7
2x
3
x
4
4x
3
x
7
+ 6x
2
4
19x
4
x
5
+ 6x
4
x
6
21x
4
x
7
+ 13x
2
5
5x
5
x
6
8x
5
x
7
2x
6
x
7
+ 13x
2
7
= 0;
x
1
x
4
x
2
x
3
x
2
x
7
2x
3
x
4
+ 47x
3
x
7
+ 2x
2
4
+ 8x
4
x
5
10x
4
x
6
+ 15x
4
x
7
26x
2
5
+ 25x
5
x
6
37x
5
x
7
3x
2
6
+ 3x
6
x
7
73x
2
7
= 0;
x
1
x
5
11x
2
x
7
x
2
3
3x
3
x
4
+ 80x
3
x
7
+ 21x
4
x
5
17x
4
x
6
+ 33x
4
x
7
48x
2
5
+ 42x
5
x
6
56x
5
x
7
5x
2
6
+ 5x
6
x
7
126x
2
7
= 0;
x
1
x
6
18x
2
x
7
5x
3
x
4
+ 99x
3
x
7
+ 29x
4
x
5
22x
4
x
6
+ 47x
4
x
7
62x
2
5
+ 54x
5
x
6
70x
5
x
7
7x
2
6
+ 8x
6
x
7
161x
2
7
= 0;
x
1
x
7
2x
2
x
7
x
3
x
7
x
2
4
+ 2x
4
x
5
+ 2x
4
x
7
x
2
5
+x
5
x
7
2x
2
7
= 0;
45
x
2
x
4
9x
2
x
7
x
2
3
+ 34x
3
x
7
2x
2
4
+ 13x
4
x
5
9x
4
x
6
+ 18x
4
x
7
23x
2
5
+ 19x
5
x
6
21x
5
x
7
2x
2
6
+ 2x
6
x
7
54x
2
7
= 0;
x
2
x
5
16x
2
x
7
x
3
x
4
+30x
3
x
7
3x
2
4
+18x
4
x
5
9x
4
x
6
+25x
4
x
7
25x
2
5
+18x
5
x
6
17x
5
x
7
2x
2
6
+ 4x
6
x
7
54x7
2
= 0;
x
2
x
6
23x
2
x
7
+ 28x
3
x
7
5x
2
4
+ 23x
4
x
5
10x
4
x
6
+ 29x
4
x
7
26x
2
5
+ 17x
5
x
6
11x
5
x
7
2x
2
6
+ 5x
6
x
7
55x
2
7
= 0;
x
3
x
5
16x
3
x
7
x
2
4
3x
4
x
5
+4x
4
x
6
5x
4
x
7
+9x
2
5
9x
5
x
6
+13x
5
x
7
+x
2
6
x
6
x
7
+24x
2
7
= 0;
x
3
x
6
23x
3
x
7
5x
4
x
5
+ 5x
4
x
6
7x
4
x
7
+ 13x
2
5
12x
5
x
6
+ 17x
5
x
7
+x
2
6
+ 35x
2
7
= 0
Genus ofX
0
(77) : 7.
Cusps:(1; 0; 0; 0; 0; 0; 0); (1=2; 1=2; 1=2; 1=2; 1=2; 1; 0); (5; 3; 3; 2; 2; 2; 1); (5; 3; 3; 2; 2; 9; 1):
X
+
=X
0
(77)
+
: genus 2 hyperelliptic curvey
2
=x
6
+ 2x
5
+ 7x
4
+ 8x
3
+ 7x
2
+ 2x + 1:
Group Structure ofJ(X
+
)(Q) :Z
5
:[Q(1;1; 0)]Z:[Q
X
+(1;1; 0)]; whereQ = (0; 1; 1)
andQ
X
+ = (1; 1; 0):
Group Structure ofGJ
0
(77)(Q) :
Z
10
:(8D
tor,1
+ 3D
tor,3
)Z
60
(31D
tor,2
8D
tor,3
)Z:D
1
where
D
tor,1
= [(1=2; 1=2; 1=2; 1=2; 1=2; 1; 0) (1; 0; 0; 0; 0; 0; 0)];
D
tor,2
= [(5; 3; 3; 2; 2; 2; 1) (1; 0; 0; 0; 0; 0; 0)];
D
tor,3
= [(5; 3; 3; 2; 2; 9; 1) (1; 0; 0; 0; 0; 0; 0)];
D
1
= [P
2
+P
2
(1=2; 1=2; 1=2; 1=2; 1=2; 1; 0) (5; 3; 3; 2; 2; 9; 1)] =
([Q
X
+ (1;1; 0)])
forP
2
defined in the table below satisfying(P
2
) =Q
X
+:
Primes used in sieve : 19,23.
There are no quadratic points onX
0
(77) that do not come fromX
0
(77)
+
(Q) via :X
0
(77)!
X
0
(77)
+
and there is no noncuspidal rational point. Moreover, we obtain the following table which
46
gives a complete list of all quadratic points up to Galois conjugacy. In addition, the elliptic curves
induced by these quadratic points do not have a 77-isogeny in a quadratic extension of the base
field.
Name w
2
Coordinates CM
P
1
-19
1
2
(w 7);
1
2
(3w 3);w;
1
2
(w + 1);w 1;
1
2
(3w 5); 1
-19
P
2
-19
1
2
(w 1);w 1;
1
2
(w 5);w 2;
1
2
(3w 1);
1
2
(3w 1); 1
-19
P
3
-7
w 2;
1
2
(3w 1);
1
2
(w 1);
1
2
(w 3);
1
2
(3w 3);
1
2
(3w 3); 1
-7
P
4
-7
w 2;
1
2
(7w 1);
1
2
(w 1);
1
2
(5w 3);
1
2
(3w 3);
1
2
(7w 3); 1
-28
Table 4.5: Noncuspidal quadratic points ofX
0
(77)
47
Chapter 5
Growth in Case E(K)[2]6=Z
2
Z
2
5.1 Certain Restrictions on Growth
LetK be a number field. We will introduce the results which are used in the classification of torsion
of quadratic twists very often. For the following lemma, we make use of the Galois property of the
Weil Pairing.
Lemma 5. LetE=K be an elliptic curve,n a positive integer andK(E[n]) the field of definition
ofE[n]. ThenK(E[n]) contains a primitiventh root of unity.
Proof. Consider a basisfe;sg ofE[n] which have coordinates inK(E[n]). Thene;s are fixed by
the action of for any2 Gal(
K=K(E[n])): By the property of the Weil pairing, it follows that
h(e);(s)i) =(he;si) =he;si: As a resulthe;si that is a primitiventh root of unity is fixed by
. This proves the claim.
Proposition 5.1. LetK be an imaginary quadratic field and letE be an elliptic curve overK: If
L is a quadratic extension ofK; then the largest odd primep such thatZ
p
Z
p
E(L) isp = 3:
Proof. Suppose Z
p
Z
p
E(L) As shown in Lemma 5, L has a primitive pth root of unity,
denoted by
p
. Then we haveQ(
p
) L and so [L : Q(
p
)][Q(
p
) : Q] = [L : Q] = 4. This
implies that [Q(
p
) :Q] divides 4: Since [Q(
p
) :Q] =p 1 for any primep andp 1 divides 4,
this is only possible whenp is 2,3 or 5. Ifp = 5,L =Q(
5
) and Gal(Q(
5
)=Q) is isomorphic to
48
Z
4
. HenceQ(
5
) has a unique intermediate fieldQ(
p
5) which is not imaginary, completing the
proof. For occurrence of full 3-torsion, see Remark 3.
The next lemma plays a key role in studying the 2-torsion part of elliptic curves.
Lemma 6 ([22]). Let E be an elliptic curve over a field k withchar(k)6= 2; 3: Suppose E is given
by
y
2
= (x)(x)(x
)
with;;
2k: ForP = (x;y) inE(k), there exists a k-rational pointQ on E such that 2Q =P
if and only ifx;x, andx
are all squares in k.
In this case if we fix the sign of the square roots ofx;x;x
; then thex-coordinate
ofQ equals either
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
If the criteria of the statement above is satisfied for a pointP; we say thatP ishalved: Now, we
will list numerous restrictions on growth in quadratic extensions of anyK inS:
Theorem 5.1. LetK be any number field inS,E an elliptic curve overK, andL any quadratic
extension ofK: Then
(i) IfE(K)
tor
=Z
4
thenZ
4
Z
8
6E(L):
(ii) IfE(K)
tor
=Z
4
thenE(L)
tor
6=Z
16
:
(iii) E(L) =Z
20
can occur only forK =Q(
p
D);D =2;11;19:
(iv) E(L)
tor
=Z
28
does not occur.
49
Proof. (i) GivenE=K as in the hypothesis, we may assume thatE has a model of the form
y
2
=f(x) =x(x +)(x +)
where ; are the roots of an irreducible quadratic polynomial over K: Hence they are defined
over the quadratic extensionL =K() =K() ofK: Without loss of generality, we may in fact
assumeL =K(
p
d) for some non-squared in K. IfZ
4
Z
4
E(L) thend =1. Note that the
2-torsion points ofE are (0; 0); (; 0) and (; 0):
LetE(K)
tor
'hP
1
i for some pointP
1
of order 4. Then 2P
1
= (0; 0): Furthermore, by Lemma
6,, are both squares inL andx(P
1
) = (
p
): By way of contradiction, supposeE(L) contains
a subgroup of the formZ
4
Z
8
: Then one of the 2-points inf(0; 0); (; 0); (; 0)g must be in
[2]E(L); which yields the following cases:
(1)9Q
1
2E(L)
tor
with 2Q
1
=P
1
where 2P
1
= (0; 0):
(2)9Q
2
2E(L)
tor
with 2Q
2
=P
2
where 2P
2
= (; 0):
(3)9Q
3
2E(L)
tor
with 2Q
3
=P
3
where 2P
3
= (; 0):
If (1) holds, thenZ
2
Z
4
E(L)=E(K) which in turn yields thatE
1
(K)[2] =Z
2
Z
2
: But
E
1
(K)[2]' E(K)[2] = Z
2
; therefore we obtain a contradiction. If (2) holds, then x(P
2
) +
is a square inL by Lemma 6. Since is a square inL by the observation above, we may write
=k
2
(a +bi)
2
for somek;a;b2K such that (a;b) = 1: We compute that
x(P
2
) + =
p
() = 2k
2
(a +bi)
p
abi
and so
p
abi must be a square inL sincea +bi;abi are coprime. Notice that 2 is a square inL: It
follows thatabi is a square inL; which implies thatabi = (x +yi)
2
for somex;y2 K: and we
derivei2K, a contradiction. If (3) holds, thenx(P
3
) + is a square inL by Lemma 6. As argued
above, we computex(P
3
) + =
p
() = 2k
2
(abi)
p
abi which similarly results in the
observation thati2K; therefore this completes the proof.
50
(ii) Theorem 5(iv) in [12].
(iii) The modular curveX
0
(20) is an elliptic curve with the model
y
2
=x
3
+x
2
+ 4x + 4:
Note thatX
0
(20)(K) has torsionZ
6
for allK2S: On the other hand,X
0
(20)(K) has rank 0
except forK =Q(
p
D);D =2;11;19: Since the torsion part consists entirely of cusps, the
claim follows. For example, whenK =Q(
p
2) one can see Table 5.1.
(iv) Suppose E(L)
tor
= Z
28
: Note that E(K)
tor
is even, otherwise E(L)
tor
would be odd.
Replacing E by a quadratic twist E
d
; if necessary, we may assume that E(K)
tor
= Z
14
: Let P
be a 4-torsion and Q a 7-torsion such that E(L) =hP;Qi and E(K) =h2P;Qi. Since 2P is
K-rational,(P ) =P orP . HencehPi andhQi are Gal(L=K)-stable. Therefore,hP;Qi forms
aK-rational cyclic 28-isogeny pointwise defined overL; contradicting Theorem 4.1.
5.2 Classification of Torsion
By Theorem 3.1, we breakS into three subsets: S
1
= fQ(
p
11);Q(
p
67);Q(
p
163)g;
S
2
=fQ(
p
2);Q(
p
19)g andS
3
=fQ(
p
7);Q(
p
43)g:
Theorem 5.2. LetK be inS,E=K an elliptic curve and letd a nonsquare inK.
1. IfK2S
1
[S
3
andE(K)
tor
'Z
15
thenE
d
(K)
tor
'Z
1
2. IfK2S
2
[S
3
andE(K)
tor
'Z
11
thenE
d
(K)
tor
'Z
1
3. IfE(K)'Z
9
when
(i)K2fQ(
p
2);Q(
p
11)g thenE
d
(K)'Z
1
orZ
3
(ii) Otherwise,E
d
(K)'Z
1
4. IfE(K)
tor
'Z
7
thenE
d
(K)'Z
1
51
5. IfE(K)
tor
'Z
5
thenE
d
(K)
tor
'Z
1
orZ
3
6. IfE(K)
tor
'Z
3
when
(i)K2fQ(
p
2);Q(
p
11)g thenE
d
(K)'Z
1
;Z
3
;Z
5
; orZ
9
(ii) Otherwise,E
d
(K)'Z
1
;Z
3
; orZ
5
7. IfE(K)
tor
'Z
1
thenE
d
(K)
tor
'Z
1
;Z
3
;Z
5
;Z
7
;Z
9
;Z
11
; orZ
15
(depending onK)
Proof. Note that ifE
d
is a quadratic twist ofE, thenE is a quadratic twist ofE
d
: Moreover, by
Remark 1, ifE(K)
tor
has odd order thenE
d
(K)
tor
should also have odd order. Then, by Theorem
3.1 bothE(K)
tor
andE
d
(K)
tor
belong to the setfZ
1
;Z
3
;Z
5
;Z
7
;Z
9
;Z
11
;Z
15
g:
Notice that if E(K)
tor
' Z
m
and E
d
(K) ' Z
m
for m 2 f5; 7; 11g, then by Lemma 1,
E(K(
p
d)) would contain fullm-torsion which is impossible by Proposition 5.1. In particular, if
one fixesE(K)
tor
'Z
5
; then there are no points of order 7, 9, and 11 inE
d
(K)
tor
since together
with a K-rational point of order 5 on E, this would give rise to a cyclic 35, 45 and 55-isogeny
over a quadratic extension ofK respectively. But this cannot happen by Theorem 4.1. Likewise,
if E(K)
tor
' Z
m
where m2f7; 11g then since there exists no cyclic 21 and 33-isogenies in a
quadratic extension ofK; it follows thatZ
3
6E
d
(K): Furthermore, whenm = 7; we can conclude
thatZ
11
6E
d
(K) due to the non-existence of a cyclic 77-isogeny over a quadratic extension of the
base field.
Having a point of order 9 in E(K)
tor
and a point of order 3 in E
d
(K) imply that E is
isogenous (over K) to an elliptic curve E
0
=K with a cyclic 27-isogeny by [29, Lemma 7]. This
is only possible when K is Q(
p
2) or Q(
p
11) by Theorem 4.1. On the other hand, if
E(K)
tor
' E
d
(K)
tor
' Z
9
then by Lemma 1, E(K(
p
d)) contains full 9-torsion which in turn
implies thatK(
p
d) has a primitive 9th root
9
of unity. This further means that [K(
p
d) :Q] = 4 is
divisible by [Q(
9
);Q]j = 6 which gives us a contradiction. IfE(K)
tor
'Z
15
andE
d
(K)
tor
'Z
3
,
then by [29, Lemma7] E is isogenous (over K) to an elliptic curve E
0
=K which has a cyclic
45-isogeny. Magma computations forN = 45 in section 3 additionally show that such a curveE
0
must be defined overK =Q(
p
11) with CM. SinceE
0
is isogenous toE; thenE is also CM with
52
E(K(
p
d))
tor
'Z
3
Z
15
which is impossible by [7]. Finally, assumeE(K)
tor
is trivial. SinceK
does not contain a primitive 3rd root of unity,E(L) doesn’t include full 3-torsion, completing the
proof.
Theorem 5.3. LetK2S andE=K an elliptic curve withE(K)[2]'Z
1
; Then, for any extension
L with [L :K] = 2;E(L)
tor
is isomorphic to one of the following groups:
Z
n
; n2f1; 3; 5; 7; 9; 11; 15g; Z
3
Z
3
; Z
3
Z
9
:
Furthermore,Z
3
Z
9
is only realized in a quadratic extension ofK =Q(
p
2) orQ(
p
11):
Proof. LetK andL be given as in the hypothesis. SupposeE=K is an elliptic curve withE(K)[2]
trivial. ThenE(L)[2] must be trivial as observed simply in Remark 1. Letp be an odd prime and
P2E(L)
tor
a point of orderp: By Theorem 3.1, the possible values ofp are 3,5,7,11,13, and 17.
Let us assume p2f13; 17g: WLOG we may assume L = K(
p
d) for some nonsquare d2 K
and following Lemma 1, we have eitherP2 E(K)[p] orP2 E
d
(K)[p]: Hence, replacingE by
E
d
if necessary, we may assumeZ
p
E(K)[p]: However, this is not possible since the groups
Z
13
andZ
17
do not arise as torsion subgroup ofE(K) for anyK2S: In the remaining cases, we
have thatE(L)
tor
'Z
m
Z
n
wherem2f3; 5; 7; 9; 11; 15g andn2f3; 5; 7; 9; 11; 15g: Note that
Proposition 5.1 eliminates the possibilities wherem;n are both primes withm =n> 3: Applying
Lemma 1 reduces the problem to showing whether E(K)
tor
' Z
m
and E
d
(K)
tor
' Z
n
can be
realized (replacingE byE
d
if necessary). We answered this question by classifying the possible
torsion structures forE
d
(K)
tor
in Theorem 5.2, proving the claim.
53
5.3 Examples of Growth
Table 5.1: Growth of Torsion overK =Q(
p
2)
E(K)
tor
E(K(
p
d))
tor
d2K Weierstrass Model [A;B] ofE; w =
p
2
Z
2
Z
2
Z
10
15 [8170875;43197806250]
Z
2
Z
6
5 [43200; 2376000]
Z
2
Z
2
1 [1;10]
Z
16
15 [61229925;138962162250]
Z
12
3 [17523;2921778]
Z
10
5 [907875; 1599918750]
Z
8
3 [2799387;1802779146]
Z
6
3 [47088;5968512]
Z
4
1 [587763;23605938]
Z
3
Z
3
Z
9
3 See Remark 3
Z
3
Z
3
3 [12096;544752]
Z
15
5 [97875; 14208750]
Z
4
Z
4
Z
4
1 [13;34]
Z
2
Z
12
15 [438075;365222250]
Z
2
Z
8
7 [13334517;14122673082]
Z
2
Z
4
65 [142587; 20723094]
Z
20
1 + 4w A =466776w + 1808757;
B =458681400w 375644250
Z
12
5 [47925; 12399750]
Z
8
5 [103707; 12854646]
Z
6
Z
3
Z
6
3 [5805;285714]
Z
2
Z
6
3 [46035;3116178]
Z
12
3 [373923; 88006878]
Z
8
Z
2
Z
8
7 [5211; 319734]
Z
16
70 [19530747; 33212905686]
Z
9
Z
3
Z
9
3 See Remark 3
Z
10
Z
2
Z
10
33 [58347; 3954150]
Z
20
1 + 4w [58347; 3954150]
Z
12
Z
2
Z
12
5 [88803;8773218]
54
Chapter 6
Growth in Case E(K)[2] =Z
2
Z
2
Let K be a noncyclotomic imaginary quadratic field of class number one and L be a quadratic
extension ofK: Given an elliptic curveE overK, in this section we present the auxiliary results to
studyE(L)
tor
whenE(K)
tor
contains full 2-torsion.
LetE=K be an elliptic curve withE[2] E(K). As known, the points of order 2 onE are
determined by the roots of the polynomial f(x) where E is defined by y
2
= f(x). Since E is
non-singular,f(x) has no repeated roots inK: ThenE has a model of the form
y
2
=x(x)(x)
where ;2 K: We denote the curve above by E(;). In studying the K-rational points of
E(;), we may make the further assumption that (;), the greatest common divisor of and,
is a square free inK becausey
2
=x(x +z
2
)(x +z
2
) is isomorphic toy
2
=x(x +)(x +)
by replacingy byy=z
3
andx byx=z
2
: Using the standard definitions, one can easily compute the
discriminant ofE(;) is = 16
2
2
()
2
and its j-invariant is
j =
256(
2
+
2
)
3
The next result is a variant of the result of Ono, see [32]).
55
Theorem 6.1 ([31]). LetK be a number field andE=K be an elliptic curve with full 2-torsion.
ThenE has a model of the formy
2
=x(x +)(x +) where;2O
K
:
1. E(K) has a point of order 4 iff; are both squares, or; are both squares, or
; are both squares inO
K
:
2. E(K) has a point of order 8 iff there exists a z2O
K
, z6= 0 and a pythagorean triple
(u;v;w) such that
=z
2
u
4
; =z
2
v
4
or we can replace; by; or; as in the first case.
3. E(K) contains a point of order 3 iff there exists a;b;z 2 O
K
with z 6= 0 and
a
b
= 2
f2;1;
1
2
; 0; 1g such that =a
3
(a + 2b)z
2
and =b
3
(b + 2a)z
2
:
In [6] Najman and Bruin proved that certain groups do not arise as torsion subgroups of elliptic
curves over any quartic extensions (not just over biquadratic extensions). Their result will be useful
in eliminating certain torsion groups.
Theorem 6.2 ([6], Theorem 7). The following groups do not occur as subgroups of elliptic curves
over quartic fields :
Z
3
Z
12
Z
3
Z
18
Z
3
Z
27
Z
3
Z
33
Z
3
Z
39
Z
4
Z
12
Z
4
Z
16
Z
4
Z
28
Z
4
Z
44
Z
4
Z
52
Z
4
Z
68
Z
8
Z
8
In order to describe our results, we introduce the following notation. Let us denote
K
(d) :=fZ
m
Z
n
:9 elliptic curveE=K :E(L)
tor
'Z
m
Z
n
for [L :K] =dg
and for a fixedG2
K
(1), we write
K
(d;G) :=fZ
m
Z
n
:9 elliptic curveE=K :E(K)
tor
'G;E(L)
tor
'Z
m
Z
n
; [L :K] =dg:
56
We are ready to state the main result of this chapter.
Theorem 6.3. GivenK =Q(
p
D) inS andE=K an elliptic curve with full 2-torsion. We may
assume thatE is in the formy
2
=x(x +)(x +) for some;2O
K
: Then,
1. IfG'Z
2
Z
12
whenD =19,43,67,163 then
K
(2;G) =fZ
2
Z
12
g:
2. IfG'Z
2
Z
10
whenD =2;11;19; then
K
(2;G) =fZ
2
Z
10
g:
3. IfG'Z
2
Z
8
when
(1) D6=7 then
K
(2;G) =fZ
2
Z
8
g:
(2) D =7 then
K
(2;G) =fZ
2
Z
8
;Z
4
Z
8
;Z
2
Z
16
g
4. IfG'Z
2
Z
6
; when
(1) D =2 then
K
(2;G) =fZ
2
Z
6
;Z
2
Z
12
g
(2) D6=2 then
K
(2;G) =fZ
2
Z
6
;Z
2
Z
12
;Z
6
Z
6
g
5. IfG'Z
2
Z
4
; then WLOG we write =s
2
and =t
2
: There is a quadratic extensionL
ofK withZ
4
Z
4
E(L)
tor
iffs
2
t
2
=z
2
for somez2K: In this caseL =K(i);
(1) D6=7 then
K
(2;G) =fZ
4
Z
4
g
(2) D =7 then
K
(2;G) =fZ
4
Z
4
;Z
4
Z
8
g
Otherwise,Z
4
Z
4
6E(L) and we have
(1) ifD =2;7;11 then
K
(2;G) =fZ
2
Z
4
;Z
2
Z
8
g
(2) IfD =19;43;67;163 then
K
(2;G) =fZ
2
Z
4
;Z
2
Z
8
;Z
2
Z
12
g
6. IfG'Z
2
Z
2
then
K
(2;G)fZ
2
Z
2n
:n = 1; 2; 3; 4; 5; 6; 8g[fZ
4
Z
4
g depending
onK:
57
6.1 Classification of Twists
One benefit of Theorem 6.1 is that any elliptic curveE=K with full 2-torsion has a modelE(;) :
y
2
= x(x +)(x +) where;2O
K
. We may also assume that gcd(;) is square-free in
O
K
: Note that thed-quadratic twistE
d
ofE is given byE
d
: y
2
= x(x +d)(x +d) whered
is a non-square inK. In particular,E
d
(K) contains full 2-torsion. IfL =K(
p
d) andE(K)
tor
is
given, thenE
d
(K)
tor
lets us bound the 2-part ofE(L)
tor
; (Proposition 3.1). The primary goal is to
classify which torsion groups occur asE
d
(K)
tor
.
Before proceeding further, we need a technical lemma to rule out 3-torsion on quadratic twists
of certain set of elliptic curves.
Lemma 7. Let K = Q(
p
D)2T and let a;b;a
0
;b
0
;c
0
;d2O
K
with d nonsquare in K and
a
b
;
a
0
b
0
= 2f2;1;
1
2
; 0; 1g Then the following system of equations have no solutions inO
K
8
>
>
<
>
>
:
da
3
(a + 2b) =a
3
0
(a
0
+ 2b
0
)c
2
0
db
3
(b + 2a) =b
3
0
(b
0
+ 2a
0
)c
2
0
(6.1)
whenD =1;2;3. Otherwise the system has infinitely many solutions.
Proof. Suppose that there exists a solution to the equations above which satisfies the hypothesis.
Note thata + 2b;a
0
+ 2b
0
6= 0; so dividing the equations yields
a
b
3
a + 2b
b + 2a
=
a
0
b
0
3
a
0
+ 2b
0
b
0
+ 2a
0
Substitutingx :=
a
0
b
0
andy :=
a
b
yields
y
3
y + 2
2y + 1
=x
3
x + 2
2x + 1
2xy
4
+y
4
+ 2xy
3
+ 2y
3
2x
4
yx
4
4yx
3
2x
3
= 0
58
The polynomial on the left hand side can be split into irredicuble polynomials as follows:
(xy)(2x
3
y + 2xy
3
+x
3
+y
3
+ 5x
2
y + 5xy
2
+ 2x
2
+ 2y
2
+ 2x
2
y
2
+ 2xy) (6.2)
Ifx =y then there is ak2K such thata =kb anda
0
=kb
0
. Plugging this into (5) implies thatd
is a square which is a contradiction. Hencex6=y andx;y = 2f2;1;1=2; 0; 1g: Then, (x;y) is
aK-rational point on the curveC defined byp
C
(x;y) = 0 where
p
C
(x;y) = 2x
3
y + 2xy
3
+x
3
+y
3
+ 5x
2
y + 5xy
2
+ 2x
2
+ 2y
2
+ 2x
2
y
2
+ 2xy
that is the second factor in (6.2). Let C denote the projective closure of C. Note that C(K)
differs fromC(K) only at finitely many points belonging to the setE =f[0; 1; 0]; [1; 0; 1]; [
p
3
1; 2; 1]; [
p
3 1; 2; 1]g: Thus, showing the existence of aK-rational point onC is equivalent
to showing the existence of aK-rational point (x;y)62E onC: Magma tells us thatC has genus 1
and is birationally equivalent (overQ) to the elliptic curveE
C
defined by the equation
y
2
+ 2xy + 2y =x
3
x
2
2x
with the map given by
:C!E
C
(x;y;z)7! (p
1
;p
2
;p
3
)
where
p
1
= 2x
2
y
2
+ 3x
2
yz + 4xy
2
zy
3
z +x
2
z
2
+x
2
z
2
+ 6xyz
2
y
2
z
2
+ 2xz
3
p
2
= 2x
2
y
2
+ 4xy
3
x
2
yz + 10xy
2
z + 3y
3
zx
2
z
2
+ 7y
2
z
2
2xz
3
+ 2yz
3
p
3
=y
4
+ 3y
3
z + 3y
2
z
2
+yz
3
59
Since theK-rational points onC map to theK-rational points onE
C
, the only potentialK-rational
points ofC belong to the union of
1
(E
C
(K)) and the set of points where the map is not defined.
We call these points as non-regular points of: SoC(K) andE
C
(K) differ at most finitely many
points. Because [0; 0; 0] cannot be in the image of , a non-regular point [x;y;z] of has the
property that p
i
(x;y;z) = 0 for all 1 i 3: Let us consider the condition p
3
(x;y;z) = 0:
If z = 0 then y = 0: If z = 1 then y = 0 or y =1: Thus a non-regular point can be written
as [x; 0; 0]; [x; 0; 1] and [x;1; 1] for some x2 K: Using the second condition p
2
(x;y;z) = 0
narrows down the set of non-regular points tof[1; 0; 0]; [0; 0; 1]; [2; 0; 1]; [1;1; 1]g: One can
quickly observe that [1; 0; 0] does not correspond to a point on the curve C and the other points
fail to recover points (x;y) onC withx;y = 2f2;1;
1
2
; 0; 1g: Thus the only non-regular points
of are [0; 0; 1]; [1;1; 1], and these are also all the singular points ofC. In order to compute
1
(E
C
(K)) we first must compute E
C
(K): By Sage, E
C
(K) has rank 0 for D =1;2;3:
See Table 6.1 below for a precise description ofE
C
(K) for allK of interest.
Table 6.1: K-rational points onE
C
K =Q(w) E
C
(K) Generators ofE
C
(K)
Q(
p
1) Z=6Z [2,-6,1]
Q(
p
2) Z=6Z [2,-6,1]
Q(
p
3) Z=6ZZ=2Z [2,-6,1], [1/2(-w+1),1/2(w-3),1]
Q(
p
7) Z=6ZZ [2,-6,1],[-2,-w+1,1]
Q(
p
11) Z=6ZZ [2,-6,1],[1/25w-17/25,-9/125w-147/125,1]
Q(
p
19) Z=6ZZ [2,-6,1],[-25/9,28/27w+16/9,1]
Q(
p
43) Z=6ZZ [2,-6,1],[-15289/13689,
153160/1601613w+1600/13689,1]
Q(
p
67) Z=6ZZ [2,-6,1],[-2398489/2350089,110531740/3602686437w+
48400/2350089, 1]
Q(
p
163) Z=6ZZ [2,-6,1],[-4220790466489/4220078644089,15276006029
948680/8669235817215083187w+711822400/
4220078644089,1]
Therefore, E
C
(K) has infinitely many points for D =7;11;19;43;67;163 and
since E
C
is birational with C, so does C(K): Equivalently the system (6.1) has infinitely many
solutions inO
K
, as claimed.
60
To compute
1
([x;y;z]), we use the birational inverse of given by
:E
C
!C
(x;y;z)7! (p
1
;p
2
;p
3
)
where
p
1
=x
4
+ 2x
3
y 4xy
2
z 2y
3
z 3x
2
z
2
y
2
z
2
2xz
3
+ 2yz
3
p
2
=x
4
8x
3
z 4x
2
yz 21x
2
z
2
14xyz
2
3y
2
z
2
22xz
3
10yz
3
8z
4
p
3
=x
4
+ 6x
3
z + 2x
2
yz + 9x
2
z
2
2xyz
2
3y
2
z
2
+ 4xz
3
4yz
3
and its non-regular points aref[0; 1; 0]; [0; 0; 1]; [1; 0; 1]; [2;6; 1]g: We observe that each point of
1
(E
C
(K)
tor
) either doesn’t correspond to a point onC, or fails to satisfyx;y = 2f2;1;
1
2
; 0; 1g:
SinceE(K) is of rank zero forD =1;2;3; we have shown thatC(K) forD =1;2;3
has no points (x;y) withx;y62f2;1;
1
2
; 0; 1g. See Table 6.2.
Table 6.2: Determination of
1
(E
C
(K)
tor
)
P2E
C
(K)
1
(P )
[0,1,0] [1,-1,1]
[-1,0,1] [-1,1,1]
[0,-2,1] [2; 0; 1]
[0,0,1] [0,1,0]
[2,-6,1] ;
[2,0,1] [0,-2,1]
[-1-
p
3,-3+
p
3,1] [
p
3 1; 2; 0] (= 2C(K))
[-1+
p
3,-3-
p
3,1] [
p
3 1; 2; 0] (= 2C(K))
[-1-
p
3,3+
p
3,1] [0,0,-1]
[-1+
p
3,3-
p
3,1] [0; 0;1]
[
1
2
(1-
p
3),
1
2
(-3+
p
3),1] [1;1; 1]
[
1
2
(1+
p
3),
1
2
(-3-
p
3),1] [1;1; 1]
[-2,1-
p
7; 1] [
1
8
(3
p
7 1);
1
4
(
p
7 + 3); 1]
[
1
25
p
11
17
25
;
9
125
p
11
147
125
; 1] [
1
24
(
p
11 43);
1
24
(19
p
11); 1]
[
25
9
;
28
27
p
19 +
16
9
; 1] [(
1
21457
(4536
p
19+8335);
1
12475
(2688
p
19+4283); 1)]
61
Similarly, one can explicitly compute the preimage point of the generator of free rank of the
Mordell-Weil groupE
C
(K) forD =43;67;163.
We can reverse the proof of Lemma 7 to produce examples of elliptic curves over certain fields.
Since the coefficients become overwhelmingly long for the remainingK’s, we didn’t include them
in Table 6.6.
Theorem 6.4. LetK =Q(
p
D)2S andE=K be an elliptic curve withE(K)[2] =Z
2
Z
2
: We
may assume thatE is in the formy
2
=x(x +)(x +) for some;2O
K
: Then,
1. IfG'Z
2
Z
12
thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
4
g:
2. IfG'Z
2
Z
10
thenT
K
(G) =fZ
2
Z
2
g:
3. IfG'Z
2
Z
8
when
(1) D6=7 thenT
K
(G) =fZ
2
Z
2
g:
(2) D =7 thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
4
;Z
2
Z
8
g:
4. IfG'Z
2
Z
6
when
(1) D =2 thenT
K
(G) =fZ
2
Z
2
g:
(2) D =7;11 thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
6
g:
(3) D =19;43;67;163 thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
4
;Z
2
Z
6
g:
5. IfG'Z
2
Z
4
when
(1) D =2;11 thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
4
g:
(2) D =7 thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
4
;Z
2
Z
8
g:
(3) D =19;43;67;163 thenT
K
(G) =fZ
2
Z
2
;Z
2
Z
4
;Z
2
Z
6
g:
6. IfG'Z
2
Z
2
thenT
K
(G)fZ
2
Z
2n
:n = 1; 2; 3; 4; 5; 6g depending onK:
62
Proof. Assume the hypothesis. Then, for any non-squared2 K; the possible torsion structures
appearing asE
d
(K)
tor
areZ
2
Z
2n
where 1n 6:
(1) Suppose now E(K)
tor
' Z
2
Z
12
: Note that this torsion structure is realized when
D =19,43,67,163: Now,Z
3
*E
d
(K)
tor
follows from Theorem 6.2. Also,Z
5
*E
d
(K)
tor
;
otherwise together with a K-rational 6-torsion point on E; there would be a K-rational cyclic
30-isogeny pointwise defined overK(
p
d) followed by Proposition 4.1. But this is not possible by
Theorem 4.1. Finally, ifZ
8
E
d
(K)
tor
thenE has two independent cyclic isogenies of order 2
and 24 (overK). From [29, Lemma 7],E is isogenous (overK) to an elliptic curveE
0
=K with a
cyclic 48-isogeny, contradicting Theorem 4.1.
(2) SupposeE(K)
tor
'Z
2
Z
10
: Note thatZ
2
Z
12
can only occur asE
d
(K)
tor
whenD =19.
We will now eliminate the casesn = 2; 3; 4; 5.
Casen = 3: AssumeE
d
(K)
tor
has a point of order 3. Since by assumptionE(K)
tor
has a point
of order 10; then by Proposition 4.1, there exists a K-rational cyclic 30-isogeny of E pointwise
defined overK(
p
d): However, this cannot happen by Theorem 4.1.
Casen2f2; 4g: In both cases,E
d
(K)
tor
contains at least a point of order 4. SinceE(K)
tor
=
Z
2
Z
10
,E has two independent cyclic isogenies of order 2 and 20 (overK). From [29, Lemma
7],E is isogenous (overK) to an elliptic curveE
0
=K with a cyclic 40-isogeny. But Theorem 4.1
asserts the non-existence of such an isogeny.
Casen = 5: Assume thatE
d
(K)
tor
has a point of order 5, then together with a point of order 5 in
E(K) and Lemma 1, the full 5-torsionZ
5
Z
5
lies inE(K(
p
d))
tor
; which impliesK(
p
d) =Q(
5
):
However, every intermediate field ofQ(
5
) is necessarily real since Gal(Q(
5
)=Q) =Z
4
, which is
a contradiction.
63
(3) Suppose E(K)
tor
' Z
2
Z
8
. Then WLOG we may assume there exists a Pythagoeran
triple (x;y;z)2O
K
withx
2
+y
2
=z
2
, such that =x
4
and =y
4
. IfE
d
(K)
tor
has a point of
order 4, then by Theorem 6.1 one of the followings is satisfied:
(i) dx
4
;dy
4
are both squares
(ii)dx
4
;dy
4
dx
4
are both squares
(iii)dy
4
;dx
4
dy
4
are both squares.
In case (i) it implies thatd is a square, a contradiction. In case (ii) or (iii), it follows thatd =u
2
for someu2O
K
: Sincedy
4
dx
4
=w
2
for somew2O
K
; then dividingd yields
y
4
x
4
=(wu
1
)
2
and multiplying by1 if necessary, we obtain x
4
y
4
= v
2
for some v 2 O
K
: Note that
x
4
;y
4
;x
4
y
4
6= 0 sinceE is smooth, sox;y;v6= 0. By [31, Proposition 11], there is no nontrivial
solution tox
4
y
4
=v
2
unlessK =Q(
p
7). Therefore,E
d
(K) does not have a point of order 4.
IfE
d
(K)
tor
has a point of order 3, then asE(K(
p
d))
tor
containsE(K)
tor
; we obtain a torsion
subgroupZ
2
Z
24
ofE(K(
p
d)): Furthermore,Z
2
andZ
24
form two independent cyclic isogenies
ofE which in turn yields a cyclic 48-isogeny of an elliptic curveE
0
=K by Lemma 7, [29]. But this
contradicts Theorem 4.1.
IfE
d
(K) has a point of order 5 which is possible ifK =Q(
p
D) forD =2;11;19; then
along with a point of order 8 inE(K)
tor
, this produces a cyclic isogenyZ
40
E(K(
p
d))
tor
which
cannot happen by Theorem 4.1. See Table 6.9 for the examples whenK =Q(
p
7):
(4) SupposeE(K)
tor
'Z
2
Z
6
. Then we may assume there existsa;b;c2O
K
withc6= 0 and
a
b
62f2;
1
2
;1; 0; 1g such that =a
3
(a + 2b)c
2
and =b
3
(b + 2a)c
2
. Furthermore, choosing
E
c
2
instead ofE we may assume that =a
3
(a+2b) and =b
3
(b+2a): By way of contradiction,
supposeZ
4
E
d
(K)
tor
then by Theorem 6.1,
64
(i) d;d are both squares
(ii)d;dd are both squares
(iii)d;dd are both squares.
WLOG we may assume case (i) holds, i.e. d = s
2
;d = t
2
for somes;t2O
K
. Note that
s;t6= 0;s
2
6=t
2
: Then
s
2
=d =da
3
(a + 2b) and t
2
=d =db
3
(b + 2a)
and multiplying by d:
ds
2
=a
3
(a + 2b)d
2
and dt
2
=b
3
(b + 2a)d
2
Taking the products and dividing both sides bya
2
b
6
along with substitutingz :=
a
b
yield
dst
a
2
b
3
2
=z(z + 2)(2z + 1)
Multiplying both sides by 4 gives
2dst
ab
3
2
= 2z(2z + 4)(2z + 1): Thus we obtain aK-rational point
on the elliptic curveE
0
defined by
y
2
=x(x + 4)(x + 1) =x
3
+ 5x
2
+ 4x
In Magma, we computeE
0
(K)'Z
2
Z
4
whenD =2;7;11: More explicitlyE
0
(K)
tor
=
f1; (0; 0); (4; 0); (1; 0); (2;2); (2;6)g. This shows that 2z =4;2;1; 0 or 2 and so
z2f2;1;
1
2
; 0; 1g which is a contradiction by the choice ofz:E
0
(K)'Z
2
Z
4
Z when
D =19;43;67;163: For example, the extra points in caseD =19 give rise to growth of
torsion, see Remark 4.
Next, we claim thatZ
5
* E
d
(K)
tor
; otherwise asZ
6
E(K), there would be a cyclic 30-
isogeny pointwise rational over the extensionK(
p
d); which cannot happen by Theorem 4.1. In
addition, ifZ
12
E
d
(K)
tor
thenZ
3
Z
12
E(K(
p
d)) which is excluded by Theorem 6.2.
65
ForE
d
(K)
tor
to have a 3-torsion point, there must exista
0
;b
0
;c
0
2O
K
and
a
0
b
0
= 2f2;1;1=2; 0; 1g
such that
d =a
3
0
(a
0
+ 2b
0
)c
2
0
; d =b
3
0
(b
0
+ 2a
0
)c
2
0
Note thatE andE
d
are isomorphic overK(
p
d) by the mapf : (x;y)7! (dx;d
3
2
y): Since an
order 3 pointP ofE(K)
tor
corresponds to an order 3 pointf(P ) ofE
d
(K(
p
d))
tor
(which is not
in K) and a 3-torsion point Q of E
d
(K)
tor
by the assumption corresponds to a point f
1
(Q) of
the same order inE(K(
p
d))
tor
, we haveZ
3
Z
3
E(K(
p
d))
tor
: ThereforeK(
p
d) must have a
primitive 3th root of unity, which implies thatd =3. Thus, we have
3a
3
(a + 2b) =a
3
0
(a
0
+ 2b
0
)c
2
0
; 3b
3
(b + 2a) =b
3
0
(b
0
+ 2a
0
)c
2
0
Since3 is a nonsquare inK, by Lemma 7 this system has solutions exceptQ(
p
2) which
establishes the assertion.
(5) SupposeE(K)
tor
'Z
2
Z
4
. ForE
d
(K)
tor
to have a point of order 4, either (a)d;d are
both squares, (b)d;dd are both squares, or (c)d;dd are both squares. Sinced
is a nonsquare, (a) is impossible. Both (b) and (c) can be argued similarly so we assume (b) holds.
Then there existsu;v2O
K
such thatds
2
=u
2
anddt
2
ds
2
=v
2
which impliesd =w
2
for
somew2O
K
. Plugging this into the second equation we obtains
2
t
2
=z
2
for somez2O
K
:
Note that1 is not a square inO
K
, otherwised is a square. As a result, if those conditions are
satisfied, thenE
d
(K)
tor
contains a point of order 4. (See Table 6.5). Ifd =1; by Lemma 17,[31],
we haveZ
4
Z
4
E(K(
p
1)): Then the possible groups forE
1
(K)
tor
areZ
2
Z
2
;Z
2
Z
4
,
Z
2
Z
8
; orZ
2
Z
12
asK varies inS. Arguing as in (3), it follows thatE
1
has no point of order 8
unlessK =Q(
p
7) One can takeE asE
1
(729; 2304) (see Table 6.9). Also, ifZ
12
E
d
(K)
tor
;
then in particularE
1
(K)
tor
has a point of order 3. SinceZ
4
Z
4
E(K(
p
d)); we deduce that
Z
4
Z
12
is a subgroup ofE(K(
p
1))
tor
which is impossible by Theorem 6.2.
66
Now, assume that the conditions in the preceding argument are not fulfilled. This leaves alone
the nontrivial cases where E
d
(K)
tor
' Z
2
Z
6
orZ
2
Z
10
: As replacing E by E
d
in (4), the
former group structure can only be realized whenD =19; 43;67; 163: For example, we may
take E(
3
;
3
) in Table 6.8. The later group is ruled out by the case n = 4 in the proof of (2)
above.
Remark 4. The extra points ofE
0
in the proof above produces explicit examples of elliptic curves.
In Table 6.8, for K = Q(
p
19); there is an elliptic curve E=K with E(K) = Z
2
Z
6
and
E
3
(K)'Z
2
Z
4
so thatE(K(
p
3))'Z
2
Z
12
:
6.2 Growth of Torsion
GivenK2S; letE : y
2
= x(x +)(x +) where and are inO
K
and (;) is square-free
inO
K
: Then for all quadratic extensionsL = K(
p
d), whered2 K is a nonsquare, the torsion
subgroupE(L)
tor
can be classified into the following series of the results.
Proposition 6.1. Let K = Q(
p
D) where D =2;11;19: If E=K is an elliptic curve with
E(K)
tor
'Z
2
Z
10
; thenE(L)
tor
'Z
2
Z
10
for any quadratic extensionL ofK:
Proof. Suppose E(K)
tor
' Z
2
Z
10
: If the torsion subgroup doesn’t grow in any quadratic
extension ofK, we are done. Let us assume there exists a quadratic extensionL ofK in which the
torsion subgroup grows. By Proposition 3.1 and Theorem 6.4, the possible group structures that
can occur asE(L)
tor
areZ
2
Z
20
orZ
4
Z
20
: Since the latter implies the former, it is enough
to prove that E(L)
tor
does not contain Z
2
Z
20
as a subgroup. By way of contradiction, let
Z
2
Z
20
E(L)
tor
: We may further assume thatL = K(
p
d) whered2 K is a nonsquare. As
known by [35, Case 4.6], the solutionss;t2K of the equation
X
1
(2; 10) :s
2
=t
3
+t
2
t
67
satisfyingt(t
2
1)(t
2
4t 1)(t
2
+t 1)6= 0 induce elliptic curves with torsionZ
2
Z
10
over
K. Magma computes thatX
1
(2; 10)(K) has rank 1 with torsionZ
6
; all torsion points satisfying
t = 0;1 or 1: (see [35] Lemma 4.9) The parametrization of induced elliptic curves with the
prescribed torsion is
E
(2;10)
P
:y
2
+ (t
3
+t
2
4t + 1)xyt(t 1)(t + 1)
3
(t
2
4t 1)y =x
3
t(t 1)(t + 1)
3
x
2
whereP = (t;s)2X
1
(2; 10) satisfying
E
(2;10) =t
5
(t
2
1)
10
(t
2
4t1)
2
(t
2
+t1)
2
: The points
P
0
= (0; 0) andQ = ((2t+s1)(s+1)t; (s+1)
2
(s+1)(t
2
2t2s
2
)) are inE(K) of order 10
and 2 respectively andhQ;P
0
i =Z
2
Z
10
: We may assume thatE isE
(2;10)
P
0
for some non-cuspidal
pointP
0
2X
1
(2; 10)(K): The torsion ofE(K) will grow fromZ
2
Z
10
toZ
2
Z
20
in a quadratic
extension ofK if and only if a solutionR
0
to 2R
0
=P
0
satisfies that [K(R
0
) :K] = 2: There are
four solutions which are conjugate to each other. So they are defined in at most two number fields.
The argument we follow is independent of the choice ofR
0
: Using the formula for the double of a
point, we obtain a necessary condition that thex-coordinate ofR
0
= (x
0
;y
0
) (withy
0
6= 0) must
be a root the polynomialq(x) =x
2
3
where
2
and
3
are the division polynomials ofE of
degree 2 and 3 respectively. Computing in Magma yields thatq(x) is in the form of
q(x) =x
4
+f(t)x
2
+g(t)x +h(t)
where
f(t) =t
10
t
9
15t
8
4t
7
+ 37t
6
+ 20t
5
25t
4
16t
3
+ 2t
2
+t
g(t) =2t
14
+8t
13
+28t
12
56t
11
222t
10
112t
9
+264t
8
+336t
7
+34t
6
152t
5
100t
4
24t
3
2t
2
h(t) =t
19
2t
18
22t
17
2t
16
+174t
15
+310t
14
62t
13
682t
12
576t
11
+250t
10
+670t
9
+314t
8
110t
7
174t
6
74t
5
14t
4
t
3
68
Adjoining a root of q(x) to K gives rise to an extension of K of degree at most 4. Hence,
K(R
0
) is a quadratic extension ofK ifq(x) is decomposed into quadratic factors inK(t)[x] as
x
4
+f(t)x
2
+g(t)x +h(t) = (x
2
+a(t)x +b(t))(x
2
a(t)x +d(t))
for somea(t);b(t);d(t)2K(t): Equating the corresponding coefficients of both sides yields
f(t) =b(t) +d(t)a
2
(t); g(t) =a(t)(d(t)b(t)); and h(t) =b(t)d(t)
By the comparison of the degrees, if such a triple (a(t);b(t);d(t)) exists then either
(deg(a(t));deg(b(t));deg(h(t))) = (4; 9; 10) or (4; 10; 9)
On the other hand, sinceb(t);d(t) are the factors ofh(t) anda
2
(t) =b(t)+d(t)f(t), it is enough
to check whetherb(t) +d(t)f(t) is a square or not for finitely many pairs of (b(t);d(t)): WLOG,
we may assumeb(t);d(t) are polynomials of degree 10 and 9 respectively. After enumerating all the
possibilities for (b(t);d(t)) in Magma, we observe that there is always one odd exponent appearing
in the decomposition ofb(t) +d(t)f(t): Therefore, there is no such polynomiala(t)2 K[t],
which is a contradiction.
Remark 5. In particular, we establish the proof above with the observation that the minimum
extension over K in which there exists a point of order 20 has degree 4. For example, let
K =Q(
p
2) andE an elliptic curve defined byy
2
+ 5xy + 66y =x
3
+ 6x
2
: We have
E(Q)
tor
'Z
10
; E(K)
tor
'Z
2
Z
10
andE(L)
tor
'Z
2
Z
20
where [L :K] = 4 generated byg(x) =x
4
330x
2
8712x26136: Note that the normal closure
b
L ofL overQ is Galois with the group isomorphic toD
4
:
69
Proposition 6.2. Let K = Q(
p
7) and let L be a quadratic extension of K: Assume E is an
elliptic curve defined overK: IfE(K)
tor
'Z
2
Z
8
thenE(L)
tor
can be isomorphic to
Z
2
Z
8
; Z
4
Z
8
; or Z
2
Z
16
:
Proof. SupposeE(K)'Z
2
Z
8
. From Theorem 6.4, we know thatE
d
(K)
tor
'Z
2
Z
2n
for
n = 1; 2 or 4: By Proposition 3.1 the possible torsion structures forE(L)
tor
are in the form of
Z
m
Z
n
; wherem = 2; 4 andn = 8; 16; 32; 64: (6.3)
Note that the occurrence ofZ
2
Z
8
comes for free sinceE(K)
tor
E(L)
tor
: As shown in Tables
6.6,6.4, the torsion structures Z
4
Z
8
and Z
2
Z
16
can be realized as E(L)
tor
: Theorem 6.2
particularly asserts thatZ
4
Z
16
cannot occur which further rules outZ
4
Z
32
andZ
4
Z
64
from
(6.3). In order to argue the nonexistence of the groupsZ
2
Z
32
andZ
2
Z
64
; we assume the
contrary and letL be a quadratic extension ofK withZ
2
Z
32
E(L)
tor
: We choose a pointQ of
order 2 and a pointP of order 32 such thatZ
2
Z
32
=hQ;PiE(L)
tor
: ThenE(K)
tor
=hQ; 4Pi:
The nontrivial element of Gal(L=K) fixesQ and sendsP to a point of order 32, which must be
of the formaP orQ +aP wherea is odd. Applying to itself once again we see that
P =
2
(P ) =a
2
P
which impliesa
2
1(mod 32) soa can only be 1, 15, 17 or 31. For all eight possibilities of(P );
maps 2P to either 2P or2P: If (2P ) = 2P then the point 2P which has order 16 would
be K-rational, contradicting the assumption. And (2P ) =2P also cannot be, otherwise the
K-rational point 4P would also be mapped to its inverse which gives a contradiction since fixes
4P:
Remark 6. In fact the argument ruling out the caseZ
2
Z
32
can be made independent of the
choice of the quadratic fieldK:
70
Proposition 6.3. Let K be a quadratic field inSnfQ(
p
7)g and E=K an elliptic curve. If
E(K)
tor
'Z
2
Z
8
: ThenE(L)
tor
'Z
2
Z
8
for any quadratic extensionL ofK:
Proof. We may assume thatE :y
2
=x(x +u
4
)(x +v
4
); where (u;v;w)2O
K
withu
2
+v
2
=w
2
and (u;v) is a square-free integer inO
K
: By Proposition 3.1 and Theorem 6.4, it follows that
E(L)
tor
is isomorphic toZ
2
Z
8
;Z
4
Z
8
;Z
2
Z
16
orZ
4
Z
16
: Let us assumeZ
16
E(L)
tor
:
Using the group law, one can obtain a pointP of order 8 inE(K) such that
P = (x;y) = (uv(u +w)(v +w);uvw(u +v)(v +w)(u +w))
with [4]P = (0; 0) and [8]P = 0; due to [25]. ThenE(K)
tor
is generated byQ = (u
4
; 0) of order
2 andP of order 8. If we defineP
0
= (x
0
;y
0
) =Q +P; we get another point of order 8 such that
P
0
= (x
0
;y
0
) = (uv(u +w)(vw);uvw(uv)(vw)(u +w)):
SinceE(K)
tor
has only two distinct cyclic subgroups of order 8, they are generated byP andP
0
:
By the assumption, if there is a pointP
0
of order 16 inE(L)
tor
then [2]P
0
2E(K)
tor
has order 8.
Therefore, either [2]P
0
=P or [2]P
0
=P
0
holds. Suppose [2]P
0
=P: By Lemma 6, this is possible
only ifx;x +u
4
orx +v
4
are squares inL: Sincex;u;v are defined inK and satisfy the curve
equationy
2
=x(x +u
4
)(x +v
4
); at least one of them must be a square inK: Ifx is a square then
we haveu;v;w;s2K such that
s
2
=x =uv(u +w)(v +w); u
2
+v
2
=w
2
(6.4)
AsC given byu
2
+v
2
=w
2
is of genus 0, the solution set is parametrized such thatfP2C(K) :
P = (2t;t
2
1;t
2
+ 1) for some t2 Kg: Note that t6= 0;1;i since our curve is smooth.
Substituting the parametrization in (6.4) yields the curve given by
C
0
:s
2
= (2t)(t
2
1)(2t +t
2
+ 1)(t
2
1 +t
2
+ 1) = 4t
3
(t + 1)
2
(t
2
1)
71
which is birational to the elliptic curve E
0
:= y
2
= x
3
x via the map (t;s)7!
t;
s
2t
2
+2t
We compute that E
0
(K) ' Z
2
Z
2
for all K 2 SnfQ(
p
7)g: More precisely E
0
(K) =
f1; (0; 0); (1; 0)g. Hence all rational points satify t = 0;1; a contradiction. If x +u
4
is a
square then there existsu;v;w;s2K such that
s
2
=x +u
4
=uw(u +v)(v +w); u
2
+v
2
=w
2
(6.5)
Arguing as above, we obtain the following curve
C
00
:s
2
= (2t)(t
2
+ 1)(2t +t
2
1)(t
2
1 +t
2
+ 1) = 4t
3
(t
2
+ 1)(t
2
+ 2t 1):
C
00
is a hyperelliptic curve of genus 2. LetJ be the Jacabian ofC
0
: In fact, for allK2S, Magma
tells us that the rank ofJ(K) is 0 and
J(K)
2-part
'Z
2
Z
2
(6.6)
J(Q)'Z
2
Z
10
Our objective is to prove thatJ(K) =J(Q): It is equivalent to show thatJ(K) has exactly of the
size 20 for any fieldK of interest.
CaseK =Q(
p
2):
Since the discriminant ofK is -8, one can quickly observe that 3 splits and 5 remains prime inO
K
:
Let p denote a prime above 3 inO
K
: The curveC
00
remains smooth mod p and 5. The reduction
of J at the good primes p and 5 has 20 and 640 points overF
3
andF
25
respectively. Since the
3-part ofJ(K)
tor
injects into the group of size 640 (modp), it is a trivial subgroup. On other hand,
the prime-to-3-part is injected into the group of size 20 (mod 3), it has size at most 20. Hence
jJ(K)
tor
j = 20 as desired.
72
CaseK =Q(
p
11):
As the discriminant of K is11; it follows that 3,5 both split inO
K
: Letp andq be primes above
3 and 5 inO
K
respectively. By Magma,J(F
3
)'Z
2
Z
10
; andJ(F
5
)' (Z
2
)
2
Z
10
: Arguing as
above, it follows thatjJ(K)
tor
j = 20:
CaseK =Q(
p
19):
Note that the discriminant ofK is -19. This implies that 3 remains prime and 5 splits inO
K
: We
computeJ(F
9
)' (Z
2
)
3
Z
10
; andJ(F
5
)' (Z
2
)
2
Z
10
: Since the prime-to-5-part ofJ(K)
tor
injects intoJ(F
5
); its size divides 2
3
:5: By (6.6) we know thatJ(K)
2-part
has size of 2
2
. Hence the
prime-to-5-part divides 20. On the other hand, under reduction mod p (where p is a prime lying
above 5) inO
K
; the 5-part ofJ(K)
tor
injects intoJ(F
9
) and so its size divides 2
4
:5: However, the
2-part is of size 2
2
. Therefore the 5-part also divides 2
2
:5: It follows thatjJ(K)
tor
j = 20:
CaseK =Q(
p
d) whered =43;67;163.
One can easily observe that 3; 5 remain prime inO
K
asK varies. We computeJ(F
9
)' (Z
2
)
3
Z
10
;
andJ(F
25
)' (Z
2
)
2
Z
4
Z
40
: The reduction map mod 5 is injective on the prime-to-5-part of
J(K)
tor
and the prime divisor ofJ(F
25
) that is coprime to 5 is 2. Similarly, the 5-part ofJ(K)
tor
is
a subgroup ofJ(F
9
) = 2
4
:5: SinceJ(K)
2-part
is of size 2
2
, as argued in the previous case, we see
both the 5-part and prime-to-5-part have size of at most 20. Therefore the claimjJ(K)
tor
j = 20
follows.
Magma represents the hyperelliptic curveC
00
in a weighted homogenous space, yielding two
points at infinity1
+
= (1; 1; 0) and1
= (1;1; 0): Now let P be a point in C
00
(K): Then
[P1
] represents a point inJ(K) which equals toJ(Q): Mumford representation that Magma
uses for the coset representative of P1
is P +1
+
1
1
+
: We enumerated the
20 coset representatives in Magma and found that the points which are in the aforementioned
form are corresponding toP2f(1; 1; 0); (0; 1; 1); (1; 0; 0); (0;1; 1)g: However, any such point
corresponds tou;v;w which contradict the hypothesis that our curve is smooth.
73
Ifx +v
4
is a square, then it yields to the same system as in (6.5) but withu andv interchanged.
Hence, using the parametrization (u;v;w) = (t
2
1; 2t;t
2
+ 1); we obtain the same hyperelliptic
curve above. Thus we have proved that [2]P
0
= P cannot happen. When [2]P
0
0
= P
0
= (x
0
0
;y
0
0
);
usingx
0
0
=uv(u +w)(vw) andu
2
+v
2
=w
2
, we obtain
x
0
0
+u
4
=uw(uv)(wv); x
0
0
+v
4
=vw(vu)(w +u)
Put differently, P
0
is related toP by substitutingv forv: Hence the rest of the argument is
exactly same as in the case [2]P
0
=P: ThereforeE(L)
tor
6'Z
2
Z
16
: The remaining possibilities
are whereE(L)'Z
4
Z
8
orZ
4
Z
16
: In either case, we particularly haveZ
4
Z
4
E(L): But
this cannot happen by Lemma 16, [31].
Prior to further processing , we prove a technical lemma.
Lemma 8. Let2K2S: If
p
(or
p
) is a square in a quadratic extensionL ofK then
or is a square inK:
Proof. Suppose [L :K] = 2 such that either
p
=z
2
(
p
=z
2
) for somez2 L: Thenz is a
root of the polynomialf(x) :=x
4
defined overK: Iff(x) is irreducible overK; then if we let
K
f
be the splitting field off; it follows that [K
f
:K] = 4: Sincef has least one root inL andL=K
is Galois,f(x) must split overL which in fact means thatL =K
f
and soL has degree 4 overK
which is contradicting the assumption. If not, it is either a product of two quadratic polynomials or
product of two linear and one quadratic polynomial since 3 does not divide the order of Gal(L=K):
We will argue both cases together. Let us writef as a product of two polynomials:
f(x) :=x
4
= (x
2
+ax +b)(x
2
+cx +d)
for a;b;c;d2 K: Then ac +b +d = 0; ad +bc = 0 and a +c = 0: Replacing a byc gives
c(bd) = 0: So eitherc = 0 orb =d: Ifc = 0 thenb =d and sobd = which implies that
=b
2
inK: Ifb =d; then =bd; so is a square inK:
74
Proposition 6.4. LetK =Q(
p
D)2S andL is a quadratic extension ofK: IfE=K is an elliptic
curve withE(K)
tor
'Z
2
Z
6
; then
(i) IfD =2 thenE(L)
tor
'Z
2
Z
6
orZ
2
Z
12
(ii) IfD =7;11 thenE(L)
tor
'Z
2
Z
6
;Z
2
Z
12
; orZ
6
Z
6
(iii) IfD =19;43;67;163 thenE(L)
tor
'Z
2
Z
6
;Z
2
Z
12
; orZ
6
Z
6
:
Proof. SupposeE(K)
tor
'Z
2
Z
6
and letL =K(
p
d) for a non-squared2K: By Theorem 6.4
and Proposition 3.1, we can list:
Case (1): The group structures which can appear asE(L)
tor
are
Z
2
Z
6
; Z
2
Z
12
; Z
4
Z
12
(6.7)
Case (2): The ones from (6.7) or the following groups
Z
2
Z
18
;Z
2
Z
36
; Z
4
Z
18
; Z
4
Z
36
; (6.8)
Z
6
Z
6
; Z
6
Z
12
:
Case (3): Either the groups listed in (6.7),(6.8) orZ
2
Z
24
orZ
4
Z
24
:
If the torsion subgroup ofE(K) does not grow, then we haveE(L)
tor
'Z
2
Z
6
: Also, Table
6.8 shows an example ofE(L)
tor
'Z
2
Z
12
for allK2S: Furthermore, the groupsZ
3
Z
12
andZ
4
Z
12
are ruled out by Theorem 6.2. It follows thatZ
6
Z
12
;Z
4
Z
24
; andZ
4
Z
36
are
eliminated from the lists above which establishes the assertion (i):
If we assumeZ
2
Z
18
E(L)
tor
; thenE(L)[9] =Z
9
: By Lemma 1, this yields
Z
9
'E(K)[9]E
d
0
(K)[9] =Z
3
E
d
0
(K)[9]
where L = K(Q(
p
d
0
)) for a nonsquare d
0
2 K: But the right hand side is a noncyclic group
whereas the left hand side is not. We derive a contradiction. So that we can narrow down our list
(6.8) by removing the groupsZ
2
Z
18
; Z
4
Z
18
; Z
2
Z
36
: We now argue whetherZ
6
Z
6
occurs
75
as a torsion subgroup over a quadratic extensionL ofK inSnfQ(
p
2)g: ForZ
6
Z
6
to appear
as torsion,E(L)[3] must be equal toZ
3
Z
3
which impliesL =K(
p
3)): Hence the quadratic
twistE
3
ofE has aK-rational point of order 3. Along with anL-rational 4-torsion point onE;
we argue as in the proof of Theorem 6.4(4), we are done by Lemma 7 (See Table 6.6 for examples).
This settles down the assertion (ii):
It remains to check if E(L)
tor
' Z
2
Z
24
when D =19;43;67;163: By way of
contradiction, assumeZ
24
E(L): Then, there exists anL-rational point of order 8 onE: WLOG
we can assume that there existss;t;z2O
L
such that =s
4
; =t
4
ands
2
+t
2
=z
2
: It follows
that either
p
or
p
is a square inL: Applying Lemma 8 either or - is a square inK: As the
same discussion applies to, either or - is a square inK: Hence there are four possibilities as
follows
(i) =s
2
0
and =t
2
0
; (iii) =s
2
0
and =t
2
0
;
(ii) =s
2
0
and =t
2
0
; (iv) =s
2
0
and =t
2
0
:
wheres
0
;t
0
6= 0 inK:
Case (i) : Note that sinceO
K
is integrally closed, ifa2O
K
is a square inK then it is a square
inO
K
: Thus, and are both squares inO
K
: It follows thatE(K) has a point of order 4 which is
contradicting the assumption.
Case (ii) : AsE(K) has a point of order 3, there area;b;c2O
K
with
a
b
= 2f2;1;1=2; 0; 1g
andc6= 0 such that = a
3
(a + 2b)c
2
and = b
3
(b + 2a)c
2
by Theorem 6.1. Together with the
assumption (ii); we obtain the system
s
2
0
=a
3
(a + 2b)c
2
t
2
0
=b
3
(b + 2a)c
2
76
Note that we must havea;b;a + 2b6= 0 so we may divide and multiply both sides by -1
s
0
t
0
2
=
a
b
3
a + 2b
b + 2a
and letting y =
s
0
t
0
and x =
a
b
we get y
2
=x
3 x+2
2x+1
which is birational to the elliptic curve
E
0
: y
2
= x
3
5x
2
+ 4x: We compute E
0
(K) has rank 0 and E
0
(K)' Z
2
Z
2
when D =
19;43;67;163: SoE
0
(K) =f(1; 0); (0; 0); (4; 0)g: But we already noted that
a
b
6= 1 and
s
0
6= 0;t
0
6= 0: Hence there is no solution to the equation above.
Case (iii) : It follows similarly as in the case (ii):
Case (iv) : The immediate observation is that1 must be a square in L: We can assume
L = K(
p
1) because K does not contain
p
1: Since E(L) has a point of order 8 so does
E
1
(L). In terms of the coefficients ofE
1
:y
2
=x(x)(x); this yields
x
4
= =s
2
0
and y
4
= =t
2
0
(6.9)
for some x;y;z2O
L
with x
2
+y
2
= z
2
: Then, particularly x
2
= is
0
oris
0
: Since the curve
defined byx
2
+y
2
= z
2
is parametrized by (1m
2
; 2m; 1 +m
2
) overL; replacingy by 2m in
(6.9), we obtainm
4
=(
t
0
4
)
2
2K: As eitherm
4
2m
2
+ 1 =is
0
oris
0
; we have
4m
4
= (is
0
+m
4
+ 1)
2
or 4m
4
= (is
0
+m
4
+ 1)
2
We can rewrite the above as
t
2
0
4
=
is
0
(
t
0
4
)
2
+ 1
2
: However, the left hand side is in K
whereas the right hand side is not inK which gives us a contradiction. Therefore, there is no cyclic
subgroup of order 24 inE(L)
tor
which rules outZ
2
Z
24
, completing the proof.
Proposition 6.5. LetK =Q(
p
D) inS andE=K be an elliptic curve withE(K)
tor
'Z
2
Z
4
;
then WLOG we may write = s
2
and = t
2
: There is a quadratic extensionL ofK such that
Z
4
Z
4
E(L)
tor
iffs
2
t
2
=z
2
for somez2K. In this caseL =K(i) and
77
(i) IfD6=7 thenE(L)
tor
'Z
4
Z
4
(ii) IfD =7 thenE(L)
tor
'Z
4
Z
4
orZ
4
Z
8
Otherwise,E(L)
tor
'Z
2
Z
4
orZ
2
Z
8
: In particular, ifD =19;43;67;163 then
E(L)
tor
'Z
2
Z
12
:
Proof. SupposeE(K)
tor
'Z
2
Z
4
: We first letK2SnfQ(
p
7)g with =s
2
and =t
2
: By
the Weil pairing, notice thatL =K(i) is the unique extension whereZ
4
Z
4
E(L)
tor
: WLOG
we may assumes
2
t
2
= z
2
: Note that under these hypothesis, we haveE
1
(K)
tor
' Z
2
Z
4
as shown in (5) of Theorem 6.4. By Proposition 3.1 along with assumption that E(L) contains
full 4-torsion, the possible torsion structures forE(L)
tor
areZ
4
Z
4
;Z
4
Z
8
; orZ
4
Z
16
: It is
enough to prove thatE(L) has no point of order 8: As argued in [25], we consider the following
three points of order 4:
Q = (st;st(s +t)); Q
0
= (s(s +z);isz(s +z))
Q +Q
0
=(t(t +iz);tz(t +iz)):
Note 2Q = (0; 0) and 2Q
0
= (s
2
; 0) so Q and Q
0
are independent order 4 points and hence
fQ;Q
0
g forms a basis forZ
4
Z
4
E(L)
tor
: If there is an order 8 pointP inE(L)
tor
, then 2P
has order 4 and so 2P = mQ +nQ
0
for some m;n2 Z
4
: Adding 2Q or 2Q
0
to both sides of
the equation if necessary, we may assume that there exists an order 8 pointP inE(L)
tor
such that
2P =Q; 2P =Q
0
or 2P =Q +Q
0
:
If Q is halved, then by Lemma 6, st;s(s +t);t(s +t) are squares in L: Since they lie in K
and L = K(i), we have st =r
2
for some r: Assume st =r
2
: On the other hand, we have
the parametrization fors
2
= t
2
+z
2
as (s;t;z) = (1 +m
2
; 2m; 1m
2
) with the condition that
m6= 0;1 becauseE is non-singular. Then we obtainr
2
= (1 +m
2
)(2m): Multiplying both
sides by 4, we get (2r)
2
= (4 + (2m))
2
(2m) which implies that there exists a K-rational
point on the elliptic curveE
1
given byE
1
: y
2
= x
3
+ 4x: We compute thatE
1
(K)'Z
4
for all
K2SnfQ(
p
7)g (of rank 1 overK =Q(
p
7)). The only points ofE
1
(K) aref(0; 0); (2;4)g
78
yieldingm = 0 orm =1 which is a contradiction. Ifst = r
2
, then using the same argument
above, the contradition follows.
IfQ
0
is halved, thens(s +t);sz;2(s +z) are squares inL: We conclude as above that
sz =r
2
inK: Ifsz =r
2
; then substituting the parametrization ofs
2
=t
2
+z
2
yields
r
2
= (1 +m
2
)(1m
2
) = 1m
4
Multiplying both sides bym
2
; we have (mr)
2
= (m
2
)
3
(m
2
) which in turn gives aK-rational
point on the elliptic curve
E
2
:y
2
=x
3
x
We compute thatE
2
(K) =Z
2
Z
2
for allK2SnfQ(
p
7)g (has rank 1 overK =Q(
p
7)).
Hence the only points ofE
2
(K) aref(0; 0); (1; 0)g which impliesm
2
= 0;1; contradicting
m6= 0;1: Ifsz =r
2
; then similarly it results in the same curveE
2
wherem
2
is replaced byx:
IfQ +Q
0
is halved, by Lemma 6 we haveitz is a square inL so thatitz = (a +bi)
2
=
a
2
b
2
+ 2abi for some a;b2 K: Then a =b and so tz =2a
2
: Using the parametrization
above, we get
a
2
=m(1m
2
)
One observes easily that it boils down to find the points on bothE
2
(K) andE
1
2
(K): But since
E
1
2
(K)
tor
= E
2
(K)
tor
;E
1
2
' E, it is enough to determine the setE
2
(K): As computed in the
previous case, E
2
(K) has rank 0 with torsionZ
2
Z
2
: In each case a = 0; which forces either
t = 0 orz = 0, a contradiction. In caseK =Q(
p
7), following Theorem 6.4 the possibilities
withZ
4
Z
4
E(L)
tor
areZ
4
Z
8
orZ
4
Z
16
: ButZ
4
Z
16
is ruled out by Theorem 6.2, which
establishes the first part of the claim.
Now given K 2 S; suppose E(K) ' Z
2
Z
4
and E(L) does not contain Z
4
Z
4
: If
D =2;7;11 then by Theorem 6.4,E(L)
tor
'Z
2
Z
4
;Z
2
Z
8
; orZ
2
Z
16
: IfE(L)
tor
'
Z
2
Z
16
for some quadratic extensionL ofK; we may assumeE(L)
tor
=hQ;Pi whereQ;P are
independent 2-torsion and 16-torsion points withE(K)
tor
'hQ; 4Pi: For the non-trivial element
79
of Gal(L=K); we have(P ) = aP or(P ) = Q +aP wherea is odd. Applying twice, we
obtain
(P ) =a
2
P
wherea
2
1 mod (16). Thena must be 1; 7; 9 or 15: For all eight possibilities of; we have either
(2P ) = 2P or(P ) =2P: The former is not possible by assumption. The later implies that
(4P ) =4P; which yields a contradiction since 4P isK-rational.
IfD =19;43;67;163 then by Theorem 6.4, there are only four possibilities: E(L)
tor
=
Z
2
Z
4
;Z
2
Z
8
;Z
2
Z
12
orZ
2
Z
24
: ReplacingE byE
d
if necessary, the groupZ
2
Z
24
is
excluded in the proof of Proposition 6.4, proving the statement.
In light of these results, we can complete the classification of the growth of torsion by proving
the main theorem.
Proof of Theorem 6.3. SupposeG'Z
2
Z
12
: By Theorem 3.1, note thatG occurs whenK =
Q(
p
D) whereD2E :=f19;43;67;163g: GivenK inE; we want to determine
K
(2;G):
For this purpose, let L = K(
p
d) be a quadratic extension of K (d2 K nonsquare) such that
G(E(L)
tor
: By Proposition 3.1 and the classification of torsion groups forE
d
(K)
tor
in Theorem
6.4,E(L)
tor
is isomorphic to one of the groups
Z
4
Z
12
; Z
2
Z
24
; Z
4
Z
24
; Z
2
Z
48
; Z
8
Z
24
orZ
4
Z
48
(6.10)
Since E(K)
tor
= Z
2
Z
12
; then by the parametrization given by [35], E is isomorphic to E
P
0
defined by
E
P
0
:y
2
+(6t
4
8t
3
+ 2t
2
+ 2t 1)xy + (12t
6
+ 30t
5
34t
4
+ 21t
3
7t
2
+t)(t 1)
5
y
=x
3
+ (12t
6
+ 30t
5
34t
4
+ 21t
3
7t
2
+t)(t 1)
2
x
2
with
E
P
0
= t
12
(t 1)
12
(2t 1)
6
(2t
2
2t + 1)
3
(3t
2
3t + 1)
4
(6t
2
6t + 1)6= 0 for some
P
0
= (t;s)2 (K) where :s
2
= 12t
4
24t
3
+ 20t
2
8t + 1 and is birationally equivalent to
80
X
1
(2; 12) (overQ): Hence, ifZ
2
Z
24
E(L)
tor
then there exists aL-rational 24-torsion pointR
0
and aK-rational 12-torsion pointP
0
onE such that 2R
0
=P
0
: Letx(R
0
) denote thex-coordinate
ofR
0
: By the duplication formula of a point,x(R
0
) must be the root ofh(x;t) =x
2
3
where
2
and
3
are the division polynomials of degree 2 and 3 associated toE: A quick computation in
Magma yields thath(x;t)2 K[t](x) has degree 4 and it factors into quadratic polynomials over
K[t] if
h(x;t) = (x
2
+a(t)x +b(t))(x
2
a(t)x +d(t)) (6.11)
whereb(t);d(t) are both of degree 15 and their leading coefficientsb
0
andd
0
satisfy the relations
b
0
+d
0
=72; b
0
d
0
= 1728
But this is not possible, which leads us to the conclusion that there does not exists a decomposition
as in (6.11) and soZ
2
Z
24
cannot appear as a subgroup of E(L): Furthermore, Theorem 6.2
asserts that Z
4
Z
12
cannot occur which eliminates all the groups in (6.10), proving (1): The
assertions (2); (3); (4) and (5) are established by Propositions 6.1, 6.2, 6.3, 6.4 and 6.5.
Finally, supposeG =Z
2
Z
2
: AsE(K(
p
d))'E
d
(K(
p
d)); it is enough to studyE
d
(K(
p
d))
tor
.
If E
d
(K)6' Z
2
Z
2
; then we may apply a previously proven case and see Tables 6.7,6.4. If
E
d
(K)'Z
2
Z
2
; then we have by Proposition 3.1 thatE
d
(K(
p
d))
tor
'Z
2
Z
2
;Z
2
Z
4
or
Z
4
Z
4
: (See Table 6.3).
6.3 Examples of Growth
Table 6.3: Example of GrowthE(K)
tor
'Z
2
Z
2
overK =Q(
p
2)
Model forE(;) E
d
(K)
tor
d2K E(K(
p
d))
tor
(1;2) Z
2
Z
2
1 Z
4
Z
4
81
Table 6.4: Example of GrowthE(K)
tor
'Z
2
Z
2
overK =Q(
p
7)
Model forE(;);w =
p
7 E
d
(K)
tor
d2K E(K(
p
d))
tor
(
1
2
(42525w 44415);
1
2
(42525w 44415) Z
2
Z
8
15 Z
2
Z
16
(85; 400) Z
2
Z
4
5 Z
2
Z
8
Table 6.5: Example of GrowthE(K)
tor
'Z
2
Z
4
overK =Q(
p
7)
Model forE(;) E
1
(K)
tor
E(K(
p
d))
tor
(68121; 69696) Z
2
Z
4
Z
4
Z
4
Table 6.6: Examples of Growth ofE(K)
tor
'Z
2
Z
6
overK =Q(
p
D)
w =
p
D Model forE(;) E
3
(K)
tor
E(K(
p
3))
tor
D =7
1
2
(21w 39);
1
2
(21w 39)
Z
2
Z
6
Z
6
Z
6
D =11 (19514w + 101438;6912w + 214272) Z
2
Z
6
Z
6
Z
6
D =19 = 10993263062353152w 37976902494666157
=10437125916000000w + 40849621725921875
Z
2
Z
6
Z
6
Z
6
Table 6.7: Example of GrowthE(K)
tor
'Z
2
Z
6
over anyK2S
Model forE(;) E
d
(K)
tor
; d = 21 E(K(
p
d))
tor
(64; 189) Z
2
Z
2
Z
2
Z
12
Table 6.8: Examples of Growth ofE(K)
tor
'Z
2
Z
6
overK =Q(
p
19)
Model forE(;); w =
p
19 E
3
(K)
tor
E(K(
p
3))
tor
=
1
806954491
(1160294229092597760w755235215206514688)
=215373816000w + 140186761425
Z
2
Z
4
Z
2
Z
12
Table 6.9: Examples of Growth ofE(K)
tor
'Z
2
Z
8
overK =Q(
p
7)
Model forE(;); w =
p
7 E
d
(K)
tor
d2K E(K(
p
d))
tor
(729; 2304) Z
2
Z
4
1 Z
4
Z
8
1
2
(93w + 449); 24w 248
Z
2
Z
8
15 Z
2
Z
16
82
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Abstract (if available)
Abstract
Let K be a noncyclotomic imaginary quadratic field of class number one and E an elliptic curve defined over K. We study the relation of Tor(E(L)) with Tor(E(K)) where L is any quadratic extension of K. If Tor(E(L)) strictly contains Tor(E(K)), we say that torsion grows. If E satisfies one of the following cases: E(K)[2] is either trivial or full 2-torsion, we determine the complete classification of possible torsion structures appearing as Tor(E(L)) for any quadratic extension L of K.
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Creator
Balçık, Irmak
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Core Title
Growth of torsion of elliptic curves over noncyclotomic imaginary quadratic fields of class number 1
School
College of Letters, Arts and Sciences
Degree
Doctor of Philosophy
Degree Program
Mathematics
Degree Conferral Date
2022-08
Publication Date
07/08/2022
Defense Date
05/02/2022
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elliptic curves,OAI-PMH Harvest,rational points,torsion subgroup
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Kamienny, Sheldon (
committee chair
), Friedlander, Eric (
committee member
), Huang, Ming-Deh A. (
committee member
)
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balcik@usc.edu,i.balcik89@gmail.com
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Tags
elliptic curves
rational points
torsion subgroup