# NON-CYCLIC TORSION OVER IMAGINARY QUADRATIC FIELDS of CLASS NUMBER 1

IRMAK BALÇIK

###### Abstract.

Let $K$ be a non-cyclotomic imaginary quadratic field of class number 1 and
$E/K$ an elliptic curve with
$E(K)[2]\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ In this
article, we determine the torsion groups that can arise as $E(L)_{\text{tor}}$
where $L$ is any quadratic extension of $K.$

##  1\. Introduction

Let $E$ be an elliptic curve over a number field $K$. The celebrated theorem
of Mordell-Weil asserts that the set of $K$-rational points $E(K)$ is a
finitely generated abelian group. Let $E(K)_{\text{tor}}$ denote the torsion
subgroup of $E(K)$. Mazur [Maz78] first completed the classification of all
possible torsion subgroups when $K=\mathbb{Q},$ which are

(1) $\begin{array}[]{ll}\mathbb{Z}/n\mathbb{Z},&1\leq n\leq 12,\ n\neq 11\\\
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z},&1\leq n\leq
4\end{array}$

Subsequently, Kamienny [Kam92] and Kenku-Momose [KM88] determined the complete
list of all the groups arising as $E(K)_{\text{tor}}$ where $E$ is an elliptic
curve over any quadratic number field $K$. These groups are:

(2) $\begin{array}[]{ll}\mathbb{Z}/n\mathbb{Z},&1\leq n\leq 18,\ n\neq 17\\\
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z},&1\leq n\leq 6\\\
\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/{3n}\mathbb{Z},&1\leq n\leq 2\\\
\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}&.\end{array}$

Over cubic fields, the full description of torsion subgroups has been settled
only last year [DEvH+21]. Over quartic fields, while we still do not have a
complete classification, analogous to (1), we know due to [JKP06] that what
torsion structures are infinitely many times occurring up to isomorphism of
elliptic curves. Also, $17$ is the largest prime dividing the torsion subgroup
over any quartic field by [DKSS17]. As a first step toward this direction, we
ask ``given an elliptic curve $E$ over any quadratic number field $K$, if
$E(K)_{\text{tor}}$ has full 2-torsion, then how does $E(L)_{\text{tor}}$
relate with $E(K)_{\text{tor}}$ where $[L:K]=2$?". If
$E(K)_{\text{tor}}\subsetneq E(L)_{\text{tor}},$ we say that torsion grows.

This leads us to a more fundamental question: whether there is a good bound
for the growth. Kwon [Kwo97] showed that for an elliptic curve $E/\mathbb{Q},$
the growth of $E(\mathbb{Q})_{\text{tor}}$ upon quadratic base change, is
controlled by the torsion structure realizing on a quadratic twist of the
curve $E,$ which can be generalized to quadratic number fields which allows us
to bound $E(L)_{\text{tor}}/E(K)_{\text{tor}}.$(See Propostion 2.1 below.) In
this note, we answer the main objection above for the set $\mathcal{S}$ of all
the non-cyclotomic imaginary quadratic fields of class number 1, that is
$\mathcal{S}=\\{\mathbb{Q}(\sqrt{-2}),\mathbb{Q}(\sqrt{-7}),\mathbb{Q}(\sqrt{-11}),\mathbb{Q}(\sqrt{-19}),\mathbb{Q}(\sqrt{-43}),\mathbb{Q}(\sqrt{-67}),\mathbb{Q}(\sqrt{-163})\\}.$
We choose to work over the base fields in $\mathcal{S}$ as these are the
remaining imaginary quadratic fields of class number 1 for which the
classification of torsion structures is known, [SS18]. The method used in
their paper is outlined by Kamienny and Najman [KN12]. In case
$E(K)[2]\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},$ we
generalize both Kwon's work in which $K=\mathbb{Q},$ and Newman's work [New17]
in which $K\in\\{\mathbb{Q}(\sqrt{-1}),\mathbb{Q}(\sqrt{-3}\\}.$ This note
will also complete the author's earlier work [Bal21] in which for given
$K\in\mathcal{S}$ and an elliptic curve $E/K$ with trivial $2$-torsion, the
growth of odd torsion subgroup $E(K)_{\text{tor}}$ is classified upon
quadratic base change.

Our methods didn't explicitly rely upon the facts that $\mathcal{O}_{K}$ 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,
(see Lemma 3) as well as used new techniques ranging from the Galois action to
the analysis of cyclic isogenies (see Theorem 3.1), thus finishing the
classification of quadratic torsion growth for the set of all imaginary
quadratic number fields of class number 1.

In order to describe our main result, we introduce the following notation. For
a fixed positive integer $d$, let us denote

$\Phi_{K}(d):=\\{\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}:\exists\
E/K:E(L)_{\text{tor}}\simeq\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}\
\text{for}\ [L:K]=d\\}$

and for a fixed $G\in\Phi_{K}(1)$, we write

$\displaystyle\Phi_{K}(d,G):=\\{\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}:\exists\
E/K:E(K)_{\text{tor}}\simeq
G,E(L)_{\text{tor}}\simeq\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z},\
[L:K]=d\\}.$

We are ready to state our main result followed by explicit examples.

######  Theorem 1.1.

Given $K=\mathbb{Q}(\sqrt{D})$ in $\mathcal{S}$ and $E/K$ an elliptic curve
with full 2-torsion. We may assume that $E$ is in the form
$y^{2}=x(x+\alpha)(x+\beta)$ for some $\alpha,\beta\in K.$ Then,

  1. (1)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}\\}.$

  2. (2)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}\\}.$

  3. (3)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$ and

    1. i.

$D\neq-7$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{8}\mathbb{Z}\\}$

    2. ii.

$D=-7$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{8}\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}\\}.$

  4. (4)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z},$ and

    1. i.

$D=-2$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{6}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}\\}$

    2. ii.

$D\neq-2$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{6}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z},\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/{6}\mathbb{Z}\\}.$

  5. (5)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},$ then there is
a quadratic extension $L$ of $K$ with
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq
E(L)_{\text{tor}}$ iff $\alpha-\beta=\pm z^{2}$ for some $z\in K.$ In this
case, $L=K(i)$ and

    1. i.

if $D\neq-7$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{4}\mathbb{Z}\\}$

    2. ii.

if $D=-7$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{4}\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}\\}.$

Otherwise, $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\not\subseteq
E(L)$ and we have

    1. i.

if $D=-2,-7,-11$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{8}\mathbb{Z}\\}$

    2. ii.

if $D=-19,-43,-67,-163$ then
$\Phi_{K}(2,G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{8}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}\\}.$

  6. (6)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ then
$\Phi_{K}(2,G)\subseteq\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z}:n=1,2,3,4,5,6,8\\}\cup\\{\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\\}.$

Table 1. Example of Growth $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\ \text{over}\ K=\mathbb{Q}(\sqrt{-7})$ Model for $E(\alpha,\beta)$ |  $E^{d}(K)_{\text{tor}}$ |  $d\in K$ |  $E(K(\sqrt{d}))_{\text{tor}}$
---|---|---|---
$\big{(}\frac{1}{2}(42525\sqrt{-7}-44415),\frac{1}{2}(-42525\sqrt{-7}-44415)\big{)}$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{8}\mathbb{Z}$ |  $-15$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}$
Table 2. Example of Growth $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\ \text{over}\ K=\mathbb{Q}(\sqrt{-2})$ Model for $E(\alpha,\beta)$ |  $E^{d}(K)_{\text{tor}}$ |  $d\in K$ |  $E(K(\sqrt{d}))_{\text{tor}}$
---|---|---|---
$(-1,-2)$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z}$ |  $d=-1$ |  $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{4}\mathbb{Z}$
$(3600,3645)$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ |  $d=-5$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$
Table 3. Example of Growth $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\ \text{over}\ K=\mathbb{Q}(\sqrt{-7})$ Model for $E(\alpha,\beta)$ |  $E^{-1}(K)_{\text{tor}}$ |  $E(K(\sqrt{d}))_{\text{tor}}$
---|---|---
$(68121,69696)$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{4}\mathbb{Z}$ |  $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{4}\mathbb{Z}$
Table 4. Examples of Growth of $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$ over $K=\mathbb{Q}(\sqrt{D})$ $w=\sqrt{D}$ |  Model for $E(\alpha,\beta)$ |  $E^{-3}(K)_{\text{tor}}$ |  $E(K(\sqrt{-3}))_{\text{tor}}$
---|---|---|---
$D=-7$ |  $\big{(}\frac{1}{2}(21w-39),\frac{1}{2}(-21w-39)\big{)}$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$ |  $\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/{6}\mathbb{Z}$
$D=-11$ |  $(-78056w+405752,-27648w+857088)$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$ |  $\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$
$D=-19$ |  $\alpha=10993263062353152w-37976902494666157$ $\beta=-10437125916000000w+40849621725921875$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$ |  $\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$
Table 5. Example of Growth $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}\ \text{over\ any}\ K\in\mathcal{S}$ Model for $E(\alpha,\beta)$ |  $E^{d}(K)_{\text{tor}},\ d=21$ |  $E(K(\sqrt{d}))_{\text{tor}}$
---|---|---
$(64,189)$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z}$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$
Table 6. Examples of Growth of $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$ over $K=\mathbb{Q}(\sqrt{-19})$ Model for $E(\alpha,\beta),\ w=\sqrt{-19}$ |  $E^{-3}(K)_{\text{tor}}$ |  $E(K(\sqrt{-3}))_{\text{tor}}$
---|---|---
$\alpha=\frac{1}{806954491}(1160294229092597760w-755235215206514688)$ $\beta=-215373816000w+140186761425$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$
Table 7. Examples of Growth of $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$ over $K=\mathbb{Q}(\sqrt{-7})$ Model for $E(\alpha,\beta),$ $w=\sqrt{-7}$ |  $E^{d}(K)_{\text{tor}}$ |  $d\in K$ |  $E(K(\sqrt{d}))_{\text{tor}}$
---|---|---|---
$(729,2304)$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ |  $-1$ |  $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{8}\mathbb{Z}$
$\big{(}\frac{1}{2}(93w+449),24w-248\big{)}$ |  $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z}$ |  $-15$ |  $\mathbb{Z}/{2}\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}$

We now give a more detailed overview of the paper. In section 2, we present a
number of known results that narrow our focus. For example, an elliptic curve
with full $2$-torsion admits a Weierstrass model of the form
$y^{2}=x(x+\alpha)(x+\beta)$. Moreover, the existence of points of order $4$,
$8$ or $3$ is then equivalent to finding solutions to certain equations
involving $\alpha$ and $\beta$. (See Theorem 2.2). These characterizations
will be later used to eliminate torsion points upon base change.

In section 3, we develop a result that is used in section 4. Specifically, we
classify certain cyclic $N$-isogenies of elliptic curves (up to isomorphism)
defined over any non-cyclotomic quadratic imaginary field of class number 1.
The values for $N$ of interest are $N=30,40,48.$ For a given $N,$ the set of
$K$-rational points on the modular curve $X_{0}(N)$ parametrizes the
isomorphism classes of pairs $(E,C)$ where $E$ is an elliptic curve defined
over $K$ and $C\subseteq E(\overline{K})$ is a Gal$(\overline{K}/K)$-invariant
cyclic subgroup of order $N.$ If $X_{0}(N)$ has genus at least 2, then there
are finitely many $K$-rational points, by a renowned result of Faltings. In
the cases $N=30,40,48$, $X_{0}(N)$ is a hyperelliptic curve with genus 3. The
main trick for studying quadratic points on such a curve $X_{0}(N)$ is to
instead study the rational points on its symmetric square $X_{0}(N)^{(2)}.$
This comes down to finding the rational points on the Jacobian
$J_{0}(N)(\mathbb{Q})$ of $X_{0}(N).$ In the light of the fact that
$J_{0}(N)(\mathbb{Q})$ has rank 0 for the aforementioned values of $N,$ Bruin
and Najman [BN15] classified all exceptional quadratic points on $X_{0}(N).$
We complemented their work and found that there are no non-exceptional
$K$-rational points lying on these curves for any $K$ in $\mathcal{S}.$

In section 4, we prove the most important ingredient for Theorem 1.1 which is
the description of $T_{K}(G)$, the set of finite abelian groups $H$ (up to
isomorphism) such that there is an elliptic curve $E/K$ with
$E(K)_{\text{tor}}\simeq G$ and $E^{d}(K)\simeq H$ for $d\in K.$ We study the
torsion structures on quadratic twists because if $L=K(\sqrt{d})$ and $E/K$ is
an elliptic curve then $E^{d}(K)_{\text{tor}}$ sets a bound for
$E(L)_{\text{tor}}$(see Proposition 2.1). If one wants to study the relation
of $E(L)_{\text{tor}}$ with $E(K)_{\text{tor}},$ the first task is to classify
which groups occur as $E^{d}(K)_{\text{tor}}.$ To do so, we make use of the
isogeny analysis from section 3 and the equivalent characterization of
elliptic curves with the prescribed torsion
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z}\ (1\leq n\leq 6)$.
Twisting and equating parametrized models of curves give rise to systems of
Diophantine equations to solve.

In section 5, we prove our main result Theorem 1.1. With the classification of
twists in hand, we can have a finite list of possible torsion structures
arising as $E(L)_{\text{tor}}.$ Furthermore, the growth occurs over $L$ if and
only if there exists $P\in E(K)_{\text{tor}}$ with $P\in
2E(L)_{\text{tor}}\backslash\ 2E(K)_{\text{tor}}.$ The criteria of the
existence of such a point is given in Lemma 1. Applying this criteria yields
systems of Diophantine equations as in the previous section. The difficulty
arises when we deal with torsion points of high order such as 10 or 12. In
order to work around this problem, we shifted our perspective and used the
galois action on $n$-torsion $E[n].$

### Notation

Unless otherwise stated, the number field $K$ runs over the quadratic fields
in $\mathcal{S}$ and $L$ denotes a quadratic extension over $K.$ We
interchangebly represent the elements of $\mathcal{S}$ as
$K=\mathbb{Q}(\sqrt{D})$ where $D$ belongs to
$\\{-2,-7,-11,-19,-43,-67,-163\\}$. We refer to the set

$\mathcal{T}=\\{\mathbb{Q}(\sqrt{D}):D=-1,-2,-3,-7,-11,-19,-43,-67,-163\\}$

as the set of all imaginary quadratic fields of class number 1. We denote the
multiplication by $n$ map on an elliptic curve $E$ by $[n]$ and denote the
kernel of $[n]$ by $E[n].$ Given an elliptic curve $E/K$ with
$E(K)[2]\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},$ we work
with the Weierstrass model $E(\alpha,\beta):y^{2}=x(x+\alpha)(x+\beta)$ of $E$
where $\alpha,\beta\in\mathcal{O}_{K}.$

We use the database [LMF23] for rational elliptic curves. Some of the proofs
in the paper rely on computations in Magma. The code can be found $\href
https://github.com/irmak-balcik/NonCyclicTorsion$

### Acknowledgement

The author is indebted to her advisor Sheldon Kamienny for suggesting this
problem and his kind support. The author is grateful to Andreas Schweizer for
communicating a proof of the case
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{32}\mathbb{Z}$ and allowing us to
include it in the paper. We would like to heartily thank Özlem Ejder for a
multitude of helpful discussions throughout this project.

##  2\. Auxiliary Results

In [SS18], Sarma and Saika have determined the complete list of possible
groups arising as the torsion subgroups of elliptic curves defined over $K$ as
$K$ varies through in $\mathcal{S}.$ We combine their results into a
standalone theorem:

######  Theorem 2.1 ([SS18]).

Let $E$ be an elliptic curve over a fixed $K\in\mathcal{S}.$

  1. (1)

If $K=\mathbb{Q}(\sqrt{-2}),$ then $E(K)_{\text{tor}}$ is isomorphic to either
one of the groups in (1),

$\mathbb{Z}/{11}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}.$

  2. (2)

If $K=\mathbb{Q}(\sqrt{-7}),$ then $E(K)_{\text{tor}}$ is isomorphic to either
one of the groups in (1),

$\mathbb{Z}/{11}\mathbb{Z},\ \mathbb{Z}/{14}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/{15}\mathbb{Z}.$

  3. (3)

If $K=\mathbb{Q}(\sqrt{-11}),$ then $E(K)_{\text{tor}}$ is isomorphic to
either one of the groups in (1),

$\mathbb{Z}/{14}\mathbb{Z},\ \mathbb{Z}/{15}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}.$

  4. (4)

If $K=\mathbb{Q}(\sqrt{-19}),$ then $E(K)_{\text{tor}}$ is isomorphic to
either one of the groups in (1),

$\mathbb{Z}/{11}\mathbb{Z},\
\mathbb{Z}/{2}\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$

  5. (5)

If $K=\mathbb{Q}(\sqrt{-43}),$ then $E(K)_{\text{tor}}$ is isomorphic to
either one of the groups in (1),

$\mathbb{Z}/{11}\mathbb{Z},\ \mathbb{Z}/{14}\mathbb{Z},\
\mathbb{Z}/{15}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$

  6. (6)

If $K=\mathbb{Q}(\sqrt{-67}),$ then $E(K)_{\text{tor}}$ is isomorphic to
either one of the groups in (1),

$\mathbb{Z}/{14}\mathbb{Z},\ \mathbb{Z}/{15}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$

  7. (7)

If $K=\mathbb{Q}(\sqrt{-163}),$ then $E(K)_{\text{tor}}$ is isomorphic to
either one of the groups in (1),

$\mathbb{Z}/{14}\mathbb{Z},\ \mathbb{Z}/{15}\mathbb{Z}\ \ \text{or}\ \
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$

A natural question arises when analyzing the growth of torsion in a quadratic
extension is whether there is an upper bound for the growth. In [Kwo97,
Proposition 1] Kwon answered this question for elliptic curves over
$\mathbb{Q}.$ We state the generalization of his result as follows.

######  Proposition 2.1.

([Kwo97]) Let K be a quadratic number field, $L=K(\sqrt{d})$ a quadratic
extension of K and let $\sigma$ denote the generator of $Gal(L/K)$. The map

$E(L)_{\text{tor}}\xrightarrow{h}E^{d}(K)_{\text{tor}}$ $P\mapsto P-\sigma(P)$

is a homomorphism with kernel $E(K)_{\text{tor}}$ and hence induces an
injection

$E(L)_{\text{tor}}/E(K)_{\text{tor}}\hookrightarrow E^{d}(K)_{\text{tor}}.$

The next lemmas play a key role in studying the 2-part torsion of elliptic
curves. They will be repeatedly used in the rest of the paper.

######  Lemma 1.

([Kna92]) Let E be an elliptic curve over a field k with $char(k)\neq 2,3.$
Suppose E is given by

$y^{2}=(x-\alpha)(x-\beta)(x-\gamma)$

with $\alpha,\beta,\gamma\in k.$ For $P=(x,y)$ in $E(k)$, there exists a
k-rational point $Q$ on E such that 2$Q=P$ if and only if $x-\alpha,x-\beta$,
and $x-\gamma$ are all squares in k.

In this case if we fix the sign of the square roots of
$x-\alpha,x-\beta,x-\gamma,$ then the $x$-coordinate of $Q$ equals either

$\sqrt{x-\alpha}\sqrt{x-\beta}\pm\sqrt{x-\alpha}\sqrt{x-\gamma}\pm\sqrt{x-\beta}\sqrt{x-\gamma}+x$

or

$-\sqrt{x-\alpha}\sqrt{x-\beta}\pm\sqrt{x-\alpha}\sqrt{x-\gamma}\mp\sqrt{x-\beta}\sqrt{x-\gamma}+x$

If the criteria of the statement above is satisfied for a point $P,$ we say
that $P$ is halved. The next result is a variant of the result of Ono, see
[Ono96]).

######  Theorem 2.2.

([New17]) Let $K$ be a number field and $E/K$ be an elliptic curve with full
2-torsion. Then $E$ has a model of the form $y^{2}=x(x+\alpha)(x+\beta)$ where
$\alpha,\beta\in\mathcal{O}_{K}.$

  1. (1)

$E(K)$ has a point of order 4 iff $\alpha,\beta$ are both squares, or
$-\alpha,\beta-\alpha$ are both squares, or $-\beta,\alpha-\beta$ are both
squares in $\mathcal{O}_{K}.$

  2. (2)

$E(K)$ has a point of order 8 iff there exists a $z\in\mathcal{O}_{K}$, $z\neq
0$ and a pythagorean triple $(u,v,w)$ such that

$\alpha=z^{2}u^{4},\beta=z^{2}v^{4}$

or we can replace $\alpha,\beta$ by $-\alpha,\beta-\alpha$ or
$-\beta,\alpha-\beta$ as in the first case.

  3. (3)

$E(K)$ contains a point of order 3 iff there exists $a,b,z\in\mathcal{O}_{K}$
with $z\neq 0$ and $\frac{a}{b}\notin\\{-2,-1,-\frac{1}{2},0,1\\}$ such that
$\alpha=a^{3}(a+2b)z^{2}$ and $\beta=b^{3}(b+2a)z^{2}.$

The following result will be of great importance in determining odd torsion
part in quadratic extensions of a number field.

######  Lemma 2.

([GJT14],Cor 4) If n is an odd positive integer we have

$E(K(\sqrt{d}))[n]\simeq E(K)[n]\oplus E^{d}(K)[n]$

##  3\. Certain Isogenies

Given an elliptic curve $E$ over a number field $K$, we call a subgroup $C$ of
$E(\overline{K})$ is a $\mathbf{K}$-$\mathbf{rational}$
$\mathbf{N}$-$\mathbf{isogeny}$ if there exists an elliptic curve
$E^{\prime}/K$ and an isogeny $\phi:E\rightarrow E^{\prime}$ defined over $K$
such that $C=ker(\phi).$ Equivalently, a subgroup $C$ of $E(\overline{K})$ of
order $N$ is a $K$-rational $N$-isogeny if it is invariant under the action of
Gal$(\overline{K}/K).$ Note that $C$ might be $K$-rational even though it
contains points that are not.

######  Proposition 3.1.

Let $K$ be a number field and $E/K$ an elliptic curve with
$|E(K)_{\text{tor}}|=m.$ Let $d\in K$ be a non-square. If $H$ is a subgroup of
$E^{d}(K)$ of odd order $n,$ then $E$ has a $K$-rational cyclic $mn$-isogeny
pointwise defined over $K(\sqrt{d})).$

###### Proof.

With respect to a Weierstrass form fixed for $E,$ there is an isomorphism
$\tau:E^{d}\rightarrow E$ defined by $\tau((x,y))=(x,\sqrt{d}y).$ By
assumption, $H$ is a subgroup pointwise defined over $K,$ so the image
$\tau(H)$ of $H$ forms a Gal$(\overline{K}/K)$-invariant subgroup of
$E(K(\sqrt{d}))$ of odd order $n.$ On the other hand, by Lemma 2, $\tau(H)\cap
E(K)_{\text{tor}}$ is trivial, hence $G:=\tau(H)\oplus E(K)_{\text{tor}}$ is
isomorphic to a Gal$(\overline{K}/K)$-invariant subgroup of $E(K(\sqrt{d})).$
Therefore, $G$ is a $K$-rational cyclic $mn$-isogeny of $E$ pointwise defined
over $K(\sqrt{d})$ as desired. ∎

With the analysis of certain isogenies in this section, our goal is to prove
the following.

######  Theorem 3.1.

Let $K$ be in $\mathcal{S}$ and let $E/K$ be an elliptic curve. Then, $E$ has
no $K$-rational cyclic $N$-isogeny pointwise defined over a quadratic
extension of $K$ for $N=30,40$ or $48.$

## $N=30$

The modular curve $X_{0}(30)$ is a hyperelliptic curve of genus 3 with the
model

(3) $\displaystyle
y^{2}+(-x^{4}-x^{3}-x^{2})y=3x^{7}+19x^{6}+60x^{5}+110x^{4}+121x^{3}+79x^{2}+28x+4.$

which can be simplified as

(4) $\displaystyle y^{2}=f_{30}(x)$
$\displaystyle=x^{8}+14x^{7}+79x^{6}+242x^{5}+441x^{4}+484x^{3}+316x^{2}+112x+16.$

Magma computes that $X_{0}(30)$ has 8 cusps which are all rationals. Write
$J_{0}(30)$ for the Jacobian of $X_{0}(30).$ By [BN15, Proposition 3],
$J_{0}(30)(\mathbb{Q})\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}.$
Note that there are only two points at infinity on $X_{0}(30)$ which are
$\infty_{+}=[1,1,0]$ and $\infty_{-}=[1,0,0].$

Let $P\in X_{0}(30)(K)$ with $x(P)\notin\mathbb{Q}$ and let $\overline{P}$
denote its Galois conjugate. Then $P+\overline{P}-\infty_{+}-\infty_{-}$ is a
rational divisor of degree 0 and hence it defines an element in
$J_{0}(30)(\mathbb{Q}).$ It follows that $P+\overline{P}$ is linearly
equivalent to $D+\infty_{+}+\infty_{-}$ where $D$ is a divisor of degree 0
representing one of the 192 elements in $J_{0}(30)(\mathbb{Q}).$ As
$X_{0}(30)$ is hyperelliptic, for $[D]\neq 0,$ the divisor
$D+\infty_{+}+\infty_{-}$ is linearly equivalent to at most one effective
divisor of degree 2. By [BN15, Table 6], the only such divisors are sums of
two rational points, except the following quadratic points with respect to the
model (3):

$\big{(}\frac{-\sqrt{5}-3}{2},4\sqrt{5}+9\big{)},\big{(}\frac{1}{2}(\pm
w-3),\mp w-3\big{)},\big{(}\frac{1}{2}(\pm w-3),\frac{1}{2}(\pm w-15)\big{)},$
$\big{(}-\sqrt{5}-3,71\sqrt{5}+159\big{)},\big{(}\frac{1}{4}(\pm
w-3),\frac{1}{16}(\mp 5w-9)\big{)},\big{(}\frac{1}{4}(\pm
w-3),\frac{1}{32}(\pm 5w+9)\big{)}$

where $w=\sqrt{-7}$. A quick computation in Magma shows that the minimum
extension of $\mathbb{Q}(\sqrt{-7})$ in which these points produce a cyclic
30-isogeny has degree 4.

It leaves the case $x(P)\in\mathbb{Q}.$ Then $(x(P),\pm\sqrt{f_{30}(x)}),$
lies on the $d$-twist $X^{d}_{0}(30)$ where $K=\mathbb{Q}(\sqrt{d}).$ If
$K\in\mathcal{S}$, we have $d=-p$ where $p$ belongs to the set
$\\{2,7,11,19,43,67,163\\}.$ By [Ozm12, Corollary 3.5], this twist does not
have $\mathbb{Q}_{p}$-rational points, therefore this leaves alone the
rational cusps.

##  $N=40$ and $N=48$

The modular curves $X_{0}(40)$ and $X_{0}(48)$ are hyperelliptic curves of
genus 3 and admit the following equations by [GR91]:

$\displaystyle y^{2}$ $\displaystyle=f_{40}(x)=x^{8}+8x^{6}-2x^{4}+8x^{2}+1$
$\displaystyle y^{2}$ $\displaystyle=f_{48}(x)=x^{8}+14x^{4}+1.$

$X_{0}(40)$ has only rational cusps whereas $X_{0}(48)$ has quadratic cusps
defined over $\mathbb{Q}(\sqrt{-1}).$ Let $N=40$ or $48$ and let $P$ be a
quadratic point on $X_{0}(N)$ such that $P=(x,\pm\sqrt{f_{N}(x))}$ with
$x\in\mathbb{Q}.$ Since the polynomial $f_{N}(x)\ \textgreater\ 0$ for all
$x\in\mathbb{R},$ then $P$ must be defined over a real quadratic field, from
which we conclude that there are no such quadratic points on $X_{0}(N)(K)$ for
any $K\in\mathcal{S}.$ On the other hand, the quadratic points with a non-
rational $x$-coordinate are completely classified in [BN15, Table 11,15]. All
such quadratic points of $X_{0}(40)$ are defined over $\mathbb{Q}(\sqrt{-1})$.
By Magma each gives rise to a cyclic 40-isogeny pointwise defined in an
extension of $\mathbb{Q}(\sqrt{-1})$ of degree at least 8. In case
$X_{0}(48)$, all such quadratic points are defined over
$\mathbb{Q}(\sqrt{-1})$ but they are entirely cuspidal points. Therefore,
there are no $K$-rational cyclic $40$ and $48$-isogeny for any $K$ in
$\mathcal{S}.$

##  4\. Classification of Twists

One benefit of Theorem 2.2 is that any elliptic curve $E/K$ with full
2-torsion can be defined by a model
$E(\alpha,\beta):y^{2}=x(x+\alpha)(x+\beta)$ where
$\alpha,\beta\in\mathcal{O}_{K}$. Note that $E(\alpha,\beta)$ is isomorphic to
$E(-\alpha,\beta-\alpha)$ and $E(\beta,\alpha-\beta)$ by the isomorphisms
$(x,y)\mapsto(x+\alpha,y)\ \text{and}\ (x,y)\mapsto(x+\beta,y)$ respectively.
Moreover, we may assume gcd$(\alpha,\beta)$ to be square-free in $K$ since
$y^{2}=x(x+z^{2}\alpha)(x+z^{2}\beta)$ is isomorphic to $E(\alpha,\beta)$ by
replacing $y$ by $y/z^{3}$ and $x$ by $x/z^{2}.$ Note that the $d$-quadratic
twist $E^{d}$ of $E$ is given by $E^{d}:y^{2}=x(x+d\alpha)(x+d\beta)$ where
$d$ is a non-square in $K$ and in particular $E^{d}(K)$ contains full
2-torsion. Throughout this section, we classify the possible torsion groups
that occur as $E^{d}(K)_{\text{tor}}$.

Before proceeding further, we need a technical lemma which is the variant of
[New17, Proposition 14]. We prove without using Grobner basis so that the
proof is reversible to recover examples of elliptic curves with certain
torsion structures.

######  Lemma 3.

Let $K=\mathbb{Q}(\sqrt{D})\in\mathcal{T}$ and let
$a,b,a_{0},b_{0},c_{0},d\in\mathcal{O}_{K}$ with $d$ non-square in $K$ and
$\frac{a}{b},\frac{a_{0}}{b_{0}}\notin\\{-2,-1,-\frac{1}{2},0,1\\}$ Then the
following system of equations have no solutions in $\mathcal{O}_{K}$

(5)
$\displaystyle\begin{cases}da^{3}(a+2b)=a_{0}^{3}(a_{0}+2b_{0})c_{0}^{2}\\\
db^{3}(b+2a)=b_{0}^{3}(b_{0}+2a_{0})c_{0}^{2}\end{cases}$

when $D=-1,-2,-3$. Otherwise, the system has infinitely many solutions except
for $K=\mathbb{Q}(\sqrt{-163}).$

###### Proof.

Suppose that there exists a solution to the system of equations in (5)
satisfying the hypothesis. Note that $a+2b,a_{0}+2b_{0}\neq 0,$ so dividing
the equations yields

$\displaystyle\left(\frac{a}{b}\right)^{3}\frac{a+2b}{b+2a}=\left(\frac{a_{0}}{b_{0}}\right)^{3}\frac{a_{0}+2b_{0}}{b_{0}+2a_{0}}$

Substituting $x:=\frac{a_{0}}{b_{0}}$ and $y:=\frac{a}{b}$ gives

$y^{3}\frac{y+2}{2y+1}=x^{3}\frac{x+2}{2x+1}$
$2xy^{4}+y^{4}+2xy^{3}+2y^{3}-2x^{4}y-x^{4}-4yx^{3}-2x^{3}=0$

The polynomial on the left hand side can be split into irreducible polynomials
as follows:

(6)
$\displaystyle(x-y)(2x^{3}y+2xy^{3}+x^{3}+y^{3}+5x^{2}y+5xy^{2}+2x^{2}+2y^{2}+2x^{2}y^{2}+2xy)$

If $x=y$ then $a=kb$ and $a_{0}=kb_{0}$ for some $k\in K.$ Plugging this in
the system (5) implies $d$ is a square, a contradiction. Hence, $x\neq y$ with
$x,y\notin\\{-2,-1,-1/2,0,1\\}.$ Then, $(x,y)$ is a $K$-rational point lying
on the curve $C$ defined by $p_{C}(x,y)=0$ where $p_{C}(x,y)$ denotes the
second factor of (6). Write $\overline{C}$ for the projective closure of $C$.
Since $\overline{C}(K)$ differs from $C(K)$ at only finitely many points, it
suffices to show the existence of a $K$-rational point
$(x,y)\not\in\mathcal{E}$ on $\overline{C}.$ By Magma, $\overline{C}$ is
birationally equivalent (over $\mathbb{Q}$) to the elliptic curve $E_{C}$
defined by

$y^{2}+2xy+2y=x^{3}-x^{2}-2x$

with the map $\phi$ given by

$\displaystyle\phi:\overline{C}$ $\displaystyle\rightarrow E_{C}$
$\displaystyle[x,y,z]$
$\displaystyle\mapsto[p_{1}^{\phi},p_{2}^{\phi},p_{3}^{\phi}]$

where

$\displaystyle p_{1}^{\phi}$
$\displaystyle=2x^{2}y^{2}+3x^{2}yz+4xy^{2}z-y^{3}z+x^{2}z^{2}+x^{2}z^{2}+6xyz^{2}-y^{2}z^{2}+2xz^{3}$
$\displaystyle p_{2}^{\phi}$
$\displaystyle=2x^{2}y^{2}+4xy^{3}-x^{2}yz+10xy^{2}z+3y^{3}z-x^{2}z^{2}+7y^{2}z^{2}-2xz^{3}+2yz^{3}$
$\displaystyle p_{3}^{\phi}$ $\displaystyle=y^{4}+3y^{3}z+3y^{2}z^{2}+yz^{3}$

Since $\overline{C}(K)$ maps to $E_{C}(K)$, the only potential points in
$\overline{C}(K)$ belong to the union of $\phi^{-1}(E_{C}(K))$ and the set of
non-regular points of $\phi.$ So $\overline{C}(K)$ and $E_{C}(K)$ must differ
at most finitely many points. If $z=0$ and $p_{3}^{\phi}(x,y,z)=0$ 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 $x\in K.$ Setting
$p_{2}^{\phi}(x,y,z)=0$ narrows down the set of non-regular points to
$\\{[1,0,0],[0,0,1],[-2,0,1],[-1,-1,1]\\}.$

One can quickly observe that $[1,0,0]$ does not correspond to a point on $C$
and the other points fail to recover points $(x,y)$ on $C$ with
$x,y\notin\\{-2,-1,-\frac{1}{2},0,1\\}.$ Hence the only non-regular points of
$\phi$ are $[0,0,1],[-1,-1,1]$ which are in fact all the singular points of
$C$. In order to determine $\phi^{-1}(E_{C}(K))$, Magma computes $E_{C}(K)$
has rank 0 for only $D=-1,-2,-3.$ As a result, $E_{C}(K)$ has infinitely many
points for $D=-7,-11,-19,-43,-67$ so does $\overline{C}(K).$ Note that Magma
is not able to determine the rank of $E(\mathbb{Q}(\sqrt{-163})).$
Equivalently the system (5) has infinitely many solutions in
$\mathcal{O}_{K}$, as claimed.

To compute $\phi^{-1}([x,y,z])$, we use the birational inverse $\psi$ of
$\phi$ defined by

$\displaystyle\psi:E_{C}$ $\displaystyle\rightarrow\overline{C}$
$\displaystyle[x,y,z]$
$\displaystyle\rightarrow[p_{1}^{\psi},p_{2}^{\psi},p_{3}^{\psi}]$

such that

$\displaystyle p_{1}^{\psi}$
$\displaystyle=x^{4}+2x^{3}y-4xy^{2}z-2y^{3}z-3x^{2}z^{2}-y^{2}z^{2}-2xz^{3}+2yz^{3}$
$\displaystyle p_{2}^{\psi}$
$\displaystyle=-x^{4}-8x^{3}z-4x^{2}yz-21x^{2}z^{2}-14xyz^{2}-3y^{2}z^{2}-22xz^{3}-10yz^{3}-8z^{4}$
$\displaystyle p_{3}^{\psi}$
$\displaystyle=x^{4}+6x^{3}z+2x^{2}yz+9x^{2}z^{2}-2xyz^{2}-3y^{2}z^{2}+4xz^{3}-4yz^{3}$

and its non-regular points are $\\{[0,1,0],[0,0,1],[-1,0,1],[2,-6,1]\\}.$ A
quick computation in Magma shows that each point of
$\phi^{-1}(E_{C}(K)_{\text{tor}})$ either doesn't correspond to a point on
$C$, or fails to satisfy $x,y\notin\\{-2,-1,\frac{-1}{2},0,1\\}.$ Therefore,
we have proved that $C(K)$ has no points $(x,y)$ with
$x,y\not\in\\{-2,-1,\frac{-1}{2},0,1\\}$ when $D=-1,-2,-3.$ See Table 8.

Table 8. Determination of $\phi^{-1}(E_{C}(K)_{\text{tor}})$ $P\in E_{C}(K)$ |  $\phi^{-1}(P)$
---|---
[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] |  $\emptyset$
[2,0,1] |  [0,-2,1]
[-1-$\sqrt{-3}$,-3+$\sqrt{-3}$,1] |  $[\sqrt{-3}-1,2,0]$ ($\notin C(K)$)
[-1+$\sqrt{-3}$,-3-$\sqrt{{}_{-}3}$,1] |  $[-\sqrt{-3}-1,2,0]$ ($\notin C(K)$)
[-1-$\sqrt{-3}$,3+$\sqrt{-3}$,1] |  [0,0,-1]
[-1+$\sqrt{-3}$,3-$\sqrt{-3}$,1] |  $[0,0,-1]$
[$\frac{1}{2}$(1-$\sqrt{-3}$),$\frac{1}{2}$(-3+$\sqrt{-3}$),1] |  $[-1,-1,1]$
[$\frac{1}{2}$(1+$\sqrt{-3}$),$\frac{1}{2}$(-3-$\sqrt{-3}$),1] |  $[-1,-1,1]$
[-2,1-$\sqrt{-7},1]$ |  $[\frac{1}{8}(-3\sqrt{-7}-1),\frac{1}{4}(\sqrt{-7}+3),1]$
$[\frac{1}{25}\sqrt{-11}-\frac{17}{25},-\frac{9}{125}\sqrt{-11}-\frac{147}{125},1]$ |  $[\frac{1}{24}(\sqrt{-11}-43),\frac{1}{24}(19-\sqrt{-11}),1]$
$[-\frac{25}{9},\frac{28}{27}\sqrt{-19}+\frac{16}{9},1]$ |  $[(\frac{1}{21457}(4536\sqrt{-19}+8335),\frac{1}{12475}(-2688\sqrt{-19}+4283),1)]$

∎

We can work backwards in the proof of Lemma 3 to produce examples of elliptic
curves over certain fields. Since the coefficients become overwhelmingly long
for the remaining $K$'s, we didn't include them in Table 4.

######  Theorem 4.1.

Let $K=\mathbb{Q}(\sqrt{D})\in\mathcal{S}$ and $E/K$ be an elliptic curve with
$E(K)[2]=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ We may assume
that $E$ is in the form $y^{2}=x(x+\alpha)(x+\beta)$ for some
$\alpha,\beta\in\mathcal{O}_{K}.$ Then,

  1. (1)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\\}.$

  2. (2)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\\}.$

  3. (3)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$ when

    1. i.

$D\neq-7$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\\}.$

    2. ii.

$D=-7$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}\\}.$

  4. (4)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$ when

    1. i.

$D=-2$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\\}.$

    2. ii.

$D=-7,-11$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}\\}.$

    3. iii.

$D=-19,-43,-67,-163$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}\\}.$

  5. (5)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ when

    1. i.

$D=-2,-11$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\\}.$

    2. ii.

$D=-7$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}\\}.$

    3. iii.

$D=-19,-43,-67,-163$ then
$T_{K}(G)=\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}\\}.$

  6. (6)

If $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ then
$T_{K}(G)\subseteq\\{\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z}:n=1,2,3,4,5,6\\}.$

###### Proof.

Note that the possible torsion structures appearing as $E^{d}(K)_{\text{tor}}$
are $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z}$ where $1\leq n\leq
6.$

(1) Suppose
$E(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$
Notice that $\mathbb{Z}/3\mathbb{Z}\nsubseteq E^{d}(K)_{\text{tor}}$ follows
from Theorem [BN16, Theorem 7]. We also have $\mathbb{Z}/5\mathbb{Z}\nsubseteq
E^{d}(K)_{\text{tor}},$ otherwise together with a $K$-rational $6$-torsion
point, $E$ would have a $K$-rational cyclic 30-isogeny pointwise defined over
$K(\sqrt{d})$ due to Proposition 3.1. But, this is impossible by Theorem 3.1.
Finally, if $\mathbb{Z}/{8}\mathbb{Z}\subseteq E^{d}(K)_{\text{tor}}$ then
replacing $E$ by $E^{d}$ in Proposition 3.1 if necessary, we obtain that $E$
has a $K$-rational cyclic isogenies of order $2$ and $24$. From [Naj16, Lemma
7], $E$ is $K$-isogenous to an elliptic curve $E^{\prime}/K$ with a cyclic
$48$-isogeny, contradicting Theorem 3.1.

(2) Suppose
$E(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}.$
We will now eliminate the cases $n=2,3,4,5$.

Case $n=3.$ Assume $E^{d}(K)_{\text{tor}}$ has a point of order $3$. Since
$E(K)_{\text{tor}}$ has a $10$-torsion point and by Proposition 3.1, there
exists a $K$-rational cyclic $30$-isogeny of $E$ pointwise defined over
$K(\sqrt{d}).$ However, this cannot happen by Theorem 3.1.

Case $n\in\\{2,4\\}.$ In both cases, we have $E^{d}(K)_{\text{tor}}$ has a
$4$-torsion point. Since
$E(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}$
by assumption, replacing $E$ with $E^{d}$ if necessary in Proposition 3.1, we
can assume that $E$ has two independent $K$-rational cyclic isogenies of order
2 and 20. From [Naj16, Lemma 7], we conclude that $E$ is isogenous over $K$ to
an elliptic curve $E^{\prime}/K$ with a cyclic 40-isogeny, contradicting
Theorem 3.1.

Case $n=5.$ Assume that $E^{d}(K)_{\text{tor}}$ has a point of order $5$, then
together with a $5$-torsion point in $E(K)$ and Lemma 2, the full $5$-torsion
lies in $E(K(\sqrt{d}))_{\text{tor}},$ which implies
$K(\sqrt{d})=\mathbb{Q}(\xi_{5}).$ However, we know
Gal$(\mathbb{Q}(\xi_{5})/\mathbb{Q})=\mathbb{Z}/{4}\mathbb{Z}$ and so every
intermediate field of $\mathbb{Q}(\xi_{5})$ is totally real, a contradiction.

(3) Suppose
$E(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$.
Without loss of generality, we may assume there exists a triple
$(x,y,z)\in\mathcal{O}_{K}$ with $x^{2}+y^{2}=z^{2}$, such that $\alpha=x^{4}$
and $\beta=y^{4}$. If $E^{d}(K)_{\text{tor}}$ has a point of order 4, then by
Theorem 2.2 one of the following is satisfied:

  1. (i)

$dx^{4},$ $dy^{4}$ are both squares

  2. (ii)

$-dx^{4},$ $dy^{4}-dx^{4}$ are both squares

  3. (iii)

$-dy^{4},$ $dx^{4}-dy^{4}$ are both squares.

In case (i), then $d$ is a square, a contradiction. In case (ii) or (iii), it
follows that $d=-u^{2}$ for some $u\in\mathcal{O}_{K}.$ Since
$dy^{4}-dx^{4}=\pm w^{2}$ for some $w\in\mathcal{O}_{K},$ then dividing by $d$
yields

$y^{4}-x^{4}=\mp(wu^{-1})^{2}$

and multiplying by $-1$ if necessary, we obtain $x^{4}-y^{4}=v^{2}$ for some
$v\in\mathcal{O}_{K}.$ Note that $x^{4},y^{4},x^{4}-y^{4}\neq 0$ since $E$ is
smooth, so $x,y,v\neq 0$. By [New17, Proposition 11], there is no nontrivial
solution to $x^{4}-y^{4}=v^{2}$ unless $K=\mathbb{Q}(\sqrt{-7})$. Therefore,
$E^{d}(K)$ does not have a point of order 4.

If $E^{d}(K)_{\text{tor}}$ has a point of order 3, then $E(K(\sqrt{d}))$
contains the group $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}$.
Furthermore, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/{24}\mathbb{Z}$ form two
independent cyclic isogenies of $E$ by Proposition 3.1, which, in turn, yields
a cyclic 48-isogeny of an elliptic curve $E^{\prime}/K$ by [Naj16, Lemma 7].
But this contradicts Theorem 3.1.

If $E^{d}(K)$ has a point of order 5 then along with a $K$-rational
$8$-torsion point on $E$, there exists a cyclic $40$-isogeny of $E$ which is
defined over $K(\sqrt{d})).$ But this cannot happen by Theorem 3.1. See Table
7, when $K=\mathbb{Q}(\sqrt{-7}).$

(4) Suppose
$E(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$.
Then, we may assume there exists $a,b,c\in\mathcal{O}_{K}$ with $c\neq 0$ and
$\frac{a}{b}\not\in\\{-2,-1,-\frac{1}{2},0,1\\}$ such that
$\alpha=a^{3}(a+2b)c^{2}$ and $\beta=b^{3}(b+2a)c^{2}$. Furthermore, choosing
$E^{c^{2}}$ instead of $E,$ we may assume $\alpha=a^{3}(a+2b)$ and
$\beta=b^{3}(b+2a).$ By way of contradiction, suppose
$\mathbb{Z}/4\mathbb{Z}\subseteq E^{d}(K)_{\text{tor}}$. By Theorem 2.2, we
have

  1. (i)

$d\alpha,d\beta$ are both squares

  2. (ii)

$-d\alpha,d\beta-d\alpha$ are both squares

  3. (iii)

$-d\beta,d\alpha-d\beta$ are both squares.

With no loss of generality, we assume (i) holds, i.e.
$d\alpha=s^{2},d\beta=t^{2}$ for some $s,t\in\mathcal{O}_{K}$. Note that
$s,t\neq 0,s^{2}\neq t^{2}.$ Then

$s^{2}=d\alpha=da^{3}(a+2b)\ \text{and}\ \ t^{2}=d\beta=db^{3}(b+2a)$

and multiplying by $d$:

$ds^{2}=a^{3}(a+2b)d^{2}\ \text{and}\ \ dt^{2}=b^{3}(b+2a)d^{2}$

Taking the products and dividing both sides by $d^{4}a^{2}b^{6}$ along with
substituting $z:=\frac{a}{b}$ yield

$\left(\frac{st}{dab^{3}}\right)^{2}=z(z+2)(2z+1)$

Multiplying by $4$ gives
$\left(\frac{2st}{dab^{3}}\right)^{2}=2z(2z+4)(2z+1).$ So that we obtain a
$K$-rational point on the elliptic curve $E_{0}$ defined by

$y^{2}=x(x+4)(x+1)=x^{3}+5x^{2}+4x$

By Magma,
$E_{0}(K)\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}=\\{\infty,(0,0),(-4,0),(-1,0),(-2,\pm
2),(2,\pm 6)\\}$ when $D=-2,-7,-11,-163.$ This shows $2z=-4,-2,-1,0$ or $2,$
and so $z\in\\{-2,-1,-\frac{1}{2},0,1\\}$, contradicting the choice of $z.$ On
the other hand,
$E_{0}(K)\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}$
when $D=-19,-43,-67.$ For example, the extra points on $E_{0}$ in case $D=-19$
give rise to growth of torsion, see Table 6.

Next, we claim that $\mathbb{Z}/5\mathbb{Z}\nsubseteq E^{d}(K)_{\text{tor}},$
otherwise as $\mathbb{Z}/6\mathbb{Z}\subseteq E(K)$, there would be a cyclic
30-isogeny pointwise rational over $K(\sqrt{d}),$ which is impossible by
Theorem 3.1. Also, if $\mathbb{Z}/{12}\mathbb{Z}\subseteq
E^{d}(K)_{\text{tor}},$ then by Lemma 2, we have
$\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}\subseteq
E(K(\sqrt{d}))$ which is excluded by [BN16, Theorem 7].

For $E^{d}(K)_{\text{tor}}$ to have a $3$-torsion point, there must exist
$a_{0},b_{0},c_{0}\in\mathcal{O}_{K}$ and
$\frac{a_{0}}{b_{0}}\notin\\{-2,-1,\frac{-1}{2},0,1\\}$ such that

$\displaystyle d\alpha=a_{0}^{3}(a_{0}+2b_{0})c_{0}^{2},\ \
d\beta=b_{0}^{3}(b_{0}+2a_{0})c_{0}^{2}$

Note that $E(K(\sqrt{d}))\simeq E^{d}(K(\sqrt{d}))$ via
$f:(x,y)\mapsto(dx,d^{\frac{3}{2}}y).$ Since the $3$-torsion point $P$ in
$E(K)_{\text{tor}}$ corresponds to the $3$-torsion point $f(P)$ in
$E^{d}(K(\sqrt{d}))_{\text{tor}}$ (which is not in $K$) and the order
$3$-point $Q$ in $E^{d}(K)_{\text{tor}}$ corresponds to a point $f^{-1}(Q)$ of
the same order in $E(K(\sqrt{d}))_{\text{tor}}$, then
$E(K(\sqrt{d}))_{\text{tor}}$ contains full 3-torsion. By the Weil paiting,
$K(\sqrt{d})$ must have a primitive 3th root of unity, which implies $d=-3$.
So, we have

$-3a^{3}(a+2b)=a_{0}^{3}(a_{0}+2b_{0})c_{0}^{2},\ \
-3b^{3}(b+2a)=b_{0}^{3}(b_{0}+2a_{0})c_{0}^{2}$

Since $-3$ is a non-square in $K$, by Lemma 3 this system has solutions except
$K=\mathbb{Q}(\sqrt{-2})$ which establishes the assertion.

(5) Suppose
$E(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$.
For $E^{d}(K)_{\text{tor}}$ to have an order $4$-point, either (a)
$d\alpha,d\beta$ are both squares, (b) $-d\alpha,d\beta-d\alpha$ are both
squares, or (c) $-d\beta,d\alpha-d\beta$ are both squares. Since $d$ is a
nonsquare, (a) is impossible. Both (b) and (c) can be argued similarly so we
assume (b) holds. Then there exists $u,v\in\mathcal{O}_{K}$ such that
$-ds^{2}=u^{2}$ and $dt^{2}-ds^{2}=v^{2}$ which implies $d=-w^{2}$ for some
$w\in\mathcal{O}_{K}$. Plugging this into the second equation we obtain
$s^{2}-t^{2}=z^{2}$ for some $z\in\mathcal{O}_{K}.$ Note that $-1$ is not a
square in $\mathcal{O}_{K}$, otherwise $d$ is a square. As a result, if those
conditions are satisfied, then $E^{d}(K)_{\text{tor}}$ contains a point of
order $4$. (See Table 3). If $d=-1,$ by [New17, Lemma 17], we have
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq E(K(\sqrt{-1})).$
Then the possible groups for $E^{-1}(K)_{\text{tor}}$ are

$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/{2}\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}\
\text{or}\ \ \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$

with $K$ in $\mathcal{S}$. Arguing as in (3) implies that $E^{-1}$ has no
point of order 8 unless $K=\mathbb{Q}(\sqrt{-7}).$ One can take $E$ as
$E^{-1}(729,2304)$ (see Table 7). Also, if $\mathbb{Z}/{12}\mathbb{Z}\subseteq
E^{d}(K)_{\text{tor}},$ then in particular $E^{-1}(K)_{\text{tor}}$ has a
$3$-torsion point. Since
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq E(K(\sqrt{d})),$
we deduce that $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$ is a
subgroup of $E(K(\sqrt{-1}))_{\text{tor}}$ which is impossible by Theorem
[BN16, Theorem 7].

Assume that the conditions in the preceding argument are not fulfilled. This
leaves alone the nontrivial cases where
$E^{d}(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}$
or $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}.$ As replacing $E$
by $E^{d}$ in (4), the former group can only be realized when $D=-19,43,-67.$
For example, we may take $E(\frac{-\alpha}{3},\frac{-\beta}{3})$ in Table 6.
The later group is ruled out by the case $n=4$ in the proof of (2) above.

∎

##  5\. Classification of Torsion

Prior to the proof of the main theorem, we first prove a technical lemma.

######  Lemma 4.

Let $\alpha\in K\in\mathcal{S}.$ If $\sqrt{\alpha}$ (or $-\sqrt{\alpha})$ is a
square in a quadratic extension $L$ of $K$ then $\alpha$ or $-\alpha$ is a
square in $K.$

###### Proof.

Suppose $[L:K]=2$ such that either $\sqrt{\alpha}=z^{2}$
($\sqrt{\alpha}=-z^{2})$ for some $z\in L.$ Then $z$ is a root of the
polynomial $f(x):=x^{4}-\alpha$ defined over $K.$ If $f(x)$ is irreducible
over $K,$ then if we let $K_{f}$ be the splitting field of $f,$ it follows
that $[K_{f}:K]=4.$ Since $f$ has least one root in $L$ and $L/K$ is Galois,
$f(x)$ must split over $L$ which in fact means that $L=K_{f}$ and so $L$ has
degree 4 over $K$ 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 write $f$ as a product
of two polynomials:

$f(x):=x^{4}-\alpha=(x^{2}+ax+b)(x^{2}+cx+d)$

for $a,b,c,d\in K.$ Then $ac+b+d=0,$ $ad+bc=0$ and $a+c=0.$ Replacing $a$ by
$-c$ gives $c(b-d)=0.$ So either $c=0$ or $b=d.$ If $c=0$ then $b=-d$ and so
$bd=-\alpha$ which implies that $\alpha=b^{2}$ in $K.$ If $b=d,$ then
$\alpha=-bd,$ so $-\alpha$ is a square in $K.$ ∎

### Proof of Theorem 1.1

Let $K\in\mathcal{S}$ and $G\in\phi_{K}(1).$ We want to determine the set
$\Phi_{K}(2,G).$ If $G$ doesn't grow upon quadratic base change, then we are
done. Let us assume there exists at least one quadratic extension $L$ of $K$
such that $G\varsubsetneq E(L)_{\text{tor}}.$

$(1)$ Suppose $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$
By Proposition 2.1 and Theorem 4.1, $E(L)_{\text{tor}}$ is isomorphic to one
of the following groups

$\displaystyle\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{48}\mathbb{Z},\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}\
\text{or}\ \mathbb{Z}/{4}\mathbb{Z}\oplus\mathbb{Z}/{48}\mathbb{Z}.$

Since [BN16, Theorem 7] rules out the subgroup
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$, this eliminates all
the these groups except $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/24\mathbb{Z}$
and $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/48\mathbb{Z}.$ If
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/24\mathbb{Z}$
then fix a $2$-torsion point $Q$ and a $24$-torsion point $P$ generating
$E(L)_{\text{tor}}.$ In particular, we have $E(K)[2]=\langle Q,12P\rangle.$
Let $\sigma$ be the non-trivial element of Gal$(L/K)$. Notice that $\sigma$
maps $P$ to $aP+Q$ or $aP$ with $a\in(\mathbb{Z}/24\mathbb{Z})^{\times}$.
Since $\sigma^{2}(P)=P$ and $\sigma(Q)=Q$, we have $\sigma(P)=aP$ for some
$a\in(\mathbb{Z}/24\mathbb{Z})^{\times}.$ Hence, $\langle Q\rangle,\langle
P\rangle\subseteq E(L)_{\text{tor}}$ form two independent $K$-rational cyclic
isogenies of $E$ with order $2$ and $24$. It follows by [Naj16, Lemma 7] that
$E$ is isogenous (over $K$) to an elliptic curve $E^{\prime}/K$ with a cyclic
$48$-isogeny, contradicting Theorem 3.1.

If
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/48\mathbb{Z},$
then $E$ has a $K$-rational cyclic $48$-isogeny, contradicting Theorem 3.1.
Therefore, we prove the assertion $(1).$

$(2)$ Suppose $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}.$
Again by Proposition 2.1 and Theorem 4.1, the only group structures that can
occur as $E(L)_{\text{tor}}$ are
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{20}\mathbb{Z}$ or
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{20}\mathbb{Z}.$

If
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{20}\mathbb{Z},$
then we fix a $2$-torsion point $Q$ and a $20$-torsion point $P$ in
$E(L)_{\text{tor}}$ such that $E(L)_{\text{tor}}=\langle Q,P\rangle$. Let
$\sigma$ be the non-trivial element of Gal$(L/K)$. Notice that $\sigma$ maps
$P$ to $aP+Q$ or $aP$ with $a\in(\mathbb{Z}/20\mathbb{Z})^{\times}.$ Since
$\sigma^{2}(P)=P$ and $\sigma(Q)=Q$, we have $\sigma(P)=aP$ for some
$a\in\\{1,7,11,19\\}.$ Hence, $\langle Q\rangle,\langle P\rangle\subseteq
E(L)_{\text{tor}}$ form two independent $K$-rational cyclic isogenies of $E$
with order $2$ and $20$. Following [Naj16, Lemma 7] again, $E$ is
$K$-isogenous to an elliptic curve $E^{\prime}/K$ with a cyclic $40$-isogeny,
contradicting Theorem 3.1.

If
$E(L)_{\text{tor}}\simeq\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/20\mathbb{Z}$,
then choose a $4$-torsion point $Q$ and a $20$-torsion point $P$ generating
$E(L)_{\text{tor}}.$ Since $E(K)$ has no $4$-torsion point, $\sigma$ maps $Q$
to $3Q,3Q+5P$ or $3Q+15P.$ If $\sigma(Q)=3Q+5P$ or $3Q+15P$, then by
Proposition 2.1, $Q-\sigma(Q)$ would be a $4$-torsion point on the $L$-twist
of $E$, which is impossible by Theorem 4.1. On the other hand, $\sigma$ maps
$P$ to $aP+bQ$ with $a\in\\{1,3,7,9,11,13,17,19\\}$ and $b\in\\{0,1,3\\}.$
From $\sigma^{2}(P)=P$ and $P-\sigma(P)$ can have order at most 2, we conclude
that $\sigma(P)=P$ or $11P$. Hence, $\langle 2Q\rangle,\langle
P\rangle\subseteq E(L)_{\text{tor}}$ must form two independent $K$-rational
cyclic isogenies of order $2$ and $20$. Then $E$ is $K$-isogenous to an
elliptic curve with a cyclic $40$-isogeny, which contradicts Theorem 3.1.

$(3)$ First, assume $K=\mathbb{Q}(\sqrt{-7})$ and
$G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$. By Theorem 4.1,
we have
$E^{d}(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{2n}\mathbb{Z}$
for $n=1,2$ or $4.$ Following Proposition 2.1, the torsion subgroup of $E(L)$
is in the form $\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}$ where

(7) $m=2,4\ \text{and}\ n=8,16,32,64\ \ \text{or}\ \ m=8\ \text{and}\ n=8,16.$

As shown in Table 7, the groups
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$ and
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}$ can be realized as
$E(L)_{\text{tor}}.$ On the other hand, [BN16, Theorem 7] excludes the
subgroups
$\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}$
from which we further rule out
$\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/16\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{32}\mathbb{Z}$
and $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{64}\mathbb{Z}$ from (7).

In order to argue the possibility of the groups
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{32}\mathbb{Z}$ and
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{64}\mathbb{Z},$ it suffices to show
that $\mathbb{Z}/{2}\mathbb{Z}\oplus\mathbb{Z}/{32}\mathbb{Z}\subsetneq
E(L)_{\text{tor}}.$ Suppose
$E(L)[32]\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{32}\mathbb{Z}.$ We
choose a point $Q$ of order 2 and a point $P$ of order 32 such that
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{32}\mathbb{Z}=\langle
Q,P\rangle\subseteq E(L)_{\text{tor}}.$ Then $E(K)_{\text{tor}}=\langle
Q,4P\rangle.$ The non-trivial element $\sigma$ of Gal($L/K)$ fixes $Q$ and
sends $P$ to a point of order 32, which must be of the form $aP$ or $Q+aP$
where $a$ is odd. Applying $\sigma$ to itself once again we see that

$P=\sigma^{2}(P)=a^{2}P$

which implies $a^{2}\equiv 1(\text{mod}\ 32)$ so $a$ can only be 1, 15, 17 or
31. For all eight possibilities of $\sigma(P),$ $\sigma$ maps $2P$ to either
$2P$ or $-2P.$ If $\sigma(2P)=2P$ then the point $2P$ which has order 16 would
be $K$-rational, contradicting the assumption. And $\sigma(2P)=-2P$ also
cannot be, otherwise the $K$-rational point $4P$ would also be mapped to its
inverse which gives a contradiction since $\sigma$ fixes $4P.$

Let $K$ be in $\mathcal{S}\backslash\\{\mathbb{Q}(\sqrt{-7})\\}$ and
$G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}.$ We may assume
$\alpha=u^{4},\beta=v^{4},$ where $(u,v,w)\in\mathcal{O}_{K}$ with
$u^{2}+v^{2}=w^{2}$ and $u,v$ are square-free integers in $\mathcal{O}_{K}.$
By Proposition 2.1 and Theorem 4.1, it follows that $E(L)_{\text{tor}}$ is
isomorphic to
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}$
or $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$

By [BN16, Theorem 7], it suffices to show that
$E(L)_{\text{tor}}\not\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$
By way of contradiction, assume
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$
Using the group law on an elliptic curve, one can obtain a point $P$ of order
8 in $E(K)$ such that

$P=(x,y)=(uv(u+w)(v+w),uvw(u+v)(v+w)(u+w))$

with $[4]P=(0,0)$ due to [Kwo97]. Then $E(K)_{\text{tor}}$ is generated by
$Q=(-u^{4},0)$ of order 2 and $P$ of order 8. Notice that
$P^{\prime}=(x^{\prime},y^{\prime})=Q+P$ is another point of order 8 such that

$P^{\prime}=(x^{\prime},y^{\prime})=(uv(u+w)(v-w),uvw(u-v)(v-w)(u+w)).$

Since $E(K)_{\text{tor}}$ has only two distinct cyclic subgroups of order 8,
one is generated by $P$ and the other is generated by $P^{\prime}.$ By
assumption, if $P_{0}\in E(L)_{\text{tor}}$ has order 16 then $[2]P_{0}\in
E(K)_{\text{tor}}$ has order 8. Hence either $[2]P_{0}=P$ or
$[2]P_{0}=P^{\prime}$ holds.

Suppose $[2]P_{0}=P.$ By Lemma 1, $x,x+u^{4}$ or $x+v^{4}$ must be squares in
$L.$ Since $x,u,v$ are in $K$ and satisfy the curve equation, at least one of
them must be a square in $K.$ If $x$ is a square then we have $u,v,w,s\in K$
such that

(8) $\displaystyle s^{2}=x=uv(u+w)(v+w),\ \ u^{2}+v^{2}=w^{2}$

Since the curve $C$ defined by $u^{2}+v^{2}=w^{2}$ is of genus 0, the solution
set is parametrized by $\\{P\in C(K):P=(2t,t^{2}-1,t^{2}+1)\ \text{for\ some}\
t\in K\\}.$ Note that $t\neq 0,\pm 1$ since our curve is smooth. Substituting
the parametrization in (8) yields to the following curve

$C^{\prime}:s^{2}=(2t)(t^{2}-1)(2t+t^{2}+1)(t^{2}-1+t^{2}+1)=4t^{3}(t+1)^{2}(t^{2}-1)$

which is birational to $E_{0}:=y^{2}=x^{3}-x$ via
$(t,s)\mapsto(t,\frac{s}{2t^{2}+2t}).$ By Magma,
$E_{0}(K)\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}=\\{\infty,(0,0),(\pm
1,0)\\}$ for all $K\in\mathcal{S}\backslash\\{\mathbb{Q}(\sqrt{-7})\\}.$ Hence
all rational points satify $t=0,\pm 1,$ a contradiction. If $x+u^{4}$ is a
square then there exists $u,v,w,s\in K$ such that

(9) $\displaystyle s^{2}=x+u^{4}=uw(u+v)(v+w),\ \ u^{2}+v^{2}=w^{2}$

Arguing as above we obtain a curve defined by

$C^{\prime\prime}:s^{2}=(2t)(t^{2}+1)(2t+t^{2}-1)(t^{2}-1+t^{2}+1)=4t^{3}(t^{2}+1)(t^{2}+2t-1)$

which is a hyperelliptic curve of genus 2. Let $J$ be the Jacabian of
$C^{\prime\prime}.$ For all $K\neq\mathbb{Q}(\sqrt{-7})$ in $\mathcal{S}$,
Magma computes that the rank of $J(K)$ is 0 and

(10) $\displaystyle J(K)_{\text{2-part}}$
$\displaystyle\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
$\displaystyle J(\mathbb{Q})$
$\displaystyle\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z}$

Our objective is to prove that $J(K)=J(\mathbb{Q}).$ It is equivalent to show
that $J(K)_{\text{tor}}$ has exactly size of 20.

$\textbf{Case}\ K=\mathbb{Q}(\sqrt{-2}).$

Since the discriminant of $K$ is -8, one can quickly observe that $3$ splits
and $5$ remains prime in $\mathcal{O}_{K}.$ Let $\mathfrak{p}$ denote a prime
above $3$ in $\mathcal{O}_{K}.$ Note that $C^{\prime\prime}$ remains smooth
mod $\mathfrak{p}$ and 5. A quick computation in Magma shows

$\displaystyle
J(\mathbb{F}_{3})\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z},\
\
J(\mathbb{F}_{25})\simeq(\mathbb{Z}/2\mathbb{Z})^{2}\oplus\mathbb{Z}/{4}\mathbb{Z}\oplus\mathbb{Z}/40\mathbb{Z}.$

Since the 3-part of $J(K)_{\text{tor}}$ injects into the group
$J(\mathbb{F}_{25})$ under reduction mod $5$, it is a trivial subgroup. On
other hand, the prime-to-3-part injects into the group $J(\mathbb{F}_{3})$
(modulo $\mathfrak{p}$), it is of size at most 20. As a result, we obtain
$|J(K)_{\text{tor}}|=20$.

$\textbf{Case}\ K=\mathbb{Q}(\sqrt{-11}).$

As the discriminant of K is $-11,$ it follows that both 3 and 5 split in
$\mathcal{O}_{K}.$ Let $\mathfrak{p}$ and $\mathfrak{q}$ be primes above $3$
and $5$ in $\mathcal{O}_{K}$ respectively. By Magma, we compute

$\displaystyle
J(\mathbb{F}_{3})\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{10}\mathbb{Z},\
\
J(\mathbb{F}_{5})\simeq(\mathbb{Z}/2\mathbb{Z})^{2}\oplus\mathbb{Z}/{10}\mathbb{Z}$

from which we conclude that $|J(K)_{\text{tor}}|\leq$ gcd$(20,400)=20.$
Therefore, $|J(K)_{\text{tor}}|=20.$

$\textbf{Case}\ K=\mathbb{Q}(\sqrt{-19}).$

Note that the discriminant of $K$ is -19. This implies that $3$ remains prime
and $5$ splits in $\mathcal{O}_{K}.$ We compute

$\displaystyle
J(\mathbb{F}_{9})\simeq(\mathbb{Z}/2\mathbb{Z})^{3}\oplus\mathbb{Z}/{10}\mathbb{Z},\
\
J(\mathbb{F}_{5})\simeq(\mathbb{Z}/2\mathbb{Z})^{2}\oplus\mathbb{Z}/{10}\mathbb{Z}$

Since the prime-to-5-part of $J(K)_{\text{tor}}$ injects into
$J(\mathbb{F}_{5}),$ its size divides $2^{3}.5.$ By (10) we know that
$J(K)_{\text{2-part}}$ is of size $2^{2}$. Hence the prime-to-5-part divides
20. On the other hand, under reduction mod $\mathfrak{p}$ (where
$\mathfrak{p}$ is a prime lying above 3) in $\mathcal{O}_{K},$ the 5-part of
$J(K)_{\text{tor}}$ injects into $J(\mathbb{F}_{9})$ 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 that $|J(K)_{tor}|=20.$ By a similar
argumentation above, the claim holds for $K=\mathbb{Q}(\sqrt{D})$ where
$D=-43,-67,-163$.

Note that the points $\infty_{+}=(1,1,0)$ and $\infty_{-}=(1,-1,0)$ lie on the
projective model of $C^{\prime\prime}.$ Let $P$ be in $C^{\prime\prime}(K)$
for $K\in\mathcal{S}\backslash\\{\mathbb{Q}(\sqrt{-7})\\}$. Then
$[P-\infty_{+}]$ represents a point in $J(K)$ which equals to $J(\mathbb{Q}).$
We enumerated the 20 coset representatives in Magma and found that any such
point $P$ belongs to the set $\\{(0,0,1),(0,-1,1),(-1,1,0)\\}.$ However, each
gives rise to a value of $u,v,w$ such that the smoothness of our curve fails.

If $x+v^{4}$ is a square, then it yields to the same system (9) but with $u$
and $v$ interchanged. Using the parametrization
$(u,v,w)=(t^{2}-1,2t,t^{2}+1),$ we obtain the same hyperelliptic curve as
above. Therefore, we exclude the case $[2]P_{0}=P.$

When $[2]P_{0}^{\prime}=P^{\prime}=(x^{\prime},y^{\prime}),$ using
$x^{\prime}=uv(u+w)(v-w)$ and $u^{2}+v^{2}=w^{2}$, we obtain

$x^{\prime}+u^{4}=uw(u-v)(w-v),\ \ x^{\prime}+v^{4}=vw(v-u)(w+u)$

In other words, $P^{\prime}$ is obtained from $P$ by substituting $-v$ for
$v.$ Hence the rest of the argument is exactly the same as in the previous
case $[2]P_{0}=P.$ Therefore, we prove that
$E(L)_{\text{tor}}\not\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$
The remaining possibility is when
$E(L)\simeq\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$. In particular,
we have $\mathbb{Z}/{4}\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq E(L).$
But this cannot happen by [New17, Lemma 16(i)].

$(4)$ Suppose $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z}.$ By
Theorem 4.1 and Proposition 2.1, we know that

Case (i): If $D=-2$ then the torsion groups which can appear as
$E(L)_{\text{tor}}$ are

(11) $\displaystyle\mathbb{Z}/{2}\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z},\
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z},\
\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$

Case (ii): If $D=-7,-11$, then either the ones in (11) or the following groups

(12) $\displaystyle\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{18}\mathbb{Z},\ $
$\displaystyle\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{36}\mathbb{Z},\
\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{18}\mathbb{Z},\
\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{36}\mathbb{Z},$
$\displaystyle\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/6\mathbb{Z},\
\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}.$

Case (iii): If $D=-19,-43,-67,-163$ then either the ones in (11),(12) or
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}.$

Table 6 shows an example of
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$.
The torsion groups $\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$ and
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$ are ruled out by
[BN16, Theorem 7] and so are
$\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z},$
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z},$ and
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{36}\mathbb{Z}$ which establishes the
first assertion.

If $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{18}\mathbb{Z}\subseteq
E(L)_{\text{tor}},$ then $E(L)[9]=\mathbb{Z}/9\mathbb{Z}.$ By Lemma 2, this
yields

$\mathbb{Z}/9\mathbb{Z}\simeq E(K)[9]\oplus
E^{d}(K)[9]=\mathbb{Z}/3\mathbb{Z}\oplus E^{d}(K)[9]$

But the right hand side of the isomorphism is a non-cyclic group whereas the
left hand side is not, a contradiction. As a result, we narrow down our list
(12) by removing $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{18}\mathbb{Z},\
\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{18}\mathbb{Z},\
\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{36}\mathbb{Z}.$

It remains to check if
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}$
when $D=-19,-43,-67,-163.$ It is enough to show that
$\mathbb{Z}/{24}\mathbb{Z}\not\subseteq E(L).$ By way of contradiction, assume
$E(L)$ has a cyclic subgroup of order $24$. Then, there exists an $L$-rational
point of order 8 on $E.$ With no loss of generality, we can assume that there
exists $s,t,z\in\mathcal{O}_{L}$ such that $\alpha=s^{4},\ \beta=t^{4}$ and
$s^{2}+t^{2}=z^{2}.$ It follows that either $\sqrt{\alpha}$ or
$-\sqrt{\alpha}$ is a square in $L.$ Applying Lemma 4 either $\alpha$ or
$-\alpha$ is a square in $K.$ As the same discussion applies to $\beta$,
either $\beta$ or $-\beta$ is a square in $K.$ Hence there are four
possibilities as follows

$\displaystyle(i)\ \alpha=s_{0}^{2}\ \text{and}\ \beta=t_{0}^{2},\ \ \ \
(iii)\ \alpha=s_{0}^{2}\ \text{and}\ \beta=-t_{0}^{2},$ $\displaystyle(ii)\
\alpha=-s_{0}^{2}\ \text{and}\ \beta=t_{0}^{2},\ (iv)\ \alpha=-s_{0}^{2}\
\text{and}\ \beta=-t_{0}^{2}.$

where $s_{0},t_{0}\neq 0$ in $K.$

Case $(i):$ Note that since $\mathcal{O}_{K}$ is integrally closed, if
$a\in\mathcal{O}_{K}$ is a square in $K$ then it is a square in
$\mathcal{O}_{K}.$ Thus, $\alpha$ and $\beta$ are both squares in
$\mathcal{O}_{K}.$ It follows that $E(K)$ has a point of order 4 by Theorem
2.2 which is contradicting the assumption.

Case $(ii):$ As $E(K)$ has a point of order 3, there are
$a,b,c\in\mathcal{O}_{K}$ with $\frac{a}{b}\notin\\{-2,-1,-1/2,0,1\\}$ and
$c\neq 0$ such that

$\alpha=a^{3}(a+2b)c^{2}\ \ \ \ \ \beta=b^{3}(b+2a)c^{2}$

again by Theorem 2.2. Together with the assumption $(ii),$ we obtain the
system

$-s_{0}^{2}=a^{3}(a+2b)c^{2}$ $t_{0}^{2}=b^{3}(b+2a)c^{2}$

Note that we must have $a,b,a+2b\neq 0$ so we may divide and multiply both
sides by -1

$\left(\frac{s_{0}}{t_{0}}\right)^{2}=-\left(\frac{a}{b}\right)^{3}\frac{a+2b}{b+2a}$

and letting $y=\frac{s_{0}}{t_{0}}$ and $x=\frac{a}{b}$ we get
$y^{2}=-x^{3}\frac{x+2}{2x+1}$ 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
$E_{0}(K)\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ when
$D=-19,-43,-67,-163.$ So $E_{0}(K)=\\{(1,0),(0,0),(4,0)\\}.$ But we already
noted that $\frac{a}{b}\neq 1$ and $s_{0}\neq 0,t_{0}\neq 0.$ 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 that $-1$ must be a square in $L,$
from which we conclude that $L=K(i)$ because $i\not\in K$. On the other hand,
we have

(13) $\displaystyle s^{4}=\alpha=-s_{0}^{2}\ \ \text{and}\ \
t^{4}=\beta=-t_{0}^{2}$

In particular, $s^{2}=is_{0}$ or $-is_{0}.$ Since the curve defined by
$s^{2}+t^{2}=z^{2}$ is parametrized by $(1-m^{2},2m,1+m^{2})$ over $L,$
replacing $t$ by $2m$ in (13), we obtain $m^{4}=-(\frac{t_{0}}{4})^{2}\in K.$
As either $s^{2}=m^{4}-2m^{2}+1=is_{0}$ or $-is_{0},$ we have

$4m^{4}=(-is_{0}+m^{4}+1)^{2}\ \text{or}\ 4m^{4}=(is_{0}+m^{4}+1)^{2}$

We can re-write the above as

$-\frac{t_{0}^{2}}{4}=\left(\pm is_{0}-\frac{t_{0}^{2}}{16}+1\right)^{2}.$

However, the left hand side is in $K$ whereas the right hand side is not which
gives a contradiction. Therefore, there is no cyclic subgroup of order 24 in
$E(L)_{\text{tor}}$ which rules out
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}$, completing the proof.

$(5)$ Suppose $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}.$
First, let $K\in\mathcal{S}\backslash\\{\mathbb{Q}(\sqrt{-7})\\}$ with
$\alpha=s^{2}$ and $\beta=t^{2}.$ If
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq
E(L)_{\text{tor}}$ then $L=K(i)$ by the Weil pairing. With no loss of
generality, let $s^{2}-t^{2}=z^{2}.$ Note that under these hypothesis
$E^{-1}(K)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$
as shown in the proof of $(5)$ in Theorem 4.1. By Proposition 2.1 the possible
torsion groups of $E(L)$ are
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},$
or $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$ It suffices to
prove that $E(L)$ has no point of order $8.$

As argued in [Kwo97], we consider the following three points of order 4:

$Q=(st,st(s+t)),\ \ Q^{\prime}=(-s(s+z),isz(s+z))$
$Q+Q^{\prime}=(-t(t+iz),-tz(t+iz)).$

Note $2Q=(0,0)$ and $2Q^{\prime}=(-s^{2},0)$ so $Q$ and $Q^{\prime}$ are
independent order 4 points and hence $\\{Q,Q^{\prime}\\}$ forms a basis for
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq
E(L)_{\text{tor}}.$ If there is an order $8$ point $P$ in $E(L)_{\text{tor}}$,
then $2P$ has order 4 and so $2P=mQ+nQ^{\prime}$ for some
$m,n\in\mathbb{Z}/4\mathbb{Z}.$ Adding $2Q$ or $2Q^{\prime}$ to both sides of
the equation if necessary, we may assume that there exists an order 8 point
$P$ in $E(L)_{\text{tor}}$ such that $2P=Q,$ $2P=Q^{\prime}$ or
$2P=Q+Q^{\prime}.$

If $Q$ is halved, then by Lemma 1, $st,s(s+t),t(s+t)$ are squares in $L.$
Since they lie in $K$ and $L=K(i)$, we have $st=\pm r^{2}$ for some $r.$
Assume $st=-r^{2}.$ On the other hand, we have the parametrization for
$s^{2}=t^{2}+z^{2}$ as $(s,t,z)=(1+m^{2},2m,1-m^{2})$ with the condition that
$m\neq 0,\pm 1$ because $E$ is non-singular. Then we obtain the following
equation

$-r^{2}=(1+m^{2})(2m)$

Multiplying both sides by 4, we get $(2r)^{2}=(4+(-2m))^{2}(-2m)$ which
implies there exists a $K$-rational point on the elliptic curve $E_{1}$ given
by the equation

$E_{1}:y^{2}=x^{3}+4x$

We compute $E_{1}(K)\simeq\mathbb{Z}/4\mathbb{Z}$ for all
$K\in\mathcal{S}\backslash\\{\mathbb{Q}(\sqrt{-7})\\}$ (of rank 1 over
$K=\mathbb{Q}(\sqrt{-7})$). The only points of $E_{1}(K)$ are $\\{(0,0),(2,\pm
4)\\}$ yielding $m=0$ or $-1$, a contradiction. If $st=r^{2}$, then using the
same argument above, we arrive at a contradiction.

If $Q^{\prime}$ is halved, then $-s(s+t),-sz,-2(s+z)$ are squares in $L.$ We
conclude as above that $-sz=\pm r^{2}$ in $K.$ If $-sz=-r^{2},$ then
substituting the parametrization of $s^{2}=t^{2}+z^{2}$ yields

$r^{2}=(1+m^{2})(1-m^{2})=1-m^{4}$

Multiplying both sides by $m^{2},$ we have $(mr)^{2}=(-m^{2})^{3}-(-m^{2})$
which in turn gives a $K$-rational point on the elliptic curve

$E_{2}:y^{2}=x^{3}-x$

We compute $E_{2}(K)=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ for
all $K\in\mathcal{S}\backslash\\{\mathbb{Q}(\sqrt{-7})\\}$ (has rank 1 over
$K=\mathbb{Q}(\sqrt{-7})$). Hence the only points of $E_{2}(K)$ are
$\\{(0,0),(\pm 1,0)\\}$ which implies $-m^{2}=0,\pm 1,$ contradicting $m\neq
0,\pm 1.$ If $-sz=r^{2},$ then similarly it results in the same curve $E_{2}$
where $m^{2}$ is replaced by $x.$

If $Q+Q^{\prime}$ is halved, by Lemma 1 we have $-itz$ is a square in $L$ so
that $-itz=(a+bi)^{2}=a^{2}-b^{2}+2abi$ for some $a,b\in K.$ Then $a=\pm b$
and so $tz=\pm 2a^{2}.$ Using the parametrization above, we get

$\pm a^{2}=m(1-m^{2})$

One observes easily that it comes down to finding the points on both
$E_{2}(K)$ and $E_{2}^{-1}(K).$ But since
$E_{2}^{-1}(K)_{\text{tor}}=E_{2}(K)_{\text{tor}},$ $E_{2}^{-1}\simeq E$, it
is enough to determine the set $E_{2}(K).$ As computed in the previous case,
$E_{2}(K)$ has rank 0 with torsion
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ In each case $a=0,$
which forces either $t=0$ or $z=0$, a contradiction. In case
$K=\mathbb{Q}(\sqrt{-7})$, following Theorem 4.1 the possibilities with
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\subseteq
E(L)_{\text{tor}}$ are $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$ or
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$ But
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}$ is ruled out by
Theorem [BN16, Theorem 7], which establishes the first part of the claim.

Now suppose $E(L)_{\text{tor}}$ does not contain
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}.$ If $D=-2,-7,-11$ then by
Theorem 4.1,
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},$
or $\mathbb{Z}/{2}\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z}.$ If
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z},$
we may assume that $E(L)_{\text{tor}}=\langle Q,P\rangle$ where $Q$ is a
2-torsion point and $P$ is a $16$-torsion point and so
$E(K)_{\text{tor}}\simeq\langle Q,4P\rangle.$ For the non-trivial element
$\sigma$ of Gal($L/K),$ we have $\sigma(P)=aP$ or $\sigma(P)=Q+aP$ where $a$
is odd. Applying $\sigma$ twice, we obtain

$\sigma(P)=a^{2}P$

where $a^{2}\equiv 1$ mod ($16$). Then $a$ must be $1,7,9$ or $15.$ For all
eight possibilities of $\sigma,$ we have either $\sigma(2P)=2P$ or
$\sigma(P)=-2P.$ The former is not possible by assumption. The later implies
that $\sigma(4P)=-4P,$ which yields a contradiction since $4P$ is
$K$-rational.

If $D=-19,-43,-67,-163$ then by Theorem 4.1, there are only four possibilities
$E(L)_{\text{tor}}=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{12}\mathbb{Z}$
or $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}.$ Replacing $E$ by
$E^{d}$ if necessary, the group
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{24}\mathbb{Z}$ is excluded in the
proof of $(4)$ above, proving the statement.

$(6)$ Suppose $G\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ As
$E(K(\sqrt{d}))\simeq E^{d}(K(\sqrt{d})),$ it is enough to study
$E^{d}(K(\sqrt{d}))_{\text{tor}}$. If
$E^{d}(K)\not\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},$ then
we may apply a previously proven case and see Tables 5,1. If
$E^{d}(K)\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},$ then we
have by Proposition 2.1 that
$E(L)_{\text{tor}}\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z},$
$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/{16}\mathbb{Z},$ or
$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}.$ (See Table 2,1).

## References

  * [Bal21] Irmak Balçık.  Growth of odd torsion over imaginary quadratic fields of class number 1\.  arXiv:2111.11399, (2021).
  * [BN15] Peter Bruin and Filip Najman.  Hyperelliptic modular curves $X_{0}(n)$ and isogenies of elliptic curves over quadratic fields.  LMS J. Comput. Math., 18(1):578–602, 2015.
  * [BN16] Peter Bruin and Filip Najman.  A criterion to rule out torsion groups for elliptic curves over number fields.  Res. Number Theory, 2:Paper No. 3, 13, 2016.
  * [DEvH+21] Maarten Derickx, Anastassia Etropolski, Mark van Hoeij, Jackson S. Morrow, and David Zureick-Brown.  Sporadic cubic torsion.  Algebra Number Theory, 15(7):1837–1864, 2021.
  * [DKSS17] Maarten Derickx, Sheldon Kamienny, William Stein, and Michael Stoll.  Torsion points on elliptic curves over number fields of small degree.  https://arxiv.org/abs/1707.00364, 2017.
  * [GJT14] Enrique González-Jiménez and José M. Tornero.  Torsion of rational elliptic curves over quadratic fields.  Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM, 108(2):923–934, 2014.
  * [GR91] Josep Gonzàlez Rovira.  Equations of hyperelliptic modular curves.  Ann. Inst. Fourier (Grenoble), 41(4):779–795, 1991.
  * [JKP06] Daeyeol Jeon, Chang Heon Kim, and Euisung Park.  On the torsion of elliptic curves over quartic number fields.  J. London Math. Soc. (2), 74(1):1–12, 2006.
  * [Kam92] S. Kamienny.  Torsion points on elliptic curves and $q$-coefficients of modular forms.  Invent. Math., 109(2):221–229, 1992.
  * [KM88] M. A. Kenku and F. Momose.  Torsion points on elliptic curves defined over quadratic fields.  Nagoya Math. J., 109:125–149, 1988.
  * [KN12] Sheldon Kamienny and Filip Najman.  Torsion groups of elliptic curves over quadratic fields.  Acta Arith., 152(3):291–305, 2012.
  * [Kna92] Anthony W. Knapp.  Elliptic curves, volume 40 of Mathematical Notes.  Princeton University Press, Princeton, NJ, 1992.
  * [Kwo97] Soonhak Kwon.  Torsion subgroups of elliptic curves over quadratic extensions.  J. Number Theory, 62(1):144–162, 1997.
  * [LMF23] The LMFDB Collaboration.  The L-functions and modular forms database.  https://www.lmfdb.org, 2023.
  * [Maz78] B. Mazur.  Rational isogenies of prime degree (with an appendix by D. Goldfeld).  Invent. Math., 44(2):129–162, 1978.
  * [Naj16] Filip Najman.  Torsion of rational elliptic curves over cubic fields and sporadic points on $X_{1}(n)$.  Math. Res. Lett., 23(1):245–272, 2016.
  * [New17] Burton Newman.  Growth of torsion of elliptic curves with full 2-torsion over quadratic cyclotomic fields.  J. Number Theory, 173:570–601, 2017.
  * [Ono96] Ken Ono.  Euler's concordant forms.  Acta Arith., 78(2):101–123, 1996.
  * [Ozm12] Ekin Ozman.  Points on quadratic twists of $X_{0}(N)$.  Acta Arith., 152(4):323–348, 2012.
  * [SS18] Naba Kanta Sarma and Anupam Saikia.  Torsion of elliptic curves over imaginary quadratic fields of class number 1.  Rocky Mountain J. Math., 48(8):2689–2703, 2018.