Twists of Elliptic Curves

Twists of Elliptic Curves††This paper is a contribution to the Special Issue
on Modular Forms and String Theory in honor of Noriko Yui. The full collection
is available at http://www.emis.de/journals/SIGMA/modular-forms.html

Max KRONBERG †, Muhammad Afzal SOOMRO ‡ and Jaap TOP †

M. Kronberg, M.A. Soomro and J. Top

† Johan Bernoulli Institute for Mathematics and Computer Science,
† Nijenborgh 9, 9747 AG Groningen, The Netherlands<EMAIL_ADDRESS><EMAIL_ADDRESS>
‡ Quaid-e-Awam University of Engineering, Science & Technology (QUEST),
‡ Sakrand Road, Nawabshah, Sindh, Pakistan<EMAIL_ADDRESS>
Received July 10, 2017, in final form October 23, 2017; Published online
October 25, 2017

In this note we extend the theory of twists of elliptic curves as presented in
various standard texts for characteristic not equal to two or three to the
remaining characteristics. For this, we make explicit use of the
correspondence between the twists and the Galois cohomology set
$H^{1}\big{(}\operatorname{G}_{\overline{K}/K},\operatorname{Aut}_{\overline{K}}(E)\big{)}$.
The results are illustrated by examples.

elliptic curve; twist; automorphisms; Galois cohomology

11G05; 11G25; 14G17

Dedicated to Noriko Yui. The third author of this note was a postdoc with her
at Queen’s University during 1989–1990.

##  1 Introduction

Throughout this paper $K$ will be a perfect field and we always fix a
separable closure of $K$, which we denote by $\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu$. For the absolute
Galois group of $\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu$
over $K$ we write $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$. Let $E/K$ be an
elliptic curve over $K$. A twist of $E$ is an elliptic curve
$E^{\operatorname{tw}}/K$ that is isomorphic to $E$ over $\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu$. In other words, it is
an elliptic curve over $K$ with $j$-invariant $j(E)$. Two such twists are
considered equal if they are isomorphic over $K$. We denote the set of twists
by $\operatorname{Twist}(E/K)$. For the automorphism group of $E$ we write
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$. The elements of $\operatorname{Twist}(E/K)$ are in one-to-one
correspondence with the classes in $H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$ [21, Chapter
X, Section 2]. We want to remark that our notation differs from the notation
used by Silverman. He denotes the set of twists by
$\operatorname{Twist}(E/K,O)$.

Recently there has been quite some interest in twists of not only elliptic
curves, but also curves in general and even in twists of algebraic varieties
over various fields [1, 7, 8, 11, 12, 14, 24]. And besides twists of
varieties, twists of maps appear to have an increasing role in arithmetic
dynamics [9, 13, 19, 23], [20, Section 4.9].

The simplest nontrivial example of twists is provided by the case of elliptic
curves, where an early account of it was given by J.W.S. Cassels in [5, Part
II, Section 9]. We briefly recall some of this theory here; in the case that
the automorphism group of the elliptic curve is cyclic this is covered in
various standard textbooks on elliptic curves. Although it is certainly known
to most experts how to extend this theory to the cases where the automorphism
group is noncyclic (and hence not even abelian), there seems to be no adequate
reference for this and we hope to fill this gap. Note, by the way, that Ian
Connell at McGill University (Montreal) in the late 1990’s wrote extensive
unpublished lecture notes on elliptic curves, including a long chapter on
twists [6]. Inspired by those notes John Cremona implemented various
algorithms for dealing with twists of elliptic curves in sage; see the lines
557–604 of [4] for sage code related to Section 2 of the present paper, and
the lines 507–554 of [4] for sage code related to our Section 3.

For a positive integer $n$ coprime to the characteristic of $K$ we denote by
$\mu_{n}(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu)$ the
group of $n$-th roots of unity in $\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu^{\times}$ and by
$\zeta_{n}$ a generator of this group. In [21], Silverman only presents an
explicit description of the twists of an elliptic curve $E/K$ in
$\operatorname{char}{K}\neq 2,3$. The main reason is that this condition
implies $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\cong\mu_{n}(\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu)$, for some
$n\in\\{2,4,6\\}$, even as $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$-modules. In
characteristic $2$ the group $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ is either
isomorphic to $\mathbb{Z}/2\mathbb{Z}$ or to a non-abelian group of order
$24$. In characteristic three the group $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ is either equals
$\mu_{2}(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu)\cong\mathbb{Z}/2\mathbb{Z}$ or it is a non-abelian group of order $12$.
By explicitly describing $H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$ in these
remaining cases, we complete the description presented in [21].

We start by considering twists of elliptic curves with $j$-invariant equal to
zero in characteristic three and two. We then consider the twists
corresponding to normal subgroups of $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$. The possible
subgroups correspond to quadratic, cubic and sextic twists.

The main results of this note can be found in Propositions 2.1 and 3.1 which
describe twists over finite fields of characteristic three respectively two,
provided the automorphism group of the elliptic curve is non-abelian, and in
Propositions 2.2 and 3.2, where we count the number of twists of any elliptic
curve over a finite field. Sections 2 and 3 also indicate how these twists can
be given explicitly in various non-trivial cases. Next, Propositions 4.1, 5.1
and 6.1 answer the question under what conditions a potentially quadratic or
cubic twist of a given elliptic curve is in fact still isomorphic to the curve
one starts with.

Parts of the results of this paper originate from the PhD thesis of the second
author [22, Section 2.6].

##  2 Twists in characteristic three

We start by considering elliptic curves over finite fields
$\mathbb{F}_{3^{n}}$. This is done by analysing the following central example.
By [14], the twists of an elliptic curve $E$ over a finite field
$\mathbb{F}=\mathbb{F}_{q}$ of cardinality $q$ are in one-to-one
correspondence with the Frobenius conjugacy classes in
$\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}\mkern-1.5mu}\mkern 1.5mu}(E)$. By
definition a Frobenius conjugacy class is obtained by fixing some
$\tau\in\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}\mkern-1.5mu}\mkern 1.5mu}(E)$, then its
Frobenius conjugacy class consists of all

$\displaystyle\big{\\{}\sigma^{-1}\tau\big{(}{}^{\operatorname{Fr}}\sigma\big{)}\big{\\}}.$

Here $\operatorname{Fr}$ is the field automorphism of $\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}\mkern-1.5mu}\mkern 1.5mu$ raising any
element to its $q$th power. It acts on an automorphism $\sigma$ of $E$ by
acting on the coefficients of the rational functions defining $\sigma$. We
will compute these Frobenius conjugacy classes for all possible actions of the
absolute Galois group.

######  Proposition 2.1.

The elliptic curve

$\displaystyle E/\mathbb{F}_{3}\colon\ y^{2}=x^{3}-x$

has $j(E)=0$, and it has precisely twelve automorphisms. These are given by

$\displaystyle\begin{aligned}
&&&&\Phi_{u,r}\colon&&E&&\longrightarrow&&&E,&&&&\\\
&&&&&&(x,y)&&\longmapsto&&&\big{(}u^{2}x+r,u^{3}y\big{)},&&&&\end{aligned}$

where $u^{4}=1$ and $r\in\mathbb{F}_{3}$. Let $n\geq 1$ and $q=3^{n}$. For
$u\in\mathbb{F}_{9}$ with $u^{4}=1$ and $r\in\mathbb{F}_{3}$, put

$\displaystyle
C_{u,r}=\big{\\{}\Phi_{u^{\prime},r^{\prime}}^{-1}\circ\Phi_{u,r}\circ\Phi_{u^{\prime
q},r^{\prime
q}}\,|\,(u^{\prime})^{4}=1,\,r^{\prime}\in\mathbb{F}_{3}\big{\\}}$

$($this is the Frobenius conjugacy class of $\Phi_{u,r}$ over
$\mathbb{F}_{q})$.

If $n$ is odd, then the Frobenius conjugacy classes of $E/\mathbb{F}_{q}$ are

$\displaystyle C_{1,0}=\\{\Phi_{1,0},\Phi_{-1,0}\\},\qquad
C_{1,1}=\\{\Phi_{1,1},\Phi_{-1,-1}\\},\qquad
C_{1,-1}=\\{\Phi_{1,-1},\Phi_{-1,1}\\},$ $\displaystyle
C_{i,0}=\\{\Phi_{u,r}\,|\,u^{2}=-1,\,r\in\mathbb{F}_{3}\\}.$

In particular there are precisely three non-trivial twists of $E$ over
$\mathbb{F}_{q}$ in case $n$ is odd, which are given as follows.

Consider the cocycle defined by $\operatorname{Fr}\mapsto\Phi_{u,0}$. The
corresponding twist is given by

$\displaystyle E^{\operatorname{tw}}\colon\ y^{2}=x^{3}+x$

and the corresponding isomorphism is defined over a quadratic extension.

Analogously the cocycle $\operatorname{Fr}\mapsto\Phi_{1,1}$ corresponds to
the twist

$\displaystyle E^{\operatorname{tw}}\colon\ y^{2}=x^{3}-x-1$

and the cocycle $\operatorname{Fr}\mapsto\Phi_{1,-1}$ corresponds to the twist

$\displaystyle E^{\operatorname{tw}}\colon\ y^{2}=x^{3}-x+1,$

where both isomorphisms are defined over a cubic extension.

In case $n$ is even, fix $i\in\mathbb{F}_{q}$ with $i^{2}=-1$. The Frobenius
conjugacy classes of $E/\mathbb{F}_{q}$ are

$\displaystyle C_{1,0}=\\{\Phi_{1,0}\\},\qquad
C_{-1,0}=\\{\Phi_{-1,0}\\},\qquad C_{1,1}=\\{\Phi_{1,1},\Phi_{1,-1}\\},$
$\displaystyle C_{-1,1}=\\{\Phi_{-1,1},\Phi_{-1,-1}\\},\qquad
C_{i,0}=\\{\Phi_{i,r}\,|\,r\in\mathbb{F}_{3}\\},\qquad
C_{-i,0}=\\{\Phi_{-i,r}\,|\,r\in\mathbb{F}_{3}\\}.$

The corresponding isomorphisms are defined over a field extension of degree
$1$, $2$, $3$, $6$, $4$, $4$, respectively.

Equivalently, since we are considering here the number of
$\mathbb{F}_{3^{n}}$-isomorphism classes of elliptic curves with $j$-invariant
$0$, i.e., of supersingular elliptic curves over $\mathbb{F}_{3^{n}}$, this
shows there are $4$ supersingular curves when $n$ is odd and $6$ such curves
when $n$ is even. This is of course well known; it is consistent with the
tables presented in [15].

###### Proof.

The statements about the $j$-invariant and about the number of automorphisms
are easy; compare, e.g., [21, Appendix A, Proposition 1.2].

Since $\Phi_{u,r}^{-1}=\Phi_{u^{-1},-u^{2}r}$ and
${}^{\operatorname{Fr}}\Phi_{u,r}=\Phi_{u^{3},r^{3}}=\Phi_{u^{-1},r}$, one can
directly calculate the Frobenius conjugacy class of $\Phi_{u,r}$, depending on
$q$ being an even or an odd power of $3$.

To verify that indeed the curves presented in the statement of the proposition
correspond to the given Frobenius conjugacy classes, one needs to use an
isomorphism $\psi\colon E\to E^{\operatorname{tw}}$ and check that
$\big{(}^{\operatorname{Fr}}\psi\big{)}^{-1}\circ\psi$ is in the Frobenius
conjugacy class. For example, with $E^{\operatorname{tw}}\colon y^{2}=x^{3}+x$
one can use $\psi\colon E\to E^{\operatorname{tw}}$ is given by
$(x,y)\mapsto(ix,-iy)$ (with $i^{2}=-1$). A direct computation shows that
$\big{(}^{\operatorname{Fr}}\psi\big{)}^{-1}\circ\psi=\Phi_{i,0}$. The other
cases are done similarly. ∎

Note that we only presented explicit equations for the twists in the case of
an extension of $\mathbb{F}_{3}$ of odd degree. If the degree is even, such an
equation will in general (as expected) depend on the field $\mathbb{F}_{q}$.

Since we are considering here the number of $\mathbb{F}_{3^{n}}$-isomorphism
classes of elliptic curves with $j$-invariant $0$, i.e., of supersingular
elliptic curves over $\mathbb{F}_{3^{n}}$, the proposition shows there are $4$
supersingular curves when $n$ is odd and $6$ such curves when $n$ is even.
This is of course well known; it is consistent with the tables presented in
[15].

We now more generally consider the case that $E/K$ is an elliptic curve
defined over a field $K$ with $\operatorname{char}(K)=3$ such that
$\\#\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)=12$. This means (compare [21, Appendix A]) that $j(E)=0$ and $E$ is
given by an equation $y^{2}=x^{3}+ax+b$, where $a,b\in K$. Thus, there exists
an isomorphism $\psi\colon E\to E^{\prime}$, where $E^{\prime}\colon
y^{2}=x^{3}-x$. We are interested in the possibilities for the field extension
where the isomorphism is defined. By [21, Appendix A, Proposition 1.2], we
have $\psi(x,y)=(u^{2}x+r,u^{3}y)$, where $u^{4}=-\frac{1}{a}$ and
$r^{3}+ar+b=0$. Thus, we see that the degree of the field extension depends on
the existence of a $K$-rational $2$-torsion point on $E$.

In the case that $E[2](K)$ is trivial, we have $b\neq 0$. As
$\psi^{-1}(x,y)=(v^{2}x+w,v^{3}y)$ where $v=u^{-1}$ and $w^{3}-w-v^{6}b=0$,
both $\psi$ and $\psi^{-1}$ are defined over the Artin–Schreier extension of
$K(u)$ defined by $w^{3}-w-v^{6}b=0$.

In the case that $E[2](K)$ is non-trivial, we may assume $b=0$ and any such
isomorphism is defined over $K(u)$.

The field $K(u)$ depends in both cases only on $a$ and is a degree four
extension if $a$ is not a square in $K$.

To complete the picture in characteristic three, note that any elliptic curve
$E/K$ in characteristic $3$ with $j(E)\neq 0$ satisfies
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)=\pm 1$ and therefore $H^{1}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E))\cong
K^{\times}/{K^{\times}}^{2}$. In particular, summarizing most of the
discussion above for the special case of a finite field, one obtains the
following.

######  Proposition 2.2.

Let $q=3^{n}$ and suppose $E/\mathbb{F}_{q}$ is an elliptic curve. Then

$\displaystyle\\#\operatorname{Twist}(E/\mathbb{F}_{q})=\begin{cases}2&\text{if}\
j(E)\neq 0,\\\ 4&\text{if}\ j(E)=0\ \text{and}\ n\ \text{is odd},\\\
6&\text{if}\ j(E)=0\ \text{and}\ n\ \text{is even}.\end{cases}$

##  3 Twists in characteristic two

In order to describe twists in characteristic two, we start by considering the
central example of a supersingular elliptic curve over the field with two
elements. As in the case of characteristic three, this is done by computing
the Frobenius conjugacy classes in all possible cases for the action of
$\operatorname{G}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K}$ on $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$. After this
description we turn to isomorphisms between an arbitrary elliptic curve over a
field with characteristic two and this particular example. Just is was done
for characteristic three, the example is formulated as a proposition, as
follows.

######  Proposition 3.1.

The elliptic curve

$\displaystyle E/\mathbb{F}_{2}\colon\ y^{2}+y=x^{3}$

has $j(E)=0$ and it has exactly $24$ automorphisms. These are described as

$\displaystyle\begin{aligned}
&&&&\Phi_{u,r,t}\colon&&E&&\longrightarrow&&&E,&&&&\\\
&&&&&&(x,y)&&\longmapsto&&&\big{(}u^{2}x+r,y+u^{2}r^{2}x+t\big{)},&&&&\end{aligned}$

where $u\in\mathbb{F}^{*}_{4}$, $r\in\mathbb{F}_{4}$ and $t^{2}+t+r^{3}=0$.
The action of the Galois group $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern
1.5mu/\mathbb{F}_{q}}$ on $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E)$ is
trivial in case $n$ is even, and nontrivial if $n$ is odd.

In case $n$ is odd, there are exactly three Frobenius conjugacy classes
$C_{u,r,t}$ in $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}\mkern-1.5mu}\mkern 1.5mu_{q}}(E)$,
namely the conjugacy class $C_{1,0,0}$ containing the identity, the class
$C_{1,\omega,\omega}$ containing $\Phi_{1,\omega,\omega}$, and the class
$C_{1,\omega,\omega^{2}}$ containing $\Phi_{1,\omega,\omega^{2}}$. Here
$\omega\in\mathbb{F}_{4}$ is a primitive $3$rd root of unity. The two
Frobenius conjugacy classes corresponding to non-trivial twists of $E$ over
$\mathbb{F}_{q}$ yield twists of $E$ over $\mathbb{F}_{q}$ which are
isomorphic to $E$ over a degree eight extension of $\mathbb{F}_{q}$.

In case $n$ is even, the Frobenius conjugacy classes coincide with the usual
conjugacy classes in $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E)$, which
are $($with $C_{u,r,t}$ denoting the conjugacy class containing $\Phi_{u,r,t}$
and $\omega\in\mathbb{F}_{4}$ a chosen primitive $3$rd root of unity$)$

$\displaystyle C_{1,0,0},\ C_{1,0,1},\ C_{\omega^{2},0,1},\ C_{\omega,0,1},\
C_{\omega,0,0},\ C_{\omega^{2},0,0},\ C_{1,1,\omega}.$

The twists of $E/\mathbb{F}_{q}$ corresponding to these conjugacy classes are
isomorphic to $E$ over an extension of $\mathbb{F}_{q}$ of degree $1$, $2$,
$6$, $6$, $3$, $3$, $4$, respectively.

In particular $E/\mathbb{F}_{2^{n}}$ has two non-trivial twists if $n$ is odd,
and six non-trivial twists in case $n$ is even.

###### Proof.

The statements about $j(E)$ and $\operatorname{Aut}(E)$ are immediate; compare
[21, Appendix A].

In any automorphism $\Phi_{u,r,t}$ one has $r\in\mathbb{F}_{4}$ hence
$r^{3}=1$ if $r\neq 0$ and $r^{3}=0$ for $r=0$. So the equality
$t^{2}+t+r^{3}$ shows $t\in\mathbb{F}_{2}$ for $r=0$ and $t\in\mathbb{F}_{4}$
for $r\neq 0$. Therefore all automorphisms are defined over $\mathbb{F}_{4}$.

To obtain the Frobenius conjugacy classes we first assume $n$ is odd, and we
write $C_{u,r,t}$ for the Frobenius conjugacy class containing $\Phi_{u,r,t}$.
Let $\omega\in\mathbb{F}_{4}$ be a fixed primitive $3$rd root of unity. A
direct calculation shows

$\displaystyle
C_{1,0,0}=\big{\\{}\Phi_{u,r,t}^{-1}\Phi_{1,0,0}\Phi_{u^{2},r^{2},t^{2}}\,|\,u\in\mathbb{F}_{4}^{*},\,r\in\mathbb{F}_{4},\,t^{2}+t+r^{3}=0\big{\\}}$
$\displaystyle\hphantom{C_{1,0,0}}{}=\big{\\{}\Phi_{u^{2},ur^{2}+r,ur}\,|\,u\in\mathbb{F}_{4}^{*},\,r\in\mathbb{F}_{4}\big{\\}},$

so $C_{1,0,0}$ consists of

$\displaystyle\big{\\{}\Phi_{1,0,0},\Phi_{1,0,1},\Phi_{1,1,\omega},\Phi_{1,1,\omega^{2}},\Phi_{\omega^{2},0,0},\Phi_{\omega^{2},\omega^{2},\omega},$
$\displaystyle\hphantom{\big{\\{}}{}\Phi_{\omega^{2},\omega^{2},\omega^{2}},\Phi_{\omega^{2},0,1},\Phi_{\omega,0,0},\Phi_{\omega,\omega,\omega^{2}},\Phi_{\omega,0,1},\Phi_{\omega,\omega,\omega}\big{\\}}.$

The other two Frobenius conjugacy classes are given by

$\displaystyle
C_{1,\omega,\omega}=\big{\\{}\Phi_{1,\omega,\omega},\Phi_{\omega,\omega^{2},\omega},\Phi_{\omega,1,\omega^{2}},\Phi_{\omega,\omega,\omega},\Phi_{\omega^{2},1,\omega},\Phi_{1,\omega^{2},\omega^{2}}\big{\\}},$
$\displaystyle
C_{1,\omega,\omega^{2}}=\big{\\{}\Phi_{1,\omega,\omega^{2}},\Phi_{\omega,\omega^{2},\omega^{2}},\Phi_{\omega^{2},1,\omega^{2}},\Phi_{\omega^{2},\omega,\omega},\Phi_{\omega,1,\omega},\Phi_{1,\omega,\omega^{2}}\big{\\}}.$

To see that a twist of $E/\mathbb{F}_{q}$ corresponding to one of the latter
two Frobenius conjugacy classes is indeed isomorphic to $E$ over
$\mathbb{F}_{q^{8}}$ and not over a smaller extension of $\mathbb{F}_{q}$,
consider a cocycle defined by $\operatorname{Fr}\mapsto\Phi$ (with $\Phi$ in
one of the given classes $C_{u,r,s}$). Using the cocycle condition one finds
that the cocycle sends $\operatorname{Fr}^{j}$ (for $j\geq 1$) to
${}^{(\operatorname{id}+\operatorname{Fr}+\dots+\operatorname{Fr}^{j-1})}\Phi$.
This is a nontrivial automorphism for $j\leq 7$ and the trivial one for $j=8$.
The assertion about the twists follows.

We now consider the case $n$ is even. The conjugacy classes in
$\operatorname{Aut}(E)$ are

$\displaystyle C_{1,0,0}=\\{\Phi_{1,0,0}\\},$ $\displaystyle
C_{1,0,1}=\\{\Phi_{1,0,1}\\},$ $\displaystyle
C_{\omega^{2},0,1}=\big{\\{}\Phi_{\omega^{2},0,1},\Phi_{\omega^{2},1,\omega},\Phi_{\omega^{2},\omega^{2},\omega},\Phi_{\omega^{2},\omega,\omega}\big{\\}},$
$\displaystyle
C_{\omega,0,1}=\big{\\{}\Phi_{\omega,0,1},\Phi_{\omega,1,\omega^{2}},\Phi_{\omega,\omega^{2},\omega^{2}},\Phi_{\omega,\omega,\omega^{2}}\big{\\}},$
$\displaystyle
C_{\omega,0,0}=\big{\\{}\Phi_{\omega,0,0},\Phi_{\omega,1,\omega},\Phi_{\omega,\omega^{2},\omega},\Phi_{\omega,\omega,\omega}\big{\\}},$
$\displaystyle
C_{\omega^{2},0,0}=\big{\\{}\Phi_{\omega^{2},0,0},\Phi_{\omega^{2},1,\omega^{2}},\Phi_{\omega^{2},\omega^{2},\omega^{2}},\Phi_{\omega^{2},\omega,\omega^{2}}\big{\\}},$
$\displaystyle
C_{1,1,\omega}=\big{\\{}\Phi_{1,1,\omega},\Phi_{1,1,\omega^{2}},\Phi_{1,\omega,\omega},\Phi_{1,\omega,\omega^{2}},\Phi_{1,\omega^{2},\omega},\Phi_{1,\omega^{2},\omega^{2}}\big{\\}}.$

So indeed there are exactly $6$ non-trivial twists of $E/\mathbb{F}_{q}$ in
this case. The statement about the extension of $\mathbb{F}_{q}$ over which
they will be isomorphic to $E$ is shown exactly as the analogous statement for
the odd $n$ case (in fact in the present situation it simply refers to the
order of the elements in a certain conjugacy class). ∎

We present some comments regarding the argument above. First take $n=1$, so
$q=2^{n}=2$. Since $-1=\Phi_{1,0,1}$ is in the same Frobenius conjugacy class
as the identity, the elliptic curve $E/\mathbb{F}_{2}$ has no non-trivial
quadratic twist. Let us now consider the cubic twists of $E$. Again we can see
that the automorphisms of order $3$ are in the same conjugacy class of the
identity and thus, $E/\mathbb{F}_{2}$ has no non-trivial cubic twists (and as
the proposition states, any non-trivial twist of $E/\mathbb{F}_{2}$ will only
be isomorphic to $E$ over extensions of $\mathbb{F}_{2}$ of degree a multiple
of $8$).

Explicit equations for the two non-trivial twists of $E/\mathbb{F}_{q}$, in
the case $q=2^{n}$ with $n$ odd, are as follows. Let

$\displaystyle E_{1}\colon\ y^{2}+y=x^{3}+x$

and

$\displaystyle E_{2}\colon\ y^{2}+y=x^{3}+x+1.$

These elliptic curves indeed satisfy $j(E_{1})=j(E_{2})=j(E)=0$. Moreover
counting points over $\mathbb{F}_{2^{n}}$ (compare [21, Section V.2]) one
finds $\\#E(\mathbb{F}_{2^{n}})$ and $\\#E_{1}(\mathbb{F}_{2^{n}})$ and
$\\#E_{2}(\mathbb{F}_{2^{n}})$ are three distinct numbers whenever $n$ is odd.
So indeed the curves $E_{j}$ are distinct and nontrivial twists of
$E/\mathbb{F}_{q}$. Of course the same conclusion can be obtained by
exhibiting an explicit isomorphism $\psi\colon E\to E_{j}$ and then
determining the Frobenius conjugacy class containing
${}^{\operatorname{Fr}}\psi^{-1}\circ\psi$.

We will not try to present equations for all non-trivial twists of
$E/\mathbb{F}_{q}$ in the case $q=2^{n}$ with $n$ even. They depend on $q$.
Instead we treat one example.

Let $q=4$. In this case ${\rm Fr}\mapsto-1=\Phi_{1,0,1}$ defines a non-trivial
cocycle class. The corresponding twist is given by

$\displaystyle E^{\operatorname{tw}}\colon\ y^{2}+y=x^{3}+\omega$

(with $\omega$ a primitive $3$rd root of unity), since

$\displaystyle\psi\colon\ (x,y)\mapsto(x,y+\tau),$

with $\tau\in\mathbb{F}_{4}$ satisfying $\tau^{2}+\tau+\omega=0$, defines an
isomorphism $\psi\colon E\rightarrow E^{\operatorname{tw}}$, and

$\displaystyle\big{(}{}^{\rm Fr}\psi\big{)}^{-1}\circ\psi=-1.$

Thus, $E/\mathbb{F}_{4}$ has a non-trivial quadratic twist.

Let $K$ be an field with $\operatorname{char}(K)=2$ and consider the elliptic
curves $E\colon y^{2}+ay=x^{3}+bx+c$ with $a\neq 0$ and $E^{\prime}\colon
y^{2}+y=x^{3}$. Since $j(E)=0=j(E^{\prime})$ these elliptic curves are
isomorphic and, by Silverman [21, Appendix A, Proposition 1.2], for an
isomorphism $\psi\colon E\to E^{\prime}$ we have
$\psi(x,y)=(u^{2}x+s^{2},ay+u^{2}s+t)$, where $u^{3}=a$, $s^{4}+as+b=0$ and
$t^{2}+at+s^{6}+bs^{2}+c=0$. Moreover from the information presented in
Proposition 3.1 it follows that in case $K$ is a finite field, such $u$, $s$,
$t$ exist in an extension of degree at most $8$ resp. $6$, depending on the
action of the Galois group on the automorphism group of $E$.

Again, we summarize the main results given here for the case of a finite
field, as follows. Here as before a crucial remark is that for $E/K$ an
elliptic curve in characteristic $2$, the automorphism group over the
separable closure is $\pm 1$ unless $j(E)=0$.

######  Proposition 3.2.

Let $q=2^{n}$ and suppose $E/\mathbb{F}_{q}$ is an elliptic curve. Then

$\displaystyle\\#\operatorname{Twist}(E/\mathbb{F}_{q})=\begin{cases}2&\text{if}\
j(E)\neq 0,\\\ 3&\text{if}\ j(E)=0\ \text{and}\ n\ \text{is odd},\\\
7&\text{if}\ j(E)=0\ \text{and}\ n\ \text{is even}.\end{cases}$

##  4 Capitulation of quadratic twists

Let $K$ be a field and $\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu$ a separable closure of $K$. Given an elliptic curve $E/K$, the
inclusion $\langle-1\rangle\subset\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ induces a map

$\displaystyle H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}\longrightarrow
H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}.$

The set of quadratic twists of $E$, i.e.,

$\displaystyle QT(E)=\big{\\{}E^{\operatorname{tw}}/K\,|\,\exists\,L/K\text{
with }[L:K]=2\text{ such that
}E^{\operatorname{tw}}\cong_{L}E\big{\\}}/_{\cong_{K}}$

is a subset of $\operatorname{Twist}(E/K)$; this subset of quadratic twists
corresponds to the image of $H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}$ in $H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$ under the
map just given. Here we consider the question whether
$E^{\operatorname{tw}}\in QT(E)$ can be isomorphic to $E$ over the ground
field. In other words, when does $E^{\operatorname{tw}}$ correspond to the
trivial element in $H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$, under the
assumption that it comes from a non-trivial element in the group
$H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}=\operatorname{Hom}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\mathbb{Z}/2\mathbb{Z}\big{)}$.

This question is analogous to a similar question in algebraic number theory
and in function field arithmetic, namely the so-called capitulation (or
principalization) problem for ideals, see, e.g., [2, 3, 10, 16].

Here is a result concerning capitulation of quadratic twists.

######  Proposition 4.1.

Let $E/K$ be an elliptic curve such that $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ is abelian. Then
the map

$\displaystyle i\colon\ H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}\longrightarrow
H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$

is injective except in the case when $\operatorname{char}(K)\not\in\\{2,3\\}$,
$j(E)=12^{3}$ and $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$.

###### Proof.

We have the following long exact sequence of groups:

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\big{)}}$$\textstyle{H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}}$$\textstyle{H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\big{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\textstyle{H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}.}$

Note that $H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\langle-1\rangle\big{)}=\mathbb{Z}/2\mathbb{Z}$. By [21, Chapter III,
Corollary 10.2], we have the following automorphism groups of an elliptic
curve:

  1. 1)

$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)=\langle-1\rangle\cong\mathbb{Z}/2\mathbb{Z}$, when $j(E)\neq
0,12^{3}$;

  2. 2)

$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\cong\mu_{4}(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu)$, when $j(E)=12^{3}$ and $\operatorname{char}(K)\not\in\\{2,3\\}$;

  3. 3)

$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\cong\mu_{6}(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu)$, when $j(E)=0$ and $\operatorname{char}(K)\not\in\\{2,3\\}$.

We consider each case separately.

1\. Since $\\#\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)=2$, the Galois group
$\operatorname{G}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K}$ acts trivially on $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$. Therefore,
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}=1$. Hence, the map $i$ is injective. This
proves the proposition in this case.

2\. First, suppose $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\cong\mu_{4}=\big{\\{}1,\zeta_{4},\zeta_{4}^{2},\zeta_{4}^{3}\big{\\}}$.
Again, here we see $\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}=2$. Hence, the first four groups in the
long exact sequence presented in the first lines of this proof have order as
indicated in the following diagram

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{4\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$.

This implies that $\pi$ is surjective; therefore, $i$ is injective. The
proposition follows in this case.

Now, suppose $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$. Thus, there exists an automorphism
$\sigma\in\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ such that

$\displaystyle\sigma(1)=1,\qquad\sigma(\zeta_{4})=\zeta_{4}^{3}.$

Therefore, $\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}=2$. Now, the
action of $\sigma$ on

$\displaystyle\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\cong\big{\\{}\big{\\{}1,\zeta_{4}^{2}\big{\\}},\big{\\{}\zeta_{4},\zeta_{4}^{3}\big{\\}}\big{\\}}$

is

$\displaystyle\sigma\big{(}\big{\\{}1,\zeta_{4}^{2}\big{\\}}\big{)}=\big{\\{}1,\zeta_{4}^{2}\big{\\}},\qquad\sigma\big{(}\big{\\{}\zeta_{4},\zeta_{4}^{3}\big{\\}}\big{)}=\big{\\{}\zeta_{4},\zeta_{4}^{3}\big{\\}}.$

We conclude that $\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}=2$. Hence, the first four groups in the
long exact sequence presented at the beginning of this proof have order as
indicated in the following diagram

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$.

Thus, $\pi$ is the constant map; therefore, $\\#{\operatorname{Ker}}(i)=2$ and
$i$ is not injective. The proposition follows in this case.

3\. First, if $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\cong\mu_{6}=\langle\zeta_{6}\rangle$, then we have
$\\#H^{0}\big{(}\\!\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$ $=6$ and
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}=3$. The first four groups in the long
exact sequence from the start of this proof therefore have order

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{6\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{3\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$.

This implies that $\pi$ is surjective; therefore, because the sequence is
exact, $i$ is injective.

Now, suppose $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$. Let $\sigma\in\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$. Then we have

$\displaystyle\sigma(1)=1,\qquad\sigma(\zeta_{6})=\zeta_{6}^{5}.$

Thus we get $\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}=2$. The
action of $\sigma$ on

$\displaystyle\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\cong\big{\\{}\big{\\{}1,\zeta_{6}^{3}\big{\\}},\big{\\{}\zeta_{6},\zeta_{6}^{4}\big{\\}},\big{\\{}\zeta_{6}^{2},\zeta_{6}^{5}\big{\\}}\big{\\}}$

is given by

$\displaystyle\sigma\big{(}\big{\\{}1,\zeta_{6}^{3}\big{\\}}\big{)}=\big{\\{}1,\zeta_{6}^{3}\big{\\}},\qquad\sigma\big{(}\big{\\{}\zeta_{6},\zeta_{6}^{4}\big{\\}}\big{)}=\big{\\{}\zeta_{6}^{2},\zeta_{6}^{5}\big{\\}},\qquad\sigma\big{(}\big{\\{}\zeta_{6}^{2},\zeta_{6}^{5}\big{\\}}\big{)}=\big{\\{}\zeta_{6},\zeta_{6}^{4}\big{\\}},$

implying that $\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{\langle-1\rangle}\big{)}=1$. The first four groups in the long
exact sequence used throughout this argument have orders

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$.

We conclude that $\pi$ is surjective; hence, $i$ is injective. This completes
the proof of the proposition. ∎

The condition in Proposition 4.1 that the automorphism group of $E$ should be
abelian, means that one excludes only the cases $j(E)=0$ in
${\operatorname{char}}(K)\in\left\\{2,3\right\\}$. We briefly consider these
two excluded cases here.

Suppose $\operatorname{char}(K)=2$ and take $E/K\colon y^{2}+y=x^{3}$. As in
Section 3 one finds that $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ precisely when $\mathbb{F}_{4}\subset K$. In that case no
capitulation of quadratic twists occurs. However, when Galois acts non-
trivially on this automorphism group, then as in the case of a finite field
studied in Section 3 where we saw that $-1$ and $1$ are in the same Frobenius
conjugacy class, capitulation occurs.

Similarly, suppose $\operatorname{char}(K)=3$ and take $E\colon
y^{2}=x^{3}-x$. Again, there is capitulation of quadratic twists precisely
when the Galois action on the automorphism group is non-trivial, which happens
precisely when $\mathbb{F}_{9}\not\subset K$.

######  Example 4.2.

Take

$\displaystyle E/\mathbb{Q}\colon\ y^{2}=x^{3}-x.$

Then

$\displaystyle\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{Q}\mkern-1.5mu}\mkern 1.5mu}(E)=\\{\pm
1,\pm\iota\\},$

where $\iota\colon E\rightarrow E$ is defined by $(x,y)\mapsto(-x,iy)$ for a
fixed choice of a primitive $4$th root of unity $i\in\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{Q}\mkern-1.5mu}\mkern 1.5mu$.

For $d\in\mathbb{Q}^{*}$ write

$\displaystyle E^{(d)}\colon\ y^{2}=x^{3}-d^{2}x.$

Then $E^{(d)}$ is a twist of $E/\mathbb{Q}$, since $\psi_{d}\colon
E\rightarrow E^{(d)}$ defined as

$\displaystyle\psi_{d}(x,y)=\big{(}dx,d\sqrt{d}y\big{)}$

is an isomorphism between $E$ and $E^{(d)}$.

If $\sigma\in\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{Q}\mkern-1.5mu}\mkern 1.5mu/\mathbb{Q}}$,
then

$\displaystyle\big{(}^{\sigma}\psi_{d}\big{)}^{-1}\circ\psi_{d}=\begin{cases}\hphantom{-}1&\text{if}\
\sigma\big{(}\sqrt{d}\big{)}=\sqrt{d},\\\ -1&\text{if}\
\sigma\big{(}\sqrt{d}\big{)}=-\sqrt{d}.\end{cases}$

So $E^{(d)}$ corresponds to the cocycle class of

$\displaystyle\sigma\mapsto\frac{\sigma\big{(}\sqrt{d}\big{)}}{\sqrt{d}}\in\operatorname{Aut}_{\bar{\mathbb{Q}}}(E).$

In the case $d=-1$, this cocycle is a coboundary, since

$\displaystyle\frac{\sigma\big{(}\sqrt{-1}\big{)}}{\sqrt{-1}}=(^{\sigma}\iota)^{-1}\circ\iota.$

So $E^{(-1)}\cong E$ over $\mathbb{Q}$, which is, of course, evident from the
equation.

######  Example 4.3.

Take $q$ a power of an odd prime, and

$\displaystyle E/\mathbb{F}_{q}\colon\ y^{2}=x^{3}-x.$

The Galois group $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern
1.5mu/\mathbb{F}_{q}}$ acts non-trivially on $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E)$ if and
only if $-1$ is not a square in $\mathbb{F}_{q}$. We have

$\displaystyle\sqrt{-1}\not\in\mathbb{F}_{q}\iff q\equiv 3\pmod{4}.$

For $d\in\mathbb{F}_{q}^{*}$, define $E^{(d)}/\mathbb{F}_{q}$ as before. This
provides a quadratic twist as in Example 4.2.

If $d$ is not a square and $q\equiv 1\pmod{4}$, then $E^{(d)}$ is the (unique)
non-trivial quadratic twist of $E/\mathbb{F}_{q}.$

If $d$ is not a square and $q\equiv 3\pmod{4}$, then $-d$ is a square.
Therefore, we have $E^{(d)}=E^{(-d)}\cong E$ over $\mathbb{F}_{q}$. So for
$q\equiv 3\pmod{4}$, a non-trivial quadratic twist of $E/\mathbb{F}_{q}$ does
not exist.

##  5 Capitulation of cubic twists

Let $E/K$ be an elliptic curve such that $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ contains a subgroup
$C_{3}$ of order $3$. This implies $j(E)=0$ by [21, Chapter III, Corollary
10.2]. Thus, we restrict ourselves in this section to elliptic curves $E$ with
$j(E)=0$. Note that in the case of $\operatorname{char}(K)=2,3$ the group
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ is not abelian; thus, the considered exact sequence is an exact
sequence of pointed sets. The non-abelian cohomology needed to describe twists
in this situation, is, e.g., described in Serre’s books [18, Chapter I,
Section 5] and [17, Chapter XIII].

######  Proposition 5.1.

Let $E/K$ be an elliptic curve with $j(E)=0$. There is a unique $($and
therefore normal$)$ subgroup $C_{3}\subset{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}$ of order $3$. The
map

$\displaystyle i\colon\ H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},C_{3}\big{)}\longrightarrow H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$

is injective except possibly when $\operatorname{char}(K)=2$.

###### Proof.

1\. First we consider the case $\operatorname{char}(K)\neq 2,3$. Then
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ is cyclic of order $6$, so indeed $C_{3}$ as desired exists and is
unique. If $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ we have

$\displaystyle\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{\mu_{3}}\big{)}=2.$

In the long exact sequence

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},C_{3}\big{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\big{)}}$$\textstyle{H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)}{C_{3}}\big{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},C_{3}\big{)}}$$\textstyle{H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)\big{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\textstyle{H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{C_{3}}\big{)}}$

the orders of the first few groups are

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{3\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{6\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

and thus, $\pi$ is surjective which implies $i$ is injective.

If on the other hand $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ then since Galois fixes the $-1$-map, any
$\sigma\in\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ that acts non-
trivially has to interchange the two non-trivial elements of $C_{3}$. This
implies that $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{\mu_{3}}$ and thus,
in the long exact sequence presented earlier in this proof $\pi$ is surjective
since $\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},\mu_{3}\big{)}=1$ and
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}=1$. So again
one concludes that $i$ is injective.

2\. Let $\operatorname{char}(K)=3$. This implies $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ is a semi-direct
product $C_{3}\rtimes C_{4}$ of cyclic groups of order $3$ and $4$ (see [21,
Appendix A, Example A.1]). In particular it follows that the automorphism
group has a unique subgroup $C_{3}$ of order $3$. If $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$, then $i$ is injective. If $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ we will consider several cases.

First we consider the case that $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$C_{3}$. This implies that any non-trivially acting
$\sigma\in\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ interchanges the two
elements of order $4$ in $C_{4}$. This implies
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},C_{3}\big{)}=3$ and
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}=6$ and
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{C_{3}}\big{)}=2$.
This gives us the sequence of orders

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{3\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{6\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

from which it follows that $\pi$ is surjective and thus $i$ is injective.

Now consider the case that all elements of $C_{4}$ are fixed under the action
of $\operatorname{G}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K}$. This implies that any $\sigma\in\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acting non-trivially
on $\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ has to interchange the non-trivial elements of $C_{3}$. Therefore,
we get

$\displaystyle\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},C_{3}\big{)}=1,$
$\displaystyle\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}=4,$
$\displaystyle\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{C_{3}}\big{)}=4.$

Similar to the previous situation this implies that $i$ is injective.

In the case that neither $C_{3}$ nor $C_{4}$ are elementwise fixed under the
action of $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$, we easily get the
following sequence of orders

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

and thus again, $i$ is injective. ∎

So the result says that capitulation of cubic twists does not occur, except
possibly in characteristic $2$. In fact Proposition 3.2 implies that it also
does not occur over finite fields in characteristic two.

##  6 Capitulation of sextic twists

Let $E/K$ be an elliptic curve such that $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ has a normal
subgroup of order $6$. This implies $j(E)=0$ and $\operatorname{char}(K)\neq
2$. A result analogous to Proposition 5.1 is the following.

######  Proposition 6.1.

Suppose $\operatorname{char}(K)\neq 2$ and let $E/K$ be an elliptic curve with
$j(E)=0$. There is a unique $($and therefore normal$)$ subgroup
$C_{6}\subset{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}$ of order $6$. The
map

$\displaystyle i\colon\ H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},C_{6}\big{)}\longrightarrow H^{1}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}$

is injective except in the two cases

  1. $1)$

$\operatorname{char}(K)=3$ and $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ acts trivially on
$C_{6}$;

  2. $2)$

$\operatorname{char}(K)=3$ and the only elements in
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ fixed by $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ are $\pm 1$.

###### Proof.

In the case $\operatorname{char}(K)\neq 2,3$ we have for $E$ as above that
$\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)$ is cyclic of order $6$. So the result is trivial in this case.

Now assume $\operatorname{char}(K)=3$. In this case
$\\#\operatorname{Aut}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu}(E)=12$ and this automorphism group contains a unique subgroup $C_{6}$
of order $6$. It is generated by the unique element of order $2$ and the
subgroup $C_{3}$ of order $3$ in $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$. In the case that
$\operatorname{G}_{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K}$ acts trivially on $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ we get once again
that $i$ is injective. Thus, we now assume that the Galois action is non-
trivial.

First case: the Galois action on $C_{3}$ is trivial. Then analogously to the
case of cubic twists here $\\#H^{0}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},C_{6})=6$ and
$\\#H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}=6$.
Obviously, $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$ fixes the residue
classes modulo $C_{6}$. Thus the group $H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{C_{6}}\big{)}$ has
order $2$. This gives us the sequence of orders

$\displaystyle\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{6\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
64.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
64.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{6\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
82.50491pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 99.5pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
99.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
134.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
134.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces,$

implying in the same way as in earlier cases that $i$ is not injective, and
the map

$\displaystyle\pi\colon\ H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)\big{)}\to
H^{0}\big{(}\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{H}\big{)}$

is constant.

Second case: the Galois action fixes the points in a cyclic order $4$ subgroup
of automorphisms. Then $\\#H^{0}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},C_{6})=2$ and
$\\#H^{0}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E))=4$. Furthermore the
action on $\faktor{\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)}{C_{6}}$ is trivial,
which gives us the sequence of orders

$\displaystyle\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
64.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
64.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{4\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
82.50491pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 99.5pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
99.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
134.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
134.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

As before we conclude that $i$ is injective.

Third case: Neither $C_{3}$ nor an order $4$ subgroup $C_{4}$ are pointwise
fixed under the action of $\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K}$. Then only $\pm 1$
are fixed in $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}(E)$ and $C_{6}$.
Further, we see that the action on the quotient group again is trivial. This
implies for the orders in the long exact sequence

$\displaystyle\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
64.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
64.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
82.50491pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 99.5pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
99.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
134.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
134.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

So reasoning as before, $\pi$ is constant and $i$ is not injective. This case
concludes the proof in characteristic $3$. ∎

In fact a slightly different proof of the same result may be obtained by
observing that a sextic twist may be regarded as a cubic twist of a quadratic
one. We will not pursue this here.

##  7 Other twists

Although the techniques used in the previous sections require the (cyclic and
Galois stable) subgroup used there to be normal, also in the non-normal cases
one can draw conclusions.

We restrict ourselves to providing two examples.

######  Example 7.1.

Take $q=3^{n}$ and consider $E/\mathbb{F}_{q}$ given by $y^{2}=x^{3}-x$. The
automorphism $\Phi_{i,0}\colon(x,y)\mapsto(-x,\sqrt{-1}y)$ generates a Galois
stable subgroup $H$ of $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E)$ of
order $4$. Then $\operatorname{Fr}\mapsto\Phi_{i,0}$ defines a cocycle in
$H^{1}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},H)$ and in
$H^{1}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E))$. In
Proposition 2.1 we saw that this corresponds to a non-trivial twist.

######  Example 7.2.

Similarly we put $q=2^{n}$ and $E/\mathbb{F}_{q}$ given by $y^{2}+y=x^{3}$.
With $\omega\in\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu$ a
primitive third root of unity, the automorphism
$\Phi_{\omega^{2},0,1}\colon(x,y)\mapsto(\omega x,y+1)$ generates a Galois
stable subgroup $H$ of $\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E)$ of
order $6$, and $\operatorname{Fr}\mapsto\Phi_{\omega^{2},0,1}$ defines a
cocycle in $H^{1}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu/K},H)$ and in
$H^{1}(\operatorname{G}_{\mkern
1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern
1.5mu/K},\operatorname{Aut}_{\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}(E))$.

Proposition 3.1 shows that for odd $n$ this results in a trivial twist, and
for $n$ even one obtains a non-trivial twist.

### Acknowledgements

It is a pleasure to thank Joe Silverman, John Cremona, Nurdagül Anbar Meidl,
and Jan Steffen Müller for helpful comments, advise, and support while
preparing this paper. We are especially grateful for the many useful
suggestions of the two referees of the first draft of this, which we hope have
improved the readability of the text a lot.

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