Twists of Elliptic Curves
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 H¹(GK¯/K,AutK¯(E)...
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irk-123456789-1492772019-02-20T01:23:45Z Twists of Elliptic Curves Kronberg, M. Soomro, M.A. Top, J. 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 H¹(GK¯/K,AutK¯(E)). The results are illustrated by examples. 2017 Article Twists of Elliptic Curves / M. Kronberg, M.A. Soomro, J. Top // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 24 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 11G05; 11G25; 14G1 DOI:10.3842/SIGMA.2017.083 http://dspace.nbuv.gov.ua/handle/123456789/149277 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України |
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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 H¹(GK¯/K,AutK¯(E)). The results are illustrated by examples. |
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Kronberg, M. Soomro, M.A. Top, J. |
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Kronberg, M. Soomro, M.A. Top, J. Twists of Elliptic Curves Symmetry, Integrability and Geometry: Methods and Applications |
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Kronberg, M. Soomro, M.A. Top, J. |
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Kronberg, M. |
| title |
Twists of Elliptic Curves |
| title_short |
Twists of Elliptic Curves |
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Twists of Elliptic Curves |
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Twists of Elliptic Curves |
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Twists of Elliptic Curves |
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twists of elliptic curves |
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Інститут математики НАН України |
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2017 |
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Twists of Elliptic Curves / M. Kronberg, M.A. Soomro, J. Top // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 24 назв. — англ. |
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Symmetry, Integrability and Geometry: Methods and Applications |
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AT kronbergm twistsofellipticcurves AT soomroma twistsofellipticcurves AT topj twistsofellipticcurves |
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2025-07-12T21:14:11Z |
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2025-07-12T21:14:11Z |
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1837477249160314880 |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (2017), 083, 13 pages
Twists of Elliptic Curves
Max KRONBERG †, Muhammad Afzal SOOMRO ‡ and Jaap TOP †
† Johan Bernoulli Institute for Mathematics and Computer Science,
Nijenborgh 9, 9747 AG Groningen, The Netherlands
E-mail: m.c.kronberg@rug.nl, j.top@rug.nl
‡ Quaid-e-Awam University of Engineering, Science & Technology (QUEST),
Sakrand Road, Nawabshah, Sindh, Pakistan
E-mail: m.a.soomro@quest.edu.pk
Received July 10, 2017, in final form October 23, 2017; Published online October 25, 2017
https://doi.org/10.3842/SIGMA.2017.083
Abstract. 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 charac-
teristics. For this, we make explicit use of the correspondence between the twists and the
Galois cohomology set H1
(
GK/K ,AutK(E)
)
. The results are illustrated by examples.
Key words: elliptic curve; twist; automorphisms; Galois cohomology
2010 Mathematics Subject Classification: 11G05; 11G25; 14G17
Dedicated to Noriko Yui. The third author of
this note was a postdoc with her at Queen’s Uni-
versity 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 K. For the absolute Galois group of K over K we write GK/K . Let E/K
be an elliptic curve over K. A twist of E is an elliptic curve Etw/K that is isomorphic to E
over K. 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 Twist(E/K).
For the automorphism group of E we write AutK(E). The elements of Twist(E/K) are in
one-to-one correspondence with the classes in H1
(
GK/K ,AutK(E)
)
[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 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
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
mailto:m.c.kronberg@rug.nl
mailto:j.top@rug.nl
mailto:m.a.soomro@quest.edu.pk
https://doi.org/10.3842/SIGMA.2017.083
http://www.emis.de/journals/SIGMA/modular-forms.html
2 M. Kronberg, M.A. Soomro and J. Top
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 µn(K) the group of
n-th roots of unity in K× and by ζn a generator of this group. In [21], Silverman only presents
an explicit description of the twists of an elliptic curve E/K in charK 6= 2, 3. The main reason
is that this condition implies AutK(E) ∼= µn(K), for some n ∈ {2, 4, 6}, even as GK/K-modules.
In characteristic 2 the group AutK(E) is either isomorphic to Z/2Z or to a non-abelian group
of order 24. In characteristic three the group AutK(E) is either equals µ2(K) ∼= Z/2Z or
it is a non-abelian group of order 12. By explicitly describing H1
(
GK/K ,AutK(E)
)
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 AutK(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 F3n . This is done by analysing the
following central example. By [14], the twists of an elliptic curve E over a finite field F = Fq of
cardinality q are in one-to-one correspondence with the Frobenius conjugacy classes in AutF(E).
By definition a Frobenius conjugacy class is obtained by fixing some τ ∈ AutF(E), then its
Frobenius conjugacy class consists of all{
σ−1τ
(
Frσ
)}
.
Here Fr is the field automorphism of F raising any element to its qth power. It acts on an
automorphism σ of E by acting on the coefficients of the rational functions defining σ. We will
compute these Frobenius conjugacy classes for all possible actions of the absolute Galois group.
Proposition 2.1. The elliptic curve
E/F3 : y2 = x3 − x
has j(E) = 0, and it has precisely twelve automorphisms. These are given by
Φu,r : E −→ E,
(x, y) 7−→
(
u2x+ r, u3y
)
,
where u4 = 1 and r ∈ F3. Let n ≥ 1 and q = 3n. For u ∈ F9 with u4 = 1 and r ∈ F3, put
Cu,r =
{
Φ−1
u′,r′ ◦ Φu,r ◦ Φu′q ,r′q | (u′)4 = 1, r′ ∈ F3
}
(this is the Frobenius conjugacy class of Φu,r over Fq).
Twists of Elliptic Curves 3
If n is odd, then the Frobenius conjugacy classes of E/Fq are
C1,0 = {Φ1,0,Φ−1,0}, C1,1 = {Φ1,1,Φ−1,−1}, C1,−1 = {Φ1,−1,Φ−1,1},
Ci,0 = {Φu,r |u2 = −1, r ∈ F3}.
In particular there are precisely three non-trivial twists of E over Fq in case n is odd, which are
given as follows.
Consider the cocycle defined by Fr 7→ Φu,0. The corresponding twist is given by
Etw : y2 = x3 + x
and the corresponding isomorphism is defined over a quadratic extension.
Analogously the cocycle Fr 7→ Φ1,1 corresponds to the twist
Etw : y2 = x3 − x− 1
and the cocycle Fr 7→ Φ1,−1 corresponds to the twist
Etw : y2 = x3 − x+ 1,
where both isomorphisms are defined over a cubic extension.
In case n is even, fix i ∈ Fq with i2 = −1. The Frobenius conjugacy classes of E/Fq are
C1,0 = {Φ1,0}, C−1,0 = {Φ−1,0}, C1,1 = {Φ1,1,Φ1,−1},
C−1,1 = {Φ−1,1,Φ−1,−1}, Ci,0 = {Φi,r | r ∈ F3}, C−i,0 = {Φ−i,r | r ∈ F3}.
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 F3n-isomorphism classes of elliptic
curves with j-invariant 0, i.e., of supersingular elliptic curves over F3n, 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 Φ−1
u,r = Φu−1,−u2r and FrΦu,r = Φu3,r3 = Φu−1,r, one can directly calculate the Frobenius
conjugacy class of Φu,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 ψ : E → Etw and check
that
(Fr
ψ
)−1 ◦ ψ is in the Frobenius conjugacy class. For example, with Etw : y2 = x3 + x one
can use ψ : E → Etw is given by (x, y) 7→ (ix,−iy) (with i2 = −1). A direct computation shows
that
(Fr
ψ
)−1 ◦ ψ = Φi,0. The other cases are done similarly. �
Note that we only presented explicit equations for the twists in the case of an extension of F3
of odd degree. If the degree is even, such an equation will in general (as expected) depend on
the field Fq.
Since we are considering here the number of F3n-isomorphism classes of elliptic curves with
j-invariant 0, i.e., of supersingular elliptic curves over F3n , 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 char(K) = 3 such that # AutK(E) = 12. This means (compare [21, Appendix A]) that
j(E) = 0 and E is given by an equation y2 = x3 + ax + b, where a, b ∈ K. Thus, there exists
4 M. Kronberg, M.A. Soomro and J. Top
an isomorphism ψ : E → E′, where E′ : y2 = x3 − 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 ψ(x, y) = (u2x+ r, u3y), where u4 = − 1
a and r3 + 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 6= 0. As ψ−1(x, y) = (v2x + w, v3y) where
v = u−1 and w3 − w − v6b = 0, both ψ and ψ−1 are defined over the Artin–Schreier extension
of K(u) defined by w3 − w − v6b = 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 characte-
ristic 3 with j(E) 6= 0 satisfies AutK(E) = ±1 and therefore H1(GK/K ,AutK(E)) ∼= K×/K×
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 = 3n and suppose E/Fq is an elliptic curve. Then
# Twist(E/Fq) =
2 if j(E) 6= 0,
4 if j(E) = 0 and n is odd,
6 if j(E) = 0 and n is even.
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 GK/K on AutK(E). After this description we turn to isomorphisms between an arbi-
trary 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
E/F2 : y2 + y = x3
has j(E) = 0 and it has exactly 24 automorphisms. These are described as
Φu,r,t : E −→ E,
(x, y) 7−→
(
u2x+ r, y + u2r2x+ t
)
,
where u ∈ F∗4, r ∈ F4 and t2 + t + r3 = 0. The action of the Galois group GFq/Fq on AutFq(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 Cu,r,t in AutFq(E),
namely the conjugacy class C1,0,0 containing the identity, the class C1,ω,ω containing Φ1,ω,ω, and
the class C1,ω,ω2 containing Φ1,ω,ω2. Here ω ∈ F4 is a primitive 3rd root of unity. The two
Frobenius conjugacy classes corresponding to non-trivial twists of E over Fq yield twists of E
over Fq which are isomorphic to E over a degree eight extension of Fq.
In case n is even, the Frobenius conjugacy classes coincide with the usual conjugacy classes
in AutFq(E), which are (with Cu,r,t denoting the conjugacy class containing Φu,r,t and ω ∈ F4
a chosen primitive 3rd root of unity)
C1,0,0, C1,0,1, Cω2,0,1, Cω,0,1, Cω,0,0, Cω2,0,0, C1,1,ω.
Twists of Elliptic Curves 5
The twists of E/Fq corresponding to these conjugacy classes are isomorphic to E over an exten-
sion of Fq of degree 1, 2, 6, 6, 3, 3, 4, respectively.
In particular E/F2n 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 Aut(E) are immediate; compare [21, Appendix A].
In any automorphism Φu,r,t one has r ∈ F4 hence r3 = 1 if r 6= 0 and r3 = 0 for r = 0. So the
equality t2 + t + r3 shows t ∈ F2 for r = 0 and t ∈ F4 for r 6= 0. Therefore all automorphisms
are defined over F4.
To obtain the Frobenius conjugacy classes we first assume n is odd, and we write Cu,r,t for
the Frobenius conjugacy class containing Φu,r,t. Let ω ∈ F4 be a fixed primitive 3rd root of
unity. A direct calculation shows
C1,0,0 =
{
Φ−1
u,r,tΦ1,0,0Φu2,r2,t2 |u ∈ F∗4, r ∈ F4, t
2 + t+ r3 = 0
}
=
{
Φu2,ur2+r,ur |u ∈ F∗4, r ∈ F4
}
,
so C1,0,0 consists of{
Φ1,0,0,Φ1,0,1,Φ1,1,ω,Φ1,1,ω2 ,Φω2,0,0,Φω2,ω2,ω,
Φω2,ω2,ω2 ,Φω2,0,1,Φω,0,0,Φω,ω,ω2 ,Φω,0,1,Φω,ω,ω
}
.
The other two Frobenius conjugacy classes are given by
C1,ω,ω =
{
Φ1,ω,ω,Φω,ω2,ω,Φω,1,ω2 ,Φω,ω,ω,Φω2,1,ω,Φ1,ω2,ω2
}
,
C1,ω,ω2 =
{
Φ1,ω,ω2 ,Φω,ω2,ω2 ,Φω2,1,ω2 ,Φω2,ω,ω,Φω,1,ω,Φ1,ω,ω2
}
.
To see that a twist of E/Fq corresponding to one of the latter two Frobenius conjugacy classes
is indeed isomorphic to E over Fq8 and not over a smaller extension of Fq, consider a cocycle
defined by Fr 7→ Φ (with Φ in one of the given classes Cu,r,s). Using the cocycle condition
one finds that the cocycle sends Frj (for j ≥ 1) to (id + Fr +···+Frj−1)Φ. This is a nontrivial
automorphism for j ≤ 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 Aut(E) are
C1,0,0 = {Φ1,0,0},
C1,0,1 = {Φ1,0,1},
Cω2,0,1 =
{
Φω2,0,1,Φω2,1,ω,Φω2,ω2,ω,Φω2,ω,ω
}
,
Cω,0,1 =
{
Φω,0,1,Φω,1,ω2 ,Φω,ω2,ω2 ,Φω,ω,ω2
}
,
Cω,0,0 =
{
Φω,0,0,Φω,1,ω,Φω,ω2,ω,Φω,ω,ω
}
,
Cω2,0,0 =
{
Φω2,0,0,Φω2,1,ω2 ,Φω2,ω2,ω2 ,Φω2,ω,ω2
}
,
C1,1,ω =
{
Φ1,1,ω,Φ1,1,ω2 ,Φ1,ω,ω,Φ1,ω,ω2 ,Φ1,ω2,ω,Φ1,ω2,ω2
}
.
So indeed there are exactly 6 non-trivial twists of E/Fq in this case. The statement about
the extension of Fq 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 = 2n = 2.
Since −1 = Φ1,0,1 is in the same Frobenius conjugacy class as the identity, the elliptic curve E/F2
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,
6 M. Kronberg, M.A. Soomro and J. Top
E/F2 has no non-trivial cubic twists (and as the proposition states, any non-trivial twist of E/F2
will only be isomorphic to E over extensions of F2 of degree a multiple of 8).
Explicit equations for the two non-trivial twists of E/Fq, in the case q = 2n with n odd, are
as follows. Let
E1 : y2 + y = x3 + x
and
E2 : y2 + y = x3 + x+ 1.
These elliptic curves indeed satisfy j(E1) = j(E2) = j(E) = 0. Moreover counting points
over F2n (compare [21, Section V.2]) one finds #E(F2n) and #E1(F2n) and #E2(F2n) are three
distinct numbers whenever n is odd. So indeed the curves Ej are distinct and nontrivial twists
of E/Fq. Of course the same conclusion can be obtained by exhibiting an explicit isomorphism
ψ : E → Ej and then determining the Frobenius conjugacy class containing Frψ−1 ◦ ψ.
We will not try to present equations for all non-trivial twists of E/Fq in the case q = 2n
with n even. They depend on q. Instead we treat one example.
Let q = 4. In this case Fr 7→ −1 = Φ1,0,1 defines a non-trivial cocycle class. The corresponding
twist is given by
Etw : y2 + y = x3 + ω
(with ω a primitive 3rd root of unity), since
ψ : (x, y) 7→ (x, y + τ),
with τ ∈ F4 satisfying τ2 + τ + ω = 0, defines an isomorphism ψ : E → Etw, and(
Frψ
)−1 ◦ ψ = −1.
Thus, E/F4 has a non-trivial quadratic twist.
Let K be an field with char(K) = 2 and consider the elliptic curves E : y2 + ay = x3 + bx+ c
with a 6= 0 and E′ : y2 + y = x3. Since j(E) = 0 = j(E′) these elliptic curves are isomorphic
and, by Silverman [21, Appendix A, Proposition 1.2], for an isomorphism ψ : E → E′ we have
ψ(x, y) = (u2x+ s2, ay + u2s+ t), where u3 = a, s4 + as+ b = 0 and t2 + at+ s6 + bs2 + 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 ±1 unless j(E) = 0.
Proposition 3.2. Let q = 2n and suppose E/Fq is an elliptic curve. Then
# Twist(E/Fq) =
2 if j(E) 6= 0,
3 if j(E) = 0 and n is odd,
7 if j(E) = 0 and n is even.
Twists of Elliptic Curves 7
4 Capitulation of quadratic twists
Let K be a field and K a separable closure of K. Given an elliptic curve E/K, the inclusion
〈−1〉 ⊂ AutK(E) induces a map
H1
(
GK/K , 〈−1〉
)
−→ H1
(
GK/K ,AutK(E)
)
.
The set of quadratic twists of E, i.e.,
QT (E) =
{
Etw/K | ∃L/K with [L : K] = 2 such that Etw ∼=L E
}
/∼=K
is a subset of Twist(E/K); this subset of quadratic twists corresponds to the image of
H1
(
GK/K , 〈−1〉
)
in H1
(
GK/K ,AutK(E)
)
under the map just given. Here we consider the
question whether Etw ∈ QT (E) can be isomorphic to E over the ground field. In other words,
when does Etw correspond to the trivial element in H1
(
GK/K ,AutK(E)
)
, under the assumption
that it comes from a non-trivial element in the group H1
(
GK/K , 〈−1〉
)
= Hom
(
GK/K ,Z/2Z
)
.
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 AutK(E) is abelian. Then the map
i : H1
(
GK/K , 〈−1〉
)
−→ H1
(
GK/K ,AutK(E)
)
is injective except in the case when char(K) 6∈ {2, 3}, j(E) = 123 and GK/K acts non-trivially
on AutK(E).
Proof. We have the following long exact sequence of groups:
1 // H0
(
GK/K , 〈−1〉
)
// H0
(
GK/K ,AutK(E)
)
H0
(
GK/K ,
AutK(E)�〈−1〉
)
//
π
// H1
(
GK/K , 〈−1〉
)
H1
(
GK/K ,AutK(E)
)
//
i
// H1
(
GK/K ,
AutK(E)�〈−1〉
)
.
Note that H0
(
GK/K , 〈−1〉
)
= Z/2Z. By [21, Chapter III, Corollary 10.2], we have the
following automorphism groups of an elliptic curve:
1) AutK(E) = 〈−1〉 ∼= Z/2Z, when j(E) 6= 0, 123;
2) AutK(E) ∼= µ4(K), when j(E) = 123 and char(K) 6∈ {2, 3};
3) AutK(E) ∼= µ6(K), when j(E) = 0 and char(K) 6∈ {2, 3}.
We consider each case separately.
1. Since # AutK(E) = 2, the Galois group GK/K acts trivially on AutK(E). Therefore,
#H0
(
GK/K ,
AutK(E)�〈−1〉
)
= 1. Hence, the map i is injective. This proves the proposition in
this case.
8 M. Kronberg, M.A. Soomro and J. Top
2. First, suppose GK/K acts trivially on AutK(E) ∼= µ4 =
{
1, ζ4, ζ
2
4 , ζ
3
4
}
. Again, here we
see #H0
(
GK/K ,
AutK(E)�〈−1〉
)
= 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
1 // 2 // 4
π // 2 // .
This implies that π is surjective; therefore, i is injective. The proposition follows in this case.
Now, suppose GK/K acts non-trivially on AutK(E). Thus, there exists an automorphism
σ ∈ GK/K such that
σ(1) = 1, σ(ζ4) = ζ3
4 .
Therefore, #H0
(
GK/K ,AutK(E)
)
= 2. Now, the action of σ on
AutK(E)�〈−1〉 ∼=
{{
1, ζ2
4
}
,
{
ζ4, ζ
3
4
}}
is
σ
({
1, ζ2
4
})
=
{
1, ζ2
4
}
, σ
({
ζ4, ζ
3
4
})
=
{
ζ4, ζ
3
4
}
.
We conclude that #H0
(
GK/K ,
AutK(E)�〈−1〉
)
= 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
1 // 2 // 2
π // 2 // .
Thus, π is the constant map; therefore, #Ker(i) = 2 and i is not injective. The proposition
follows in this case.
3. First, if GK/K acts trivially on AutK(E) ∼= µ6 = 〈ζ6〉, then we have #H0
(
GK/K ,AutK(E)
)
= 6 and #H0
(
GK/K ,
AutK(E)�〈−1〉
)
= 3. The first four groups in the long exact sequence
from the start of this proof therefore have order
1 // 2 // 6
π // 3 // .
This implies that π is surjective; therefore, because the sequence is exact, i is injective.
Now, suppose GK/K acts non-trivially on AutK(E). Let σ ∈ GK/K acts non-trivially on
AutK(E). Then we have
σ(1) = 1, σ(ζ6) = ζ5
6 .
Thus we get #H0
(
GK/K ,AutK(E)
)
= 2. The action of σ on
AutK(E)�〈−1〉 ∼=
{{
1, ζ3
6
}
,
{
ζ6, ζ
4
6
}
,
{
ζ2
6 , ζ
5
6
}}
is given by
σ
({
1, ζ3
6
})
=
{
1, ζ3
6
}
, σ
({
ζ6, ζ
4
6
})
=
{
ζ2
6 , ζ
5
6
}
, σ
({
ζ2
6 , ζ
5
6
})
=
{
ζ6, ζ
4
6
}
,
implying that #H0
(
GK/K ,
AutK(E)�〈−1〉
)
= 1. The first four groups in the long exact sequence
used throughout this argument have orders
1 // 2 // 2
π // 1 // .
We conclude that π is surjective; hence, i is injective. This completes the proof of the proposi-
tion. �
Twists of Elliptic Curves 9
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 char(K) ∈ {2, 3}. We briefly consider these two
excluded cases here.
Suppose char(K) = 2 and take E/K : y2 + y = x3. As in Section 3 one finds that GK/K acts
trivially on AutK(E) precisely when F4 ⊂ 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 char(K) = 3 and take E : y2 = x3 − x. Again, there is capitulation of
quadratic twists precisely when the Galois action on the automorphism group is non-trivial,
which happens precisely when F9 6⊂ K.
Example 4.2. Take
E/Q : y2 = x3 − x.
Then
AutQ(E) = {±1,±ι},
where ι : E → E is defined by (x, y) 7→ (−x, iy) for a fixed choice of a primitive 4th root of
unity i ∈ Q.
For d ∈ Q∗ write
E(d) : y2 = x3 − d2x.
Then E(d) is a twist of E/Q, since ψd : E → E(d) defined as
ψd(x, y) =
(
dx, d
√
dy
)
is an isomorphism between E and E(d).
If σ ∈ GQ/Q, then
(σ
ψd
)−1 ◦ ψd =
{
1 if σ
(√
d
)
=
√
d,
−1 if σ
(√
d
)
= −
√
d.
So E(d) corresponds to the cocycle class of
σ 7→
σ
(√
d
)
√
d
∈ AutQ̄(E).
In the case d = −1, this cocycle is a coboundary, since
σ
(√
−1
)
√
−1
= (σι)−1 ◦ ι.
So E(−1) ∼= E over Q, which is, of course, evident from the equation.
Example 4.3. Take q a power of an odd prime, and
E/Fq : y2 = x3 − x.
The Galois group GFq/Fq acts non-trivially on AutFq(E) if and only if −1 is not a square in Fq.
We have
√
−1 6∈ Fq ⇐⇒ q ≡ 3 (mod 4).
10 M. Kronberg, M.A. Soomro and J. Top
For d ∈ F∗q , define E(d)/Fq as before. This provides a quadratic twist as in Example 4.2.
If d is not a square and q ≡ 1 (mod 4), then E(d) is the (unique) non-trivial quadratic twist
of E/Fq.
If d is not a square and q ≡ 3 (mod 4), then −d is a square. Therefore, we have E(d) =
E(−d) ∼= E over Fq. So for q ≡ 3 (mod 4), a non-trivial quadratic twist of E/Fq does not exist.
5 Capitulation of cubic twists
Let E/K be an elliptic curve such that AutK(E) contains a subgroup C3 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 char(K) = 2, 3 the group AutK(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 C3 ⊂ AutK(E) of order 3. The map
i : H1
(
GK/K , C3
)
−→ H1
(
GK/K ,AutK(E)
)
is injective except possibly when char(K) = 2.
Proof. 1. First we consider the case char(K) 6= 2, 3. Then AutK(E) is cyclic of order 6, so
indeed C3 as desired exists and is unique. If GK/K acts trivially on AutK(E) we have
#H0
(
GK/K ,
AutK(E)�µ3
)
= 2.
In the long exact sequence
1 // H0
(
GK/K , C3
)
// H0
(
GK/K ,AutK(E)
)
H0
(
GK/K ,
AutK(E)�C3
)
//
π
// H1
(
GK/K , C3
)
H1
(
GK/K ,AutK(E)
)
//
i
// H1
(
GK/K ,
AutK(E)�C3
)
the orders of the first few groups are
1 // 3 // 6
π // 2 //
and thus, π is surjective which implies i is injective.
If on the other hand GK/K acts non-trivially on AutK(E) then since Galois fixes the −1-map,
any σ ∈ GK/K that acts non-trivially has to interchange the two non-trivial elements of C3. This
implies that GK/K acts trivially on AutK(E)�µ3 and thus, in the long exact sequence presented
earlier in this proof π is surjective since #H0
(
GK/K , µ3
)
= 1 and #H0
(
GK/K ,AutK(E)
)
= 1.
So again one concludes that i is injective.
2. Let char(K) = 3. This implies AutK(E) is a semi-direct product C3 o C4 of cyclic
groups of order 3 and 4 (see [21, Appendix A, Example A.1]). In particular it follows that the
Twists of Elliptic Curves 11
automorphism group has a unique subgroup C3 of order 3. If GK/K acts trivially on AutK(E),
then i is injective. If GK/K acts non-trivially on AutK(E) we will consider several cases.
First we consider the case that GK/K acts trivially on C3. This implies that any non-
trivially acting σ ∈ GK/K interchanges the two elements of order 4 in C4. This implies
#H0
(
GK/K , C3
)
= 3 and #H0
(
GK/K ,AutK(E)
)
= 6 and #H0
(
GK/K ,
AutK(E)�C3
)
= 2.
This gives us the sequence of orders
1 // 3 // 6
π // 2 //
from which it follows that π is surjective and thus i is injective.
Now consider the case that all elements of C4 are fixed under the action of GK/K . This
implies that any σ ∈ GK/K acting non-trivially on AutK(E) has to interchange the non-trivial
elements of C3. Therefore, we get
#H0
(
GK/K , C3
)
= 1,
#H0
(
GK/K ,AutK(E)
)
= 4,
#H0
(
GK/K ,
AutK(E)�C3
)
= 4.
Similar to the previous situation this implies that i is injective.
In the case that neither C3 nor C4 are elementwise fixed under the action of GK/K , we easily
get the following sequence of orders
1 // 1 // 2
π // 1 //
and thus again, i is injective. �
So the result says that capitulation of cubic twists does not occur, except possibly in char-
acteristic 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 AutK(E) has a normal subgroup of order 6. This implies
j(E) = 0 and char(K) 6= 2. A result analogous to Proposition 5.1 is the following.
Proposition 6.1. Suppose char(K) 6= 2 and let E/K be an elliptic curve with j(E) = 0. There
is a unique (and therefore normal) subgroup C6 ⊂ AutK(E) of order 6. The map
i : H1
(
GK/K , C6
)
−→ H1
(
GK/K ,AutK(E)
)
is injective except in the two cases
1) char(K) = 3 and GK/K acts trivially on C6;
2) char(K) = 3 and the only elements in AutK(E) fixed by GK/K are ±1.
Proof. In the case char(K) 6= 2, 3 we have for E as above that AutK(E) is cyclic of order 6.
So the result is trivial in this case.
Now assume char(K) = 3. In this case # AutK(E) = 12 and this automorphism group
contains a unique subgroup C6 of order 6. It is generated by the unique element of order 2 and
the subgroup C3 of order 3 in AutK(E). In the case that GK/K acts trivially on AutK(E) we
get once again that i is injective. Thus, we now assume that the Galois action is non-trivial.
12 M. Kronberg, M.A. Soomro and J. Top
First case: the Galois action on C3 is trivial. Then analogously to the case of cubic twists
here #H0(GK/K , C6) = 6 and #H0
(
GK/K ,AutK(E)
)
= 6. Obviously, GK/K fixes the residue
classes modulo C6. Thus the group H0
(
GK/K ,
AutK(E)�C6
)
has order 2. This gives us the
sequence of orders
1 // 6 // 6
π // 2 // ,
implying in the same way as in earlier cases that i is not injective, and the map
π : H0
(
GK/K ,AutK(E)
)
→ H0
(
GK/K ,
AutK(E)�H
)
is constant.
Second case: the Galois action fixes the points in a cyclic order 4 subgroup of automor-
phisms. Then #H0(GK/K , C6) = 2 and #H0(GK/K ,AutK(E)) = 4. Furthermore the action on
AutK(E)�C6
is trivial, which gives us the sequence of orders
1 // 2 // 4
π // 2 // .
As before we conclude that i is injective.
Third case: Neither C3 nor an order 4 subgroup C4 are pointwise fixed under the action
of GK/K . Then only ±1 are fixed in AutK(E) and C6. Further, we see that the action on the
quotient group again is trivial. This implies for the orders in the long exact sequence
1 // 2 // 2
π // 2 // .
So reasoning as before, π 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 = 3n and consider E/Fq given by y2 = x3 − x. The automorphism
Φi,0 : (x, y) 7→ (−x,
√
−1y) generates a Galois stable subgroup H of AutFq(E) of order 4. Then
Fr 7→ Φi,0 defines a cocycle in H1(GK/K , H) and in H1(GK/K ,AutFq(E)). In Proposition 2.1
we saw that this corresponds to a non-trivial twist.
Example 7.2. Similarly we put q = 2n and E/Fq given by y2 +y = x3. With ω ∈ Fq a primitive
third root of unity, the automorphism Φω2,0,1 : (x, y) 7→ (ωx, y + 1) generates a Galois stable
subgroup H of AutFq(E) of order 6, and Fr 7→ Φω2,0,1 defines a cocycle in H1(GK/K , H) and in
H1(GK/K ,AutFq(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.
Twists of Elliptic Curves 13
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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1 Introduction
2 Twists in characteristic three
3 Twists in characteristic two
4 Capitulation of quadratic twists
5 Capitulation of cubic twists
6 Capitulation of sextic twists
7 Other twists
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