Twists of Elliptic Curves

arXiv:1707.02893v1 [math.NT] 10 Jul 2017

Twists of Elliptic Curves Max Kronberg, Muhammad Afzal Soomro, and Jaap Top Dedicated to Noriko Yui. The third author of this note was a postdoc with her at Queen’s University during 1989–1990. 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 all remaining cases. For this, we make explicit use ofthe correspondence  between the twists and the Galois cohomology set H 1 GK/K , AutK (E) . The results are illustrated by examples.

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 E tw /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 of Twist(E/K) are automorphism group of E we write AutK (E). The elements  

in one-to-one correspondence with the classes in H 1 GK/K , AutK (E) [Sil09, Chap. X, §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, Ø). 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 [MT10], [KP17], [Gar16], [LG16], [TV17], [GY95]. The simplest nontrivial example of twists is provided by the case of elliptic curves, where probably the first account of it was given in [Cas66, Part II, § 9]. We briefly recall some of this theory here. Although it is certainly known to most experts, there seems to be no adequate reference for it and we hope that this note fills this gap. For a positive integer n which is 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 [Sil09], Silverman only presents an explicit description of the twists of an elliptic curve E/K in char K 6= 2, 3. The main reason for this 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 1

to a non-abelian group of order 24. In characteristic three the group AutK (E) is one of the groups µ2 (K) and a non-abelian group of order 12. By explicitly  1 describing H GK/K , AutK (E) in these remaining cases, we complete the

description presented in [Sil09]. 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 the Propositions 2.2, 3.2, where we count the number of twists of any elliptic curve over a finite field. Examples 2.1 and 3.1 indicate how these twists can be given explicitly in the non-trivial cases. Moreover Propositions 4.1 and 5.1 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 started from the PhD thesis of the second author [Soo13, 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 [MT10], the twists of an elliptic curve over a finite field are in one-to-one correspondence with the Frobenius conjugacy classes in AutK (E). We will compute these classes for all possible actions of the absolute Galois group. Example 2.1. Consider the elliptic curve given by E/F3 : y 2 = x3 − x. Then j(E) = 0. By [Sil09, Appendix A, Prop. 1.2] the twelve automorphisms of E are given by Φu,r :

E (x, y)

−→ E

7−→ (u2 x + r, u3 y),

where u4 = 1 and r ∈ F3 . We have Φ−1 u,r = Φu−1 ,−u2 r and Φu−1 ,r . Thus, we obtain the Frobenius conjugacy classes

Fr

Φu,r = Φu3 ,r3 =

C1,0 ={Φ1,0 , Φ−1,0 }

C1,1 ={Φ1,1 , Φ−1,−1 }

C1,−1 ={Φ1,−1 , Φ−1,1 } Ci,0 ={Φ±i,r | r ∈ F3 }, where Cu,r = {Φ−1 u′ ,r ′ ◦Φu,r ◦Φu′3 ,r ′3 | u, r as above} denotes the class of Φu,r and i is a fixed square root of −1. As a consequence, there are precisely three nontrivial twists of E over F3 , which we now describe. First consider the cocycle 2

given by Fr 7→ Φi,0 . The corresponding twist is given by E tw : y 2 = x3 + x, where the isomorphism ψ : E → E tw is given by (x, y) 7→ (ix, −iy). A direct
−1 computation shows that Fr ψ ◦ ψ = Φi,0 . Analogously we see that the cocycle Fr 7→ Φ1,1 corresponds to the twist E tw : y 2 = x3 − x − 1 and the cocycle Fr 7→ Φ1,−1 corresponds to the twist E tw : y 2 = x3 − x + 1. Over F9 the action of the absolute Galois group on the automorphism group of E becomes trivial. This gives us that for n ∈ 2Z we have the following classes for the action of Frobenius C1,0 ={Φ1,0 }

C−1,0 ={Φ−1,0 } C1,1 ={Φ1,1 , Φ1,−1 }

C1,−1 ={Φ1,−1 , Φ−1,−1 } Ci,0 ={Φi,r | r ∈ F3 }

C−i,0 ={Φ−i,r | r ∈ F3 }.

More generally, over extensions of F3 of even degree there exist 6 twists of E and for n 6∈ 2Z the curve E/F3n has 4 twists. 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 [Sch87]. Let us consider now 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 that j(E) = 0 and E is given by an equation y 2 = x3 + ax + b, where a, b ∈ K. Thus, there exists an isomorphism ψ : E → E ′ , where E ′ : y 2 = x3 − x. We are now interested the possibilities for the field extension where the isomorphism is defined. By [Sil09, Appendix A, Prop. 1.2], we have ψ(x, y) = (u2 x + r, u3 y), where u4 = − a1 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. And by ψ −1 (x, y) = 2 (v x + w, v 3 y), where v = u−1 and w3 − w − v 6 b = 0. Thus, both ψ and ψ −1 are defined over the Artin-Schreier extension defined by w3 − w − v 6 b = 0 over K(u). In the case that E[2](K) is non-trivial, we may assume b = 0 and any such isomorphism is defined over K(u). 3

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. Note that any elliptic curve E/K in characteristic 3 with j(E) 6= 0 satisfies 2 AutK (E) = ±1 and therefore H 1 (GK/K , AutK (E)) ∼ = K × /K × . In particular, summarizing most of the discussion above for the special case of a finite field, the result is as follows. Proposition 2.2. Let q = 3n and suppose E/Fq is an elliptic curve. Then   2 if j(E) 6= 0; # Twist(E/Fq ) = 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 arbitrary elliptic curve over a field with characteristic two and this particular example. Example 3.1. Let E/F2 : y 2 + y = x3 , so j(E) = 0. By using the formulae given in [Sil09, Table 3.1], all 24 automorphisms can be described as Φu,r,t :

E (x, y)

−→ E

7−→ (u2 x + r, y + u2 r2 x + t)

,

where u ∈ F∗4 , r ∈ F4 and t2 + t + r3 = 0. Note that with r ∈ F4 one has r3 = 1 if r 6= 0 and r3 = 0 for r = 0. In particular, for given r the two possibilities for t are in F4 \F2 when r 6= 0 and in F2 otherwise. We have  AutF2 (E) = Φu,r,t |u ∈ F∗4 , r ∈ F4 and t2 + t + r3 = 0 . We now describe the set of Frobenius conjugacy classes {Cu,r,t } in AutF2 , where o n u ∈ F∗4 , r˜ ∈ F4 and t˜ ∈ F16 satisfies t˜2 + t˜ + r˜3 = 0 . Cu,r,t = Φ−1 Φ Φ ˜2 ,˜r2 ,t˜2 |˜ u,˜ ˜ r,t˜ u,r,t u

For any Φu,r,t ∈ AutF2 (E) we have Φ−1 u,r,t = Φu2 ,ur,t+r 3 . We compute the conjugacy class C1,0,0 of the identity.  C1,0,0 = Φ−1 u,r,t Φ1,0,0 Φu2 ,r 2 ,t2 |u, r and t are as above  = Φu2 ,ur2 +r,ur . So the identity class C1,0,0 consists of the following automorphisms. 4

u\r 1 ζ3 ζ32

0

1

ζ3

ζ32

Φ1,0,0 Φζ32 ,0,0 Φζ3 ,0,0

Φ1,0,1 Φζ32 ,ζ32 ,ζ3 Φζ3 ,ζ3 ,ζ32

Φ1,1,ζ3 Φζ32 ,ζ32 ,ζ32 Φζ3 ,0,1

Φ1,1,ζ32 Φζ32 ,0,1 Φζ3 ,ζ3 ,ζ3

Since −1 = Φ1,0,1 is in this class, 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, E/F2 has no non-trivial cubic twists. The other two Frobenius conjugacy classes are given by C1,ζ3 ,ζ3 = {Φ1,ζ3 ,ζ3 , Φζ3 ,ζ32 ,ζ3 , Φζ3 ,1,ζ32 , Φζ3 ,ζ3 ,ζ3 , Φζ32 ,1,ζ3 , Φ1,ζ32 ,ζ32 },

C1,ζ3 ,ζ32 = {Φ1,ζ3 ,ζ32 , Φζ3 ,ζ32 ,ζ32 , Φζ32 ,1,ζ32 , Φζ32 ,ζ3 ,ζ3 , Φζ3 ,1,ζ3 , Φ1,ζ3 ,ζ32 }. These two classes correspond to the two non-trivial twists of E over F2 . In both cases an isomorphism is defined over a degree eight extension of F2 . Now, if we consider E over F4 , then the action of GF4 /F4 on AutF4 (E) is trivial. The Frobenius conjugacy classes in this case are C1,0,0 ={Φ1,0,0 },

C1,0,1 ={Φ1,0,1 }, (Fr2 7→ 1),

Cζ32 ,0,1 ={Φζ32 ,0,1 , Φζ32 ,1,ζ3 , Φζ32 ,ζ32 ,ζ3 , Φζ32 ,ζ3 ,ζ3 }, (Fr6 7→ 1), Cζ3 ,0,1 ={Φζ3 ,0,1 , Φζ3 ,1,ζ32 , Φζ3 ,ζ32 ,ζ32 , Φζ3 ,ζ3 ,ζ32 }, (Fr6 7→ 1), Cζ3 ,0,0 ={Φζ3 ,0,0 , Φζ3 ,1,ζ3 , Φζ3 ,ζ32 ,ζ3 , Φζ3 ,ζ3 ,ζ3 }, (Fr3 7→ 1),

Cζ32 ,0,0 ={Φζ32 ,0,0 , Φζ32 ,1,ζ32 , Φζ32 ,ζ32 ,ζ32 , Φζ32 ,ζ3 ,ζ32 }, (Fr3 7→ 1),

C1,1,ζ3 ={Φ1,1,ζ3 , Φ1,1,ζ32 , Φ1,ζ3 ,ζ3 , Φ1,ζ3 ,ζ32 , Φ1,ζ32 ,ζ3 , Φ1,ζ32 ,ζ32 }, (Fr4 7→ 1).

Thus, Fr 7→ −1 = Φ1,0,1 defines a non-trivial cocycle class. The corresponding twist is given by E tw : y 2 + y = x3 + ζ3 , since ψ : (x, y) 7→ (x, y + τ ),

with τ ∈ F4 satisfying τ 2 + τ + ζ3 = 0, defines an isomorphism ψ : E → E tw , and (Fr ψ)−1 ◦ ψ = −1. Thus, E/F4 has a non-trivial quadratic twist. In general we see that E/F2n has two non-trivial twists if n is odd, and six in case n is even. Let K be an field with char(K) = 2 and consider the elliptic curves E : y 2 + ay = x3 + bx + c with a 6= 0 and E ′ : y 2 + y = x3 . Since j(E) = 0 = j(E ′ ) these elliptic curves are isomorphic and, by Silverman [Sil09, Appendix A, Prop. 5

1.2], for an isomorphism ψ : E → E ′ we have ψ(x, y) = (u2 x + s2 , ay + u2 s + t), where u3 = a, s4 + as + b = 0 and t2 + at + s6 + bs2 + c = 0. Moreover from the information presented in Example 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   2 if j(E) 6= 0; # Twist(E/Fq ) = 3 if j(E) = 0 and n is odd;  7 if j(E) = 0 and n is even.

4

Quadratic Twists

The set of quadratic twists of E, i.e.,  QT (E) = E tw /K | ∃ L/K with [L : K] = 2 s.t. E tw ∼ =L E /∼ =K

is a subset of Twist(E/K);  this subset of quadratic twistscorresponds to the 1 image of H GK/K , h−1i in the set H 1 GK/K , AutK (E) under the map in-

duced by the inclusion h−1i ⊂ AutK (E). Here we consider the question whether E tw ∈ QT (E) can be isomorphic to E over the ground field.  In other words, 

when does E tw correspond to the trivial element in H 1 GK/K , AutK (E) , under that  the assumption   it comes from  a non-trivial element in the group 1 H GK/K , h−1i = Hom GK/K , h−1i . Proposition 4.1. Let E/K be an elliptic curve such that AutK (E) is abelian, which means we exclude the case j(E) = 0 in char(K) ∈ {2, 3}. Then the map     i : H 1 GK/K , h−1i −→ H 1 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).

6

Proof. We have the following long exact sequence of groups.   / H0 G , h−1i K/K

1

  / H0 G , Aut (E) K/K K ED BC

π

@A GF

/ H0



 GK/K , AutK (E)h−1i

  / H1 G , h−1i K/K

i

GF @A

/ H1



ED BC

  AutK (E) / H1 G , K/K h−1i

 GK/K , AutK (E)

  Note that H 0 GK/K , h−1i = h−1i. By [Sil09, Chap. III, Cor. 10.2], we have the following automorphism groups of an elliptic curve. 1. AutK (E) ∼ = 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, #H 0 GK/K , AutK (E)h−1i = 1. Hence, the map i is injective. This proves the proposition in this case.  2. First, suppose GK/K acts trivially on AutK (E) ∼ = µ4 = 1, ζ4 , ζ42 , ζ43 .   Again, here we see #H 0 GK/K , AutK (E)h−1i = 2. Hence, the first four groups in the above long exact sequence have order as indicated in the following diagram. /2 /4 π /2 /. 1 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 ) = ζ43 .   Therefore, #H 0 GK/K , AutK (E) = 2. Now, the action of σ on  2  3 AutK (E) ∼ h−1i = 1, ζ4 , ζ4 , ζ4

7

is   σ( 1, ζ42 ) = 1, ζ42 ,   σ( ζ4 , ζ43 ) = ζ4 , ζ43 .   We conclude that #H 0 GK/K , AutK (E)h−1i = 2. Hence, the first four groups in the long exact sequence have order as indicated below. 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 = hζ6 i, then we have     #H 0 GK/K , AutK (E) = 6 and #H 0 GK/K , AutK (E)h−1i = 3. The first four groups in the long exact sequence 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 ) = ζ65 .   Thus we get #H 0 GK/K , AutK (E) = 2. Now, the action of σ on  2 5  3  4 AutK (E) ∼ h−1i = 1, ζ6 , ζ6 , ζ6 , ζ6 , ζ6 is given by   σ( 1, ζ63 ) = 1, ζ63 ;   σ( ζ6 , ζ64 ) = ζ62 , ζ65 ;   σ( ζ62 , ζ65 ) = ζ6 , ζ64 ,   implying that #H 0 GK/K , AutK (E)h−1i = 1. The first four groups in the above long exact sequence have orders 1

/2

/2

π

/1

/.

We conclude that π is surjective; hence, i is injective. This completes the proof of the proposition. 8

Example 4.2. Take E/Q : y 2 = x3 − x. Then AutQ (E) = {±1, ±ι} , √ where ι : E → E is defined by (x, y) 7→ (−x, −1y) for a fixed choice of √ −1 ∈ Q. For d ∈ Q∗ write E (d) : y 2 = x3 − d2 x. 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

−1

( ψd )

◦ ψd =



1 −1

√ √ d; if σ(√d) = √ if σ( d) = − d.

So E (d) corresponds to the cocycle class of √ σ( d) ∈ AutQ¯ (E). σ 7→ √ d In the case d = −1, this cocycle is a coboundary, since √ σ( −1) √ = (σ ι)−1 ◦ ι. −1 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 : y 2 = 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). 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. 9

5

Cubic Twists

Let E/K be an elliptic curve such that AutK (E) has a subgroup of order 3. This implies j(E) = 0 by [Sil09, Chap. III, Cor. 10.2]. Thus, we restrict ourselves in this section to elliptic curves E with j(E) = 0. We want to remark, 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 nonabelian cohomology needed to describe twists in this situation, is, e.g., described in Serre’s books [Ser97, Chapter I, § 5] and [Ser79, Chapter XIII]. Proposition 5.1. Let E/K be an elliptic curve with j(E) = 0. The map     i : H 1 GK/K , µ3 −→ H 1 GK/K , AutK (E) is injective except in the case when char(K) = 3 and GK/K acts non-trivially on µ4 ⊂ AutK (E) or when char(K) = 2. Proof. 1. First we consider the case char(K) 6= 2, 3. In this case we have AutK (E) ∼ = µ6 . If GK/K acts trivially on AutK (E) we have   #H 0 GK/K , AutK (E)µ3 = 2. This gives us the following sequence of group orders 1

/3

/6

π

/2 /

and thus, π is surjective. If GK/K acts non-trivially on AutK (E) we have for any σ ∈ GK/K that acts non-trivially has to exchange the two primitive third roots of unity. This implies that GK/K acts trivially on AutK (E)µ3 and thus, π is surjective since     #H 0 GK/K , µ3 = 1 and #H 0 GK/K , AutK (E) = 1. ∼ Z/3Z ⋊ µ4 2. Let char(K) = 3 and j(E) = 0. This implies AutK (E) = by [Sil09, Appendix A, Ex. A.1]. If GK/K acts trivially on AutK (E), then i is injective. If GK/K acts non-trivially on AutK (E) we have to consider several cases. First we consider the case that Z/3Z is fixed under the action of GK/K . This implies that any non-trivially acting σ ∈ GK/K has to send ζ4 to ζ43 and fixes  2 1, ζ4 = µ2 ⊂ µ4 everything else in AutK (E). This is due to the fact that  is fixed under the action of GK/K . This implies #H 0 GK/K , Z/3Z = 3 and   #H 0 GK/K , AutK (E) = 3. We now compute   H 0 GK/K , AutK (E)(Z/3Z) . For this, we write αi := ζ4i (Z/3Z) 10

for i = 0, . . . , 3. This gives AutK (E) ∼ (Z/3Z) = {αi | i ∈ {0, . . . , 3}} . Let now σ ∈ GK/K act non-trivially on AutK (E), then exactly α0 and α2 are fixed under σ. Thus,   #H 0 GK/K , AutK (E)(Z/3Z) = 2. This gives us the sequence of orders 1

/3

/3

π

/2 /

from which it follows that π is not surjective and thus, i is not injective. Now consider the case that µ4 is fixed under the action of GK/K . This implies that any σ ∈ GK/K acting non-trivially on AutK (E) has to interchange ζ3 and ζ32 . Therefore, we get   #H 0 GK/K , Z/3Z = 1;   #H 0 GK/K , AutK (E) = 4;   #H 0 GK/K , AutK (E)(Z/3Z) = 4. This implies i is injective. In the case that neither Z/3Z nor µ4 is fixed under the action of GK/K , we easily get the following sequence of orders 1

/1

/2

π

/1 /

and thus, i is injective.

6

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. In the case char(K) 6= 2, 3 we have for j(E) = 0 that # AutK (E) = 6. Thus, π is surjective and i is injective. Let us assume now that char(K) = 3. In this case # AutK (E) = 12 and the group H of order 6 is generated by the unique element of order 2 and the subgroup of order 3 is normal 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. First case: µ3 is fixed under the action. Then we have that any non-trivially acting σ ∈ GK/K interchanges the two elements of order 4 in AutK (E) and fixes 11

everything else. This implies #H 0 (GK/K , H) = 6 and #H 0 (GK/K , AutK (E)) = 6. Obviously, GK/K fixes the residue classes modulo H. Thus, we have two elements in #H 0 (GK/K , AutK (E)H ). This gives us the sequence of orders 1

/6

/6

π

/2

/ ,

implying that π is constant and i is not injective. Second case: µ4 is fixed under the Galois action. Then #H 0 (GK/K , H) = 2 and #H 0 (GK/K , AutK (E)) = 4. Furthermore the action on AutK (E)H is trivial, which gives us the following sequence of orders 1

/2

/4

π

/2

/ .

Thus, we can conclude i is injective. Third case: Neither µ3 nor µ4 are fixed under the action of GK/K . Then only ±1 are fixed in AutK (E) and H. Further, we see that the action on the quotient group again is trivial. This implies for the orders in the long exact sequence /2 π /2 / . /2 1 So, π is constant and i is not injective. This case concludes the proof in characteristic 3.

7

Other Twists

Although the techniques used in the previous sections require the (cyclic and Galois stable) subgroup H 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 y 2 = 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 H 1 (GK/K , H) and in H 1 (GK/K , AutFq (E)). In Example 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 y 2 + y = x3 . With ζ3 ∈ Fq a primitive thrird root of unity, the automorphism Φζ32 ,0,1 : (x, y) 7→ (ζ3 x, y + 1) generates a Galois stable subgroup H of AutFq (E) of order 6, and Fr 7→ Φζ32 ,0,1 defines a cocycle in H 1 (GK/K , H) and in H 1 (GK/K , AutFq (E)). The calculation presented in Example 3.1 shows that for odd n this results in a trivial twist, and for n even one obtains a non-trivial twist.

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