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 $H^1\big(\operatorname{G}_{\overline{K}/K}, \operatorname{Aut}_{\overline{K}}(E)\big)$. The results are illustrated by examples.


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 G K/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 automorphism group of E we write Aut K (E). The elements of Twist(E/K) are in one-to-one correspondence with the classes in H 1 G K/K , Aut K (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 Note, by the way, that Ian Connell at McGill University (Montreal) in the late 1990's wrote extensive unpublished lecture notes on elliptic curves, including a long chapter on twists [6].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 µ n (K) the group of n-th roots of unity in K × and by ζ n a generator of this group. In [21], Silverman only presents an explicit description of the twists of an elliptic curve E/K in char K = 2, 3. The main reason is that this condition implies Aut K (E) ∼ = µ n (K), for some n ∈ {2, 4, 6}, even as G K/K -modules. In characteristic 2 the group Aut K (E) is either isomorphic to Z/2Z or to a non-abelian group of order 24. In characteristic three the group Aut K (E) is either equals µ 2 (K) ∼ = Z/2Z or it is a non-abelian group of order 12. By explicitly describing H 1 G K/K , Aut K (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 Aut K (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].

Twists in characteristic three
We start by considering elliptic curves over finite fields 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 F = F q of cardinality q are in one-to-one correspondence with the Frobenius conjugacy classes in Aut F (E). By definition a Frobenius conjugacy class is obtained by fixing some τ ∈ Aut F (E), then its Frobenius conjugacy class consists of all 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.
has j(E) = 0, and it has precisely twelve automorphisms. These are given by where u 4 = 1 and r ∈ F 3 . Let n ≥ 1 and q = 3 n . For u ∈ F 9 with u 4 = 1 and r ∈ F 3 , put (this is the Frobenius conjugacy class of Φ u,r over F q ).
If n is odd, then the Frobenius conjugacy classes of E/F q are In particular there are precisely three non-trivial twists of E over F q in case n is odd, which are given as follows. Consider the cocycle defined by Fr → Φ u,0 . The corresponding twist is given by and the corresponding isomorphism is defined over a quadratic extension. Analogously the cocycle Fr → Φ 1,1 corresponds to the twist and the cocycle Fr → Φ 1,−1 corresponds to the twist where both isomorphisms are defined over a cubic extension. In case n is even, fix i ∈ F q with i 2 = −1. The Frobenius conjugacy classes of E/F q are 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 F 3 n -isomorphism classes of elliptic curves with j-invariant 0, i.e., of supersingular elliptic curves over 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 Φ −1 u,r = Φ u −1 ,−u 2 r and Fr Φ u,r = Φ u 3 ,r 3 = Φ u −1 ,r , one can directly calculate the Frobenius conjugacy class of Φ 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 ψ : E → E tw and check that Fr ψ −1 • ψ is in the Frobenius conjugacy class. For example, with E tw : The other cases are done similarly.
Note that we only presented explicit equations for the twists in the case of an extension of F 3 of odd degree. If the degree is even, such an equation will in general (as expected) depend on the field F q .
Since we are considering here the number of F 3 n -isomorphism classes of elliptic curves with j-invariant 0, i.e., of supersingular elliptic curves over 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 char(K) = 3 such that # Aut K (E) = 12. This means (compare [21, Appendix A]) that j(E) = 0 and E is given by an equation 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) = (u 2 x + r, u 3 y), where u 4 = − 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 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) = 0 satisfies Aut K (E) = ±1 and therefore 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/F q is an elliptic curve. Then

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 G K/K on Aut K (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.
has j(E) = 0 and it has exactly 24 automorphisms. These are described as where u ∈ F * 4 , r ∈ F 4 and t 2 + t + r 3 = 0. The action of the Galois group G Fq/Fq on Aut Fq (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 Aut Fq (E), namely the conjugacy class C 1,0,0 containing the identity, the class C 1,ω,ω containing Φ 1,ω,ω , and the class C 1,ω,ω 2 containing Φ 1,ω,ω 2 . Here ω ∈ F 4 is a primitive 3rd root of unity. The two Frobenius conjugacy classes corresponding to non-trivial twists of E over F q yield twists of E over F q which are isomorphic to E over a degree eight extension of F q .
In case n is even, the Frobenius conjugacy classes coincide with the usual conjugacy classes in Aut Fq (E), which are (with C u,r,t denoting the conjugacy class containing Φ u,r,t and ω ∈ F 4 a chosen primitive 3rd root of unity) The twists of E/F q corresponding to these conjugacy classes are isomorphic to E over an extension of F q of degree 1, 2, 6, 6, 3, 3, 4, respectively.
In particular E/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 Aut(E) are immediate; compare [21,Appendix A].
In any automorphism Φ u,r,t one has r ∈ F 4 hence r 3 = 1 if r = 0 and r 3 = 0 for r = 0. So the equality t 2 + t + r 3 shows t ∈ F 2 for r = 0 and t ∈ F 4 for r = 0. Therefore all automorphisms are defined over 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 Φ u,r,t . Let ω ∈ F 4 be a fixed primitive 3rd root of unity. A direct calculation shows The other two Frobenius conjugacy classes are given by To see that a twist of E/F q corresponding to one of the latter two Frobenius conjugacy classes is indeed isomorphic to E over F q 8 and not over a smaller extension of F q , consider a cocycle defined by Fr → Φ (with Φ in one of the given classes C u,r,s ). Using the cocycle condition one finds that the cocycle sends Fr j (for j ≥ 1) to (id + Fr +···+Fr j−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 C 1,0,0 = {Φ 1,0,0 }, So indeed there are exactly 6 non-trivial twists of E/F q in this case. The statement about the extension of 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 = Φ 1,0,1 is in the same Frobenius conjugacy class as the identity, the elliptic curve E/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/F 2 has no non-trivial cubic twists (and as the proposition states, any non-trivial twist of E/F 2 will only be isomorphic to E over extensions of F 2 of degree a multiple of 8).
Explicit equations for the two non-trivial twists of E/F q , in the case q = 2 n with n odd, are as follows. Let E 1 : y 2 + y = x 3 + x and E 2 : 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 F 2 n (compare [21, Section V.2]) one finds #E(F 2 n ) and #E 1 (F 2 n ) and #E 2 (F 2 n ) are three distinct numbers whenever n is odd. So indeed the curves E j are distinct and nontrivial twists of E/F q . Of course the same conclusion can be obtained by exhibiting an explicit isomorphism ψ : E → E j and then determining the Frobenius conjugacy class containing Fr ψ −1 • ψ.
We will not try to present equations for all non-trivial twists of E/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 Fr → −1 = Φ 1,0,1 defines a non-trivial cocycle class. The corresponding twist is given by with ω a primitive 3rd root of unity), since ψ : (x, y) → (x, y + τ ), with τ ∈ F 4 satisfying τ 2 + τ + ω = 0, defines an isomorphism ψ : E → E tw , and Thus, E/F 4 has a non-trivial quadratic twist. Let K be an field with char(K) = 2 and consider the elliptic curves E : y 2 + ay = x 3 + bx + c with a = 0 and E : y 2 + y = x 3 . 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) = (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 ±1 unless j(E) = 0.

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 ⊂ Aut K (E) induces a map The set of quadratic twists of E, i.e., is a subset of Twist(E/K); this subset of quadratic twists corresponds to the image of H 1 G K/K , −1 in H 1 G K/K , Aut K (E) under the map just given. 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 G K/K , Aut K (E) , under the assumption that it comes from a non-trivial element in the group H 1 G K/K , −1 = Hom G K/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 Aut K (E) is abelian. Then the map is injective except in the case when char(K) ∈ {2, 3}, j(E) = 12 3 and G K/K acts non-trivially on Aut K (E).
We consider each case separately.
1. Since # Aut K (E) = 2, the Galois group G K/K acts trivially on Aut K (E). Therefore, Hence, the map i is injective. This proves the proposition in this case.
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 : y 2 + y = x 3 . As in Section 3 one finds that G K/K acts trivially on Aut K (E) precisely when F 4 ⊂ 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 : Again, there is capitulation of quadratic twists precisely when the Galois action on the automorphism group is non-trivial, which happens precisely when F 9 ⊂ K. For d ∈ Q * write is an isomorphism between E and E (d) .
If σ ∈ G Q/Q , then So E (d) corresponds to the cocycle class of In the case d = −1, this cocycle is a coboundary, since So E (−1) ∼ = E over Q, which is, of course, evident from the equation. The Galois group G Fq/Fq acts non-trivially on Aut Fq (E) if and only if −1 is not a square in F q . We have For d ∈ F * q , define E (d) /F q 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/F q .
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 F q . So for q ≡ 3 (mod 4), a non-trivial quadratic twist of E/F q does not exist.

Capitulation of cubic twists
Let E/K be an elliptic curve such that Aut K (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 char(K) = 2, 3 the group Aut K (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 ⊂ Aut K (E) of order 3. The map is injective except possibly when char(K) = 2.
Proof . 1. First we consider the case char(K) = 2, 3. Then Aut K (E) is cyclic of order 6, so indeed C 3 as desired exists and is unique. If G K/K acts trivially on Aut K (E) we have In the long exact sequence the orders of the first few groups are and thus, π is surjective which implies i is injective. If on the other hand G K/K acts non-trivially on Aut K (E) then since Galois fixes the −1-map, any σ ∈ G K/K that acts non-trivially has to interchange the two non-trivial elements of C 3 . This implies that G K/K acts trivially on Aut K (E) µ 3 and thus, in the long exact sequence presented earlier in this proof π is surjective since #H 0 G K/K , µ 3 = 1 and #H 0 G K/K , Aut K (E) = 1. So again one concludes that i is injective.
2. Let char(K) = 3. This implies Aut K (E) is a semi-direct product C 3 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 G K/K acts trivially on Aut K (E), then i is injective. If G K/K acts non-trivially on Aut K (E) we will consider several cases.
First we consider the case that G K/K acts trivially on C 3 . This implies that any nontrivially acting σ ∈ G K/K interchanges the two elements of order 4 in C 4 . This implies #H 0 G K/K , C 3 = 3 and #H 0 G K/K , Aut K (E) = 6 and #H 0 G K/K , Aut K (E) C 3 = 2. This gives us the sequence of orders from which it follows that π is surjective and thus i is injective. Now consider the case that all elements of C 4 are fixed under the action of G K/K . This implies that any σ ∈ G K/K acting non-trivially on Aut K (E) has to interchange the non-trivial elements of C 3 . Therefore, we get #H 0 G K/K , C 3 = 1, 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 G K/K , we easily get the following sequence of orders 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.

Capitulation of sextic twists
Let E/K be an elliptic curve such that Aut K (E) has a normal subgroup of order 6. This implies j(E) = 0 and char(K) = 2. A result analogous to Proposition 5.1 is the following. Proposition 6.1. Suppose char(K) = 2 and let E/K be an elliptic curve with j(E) = 0. There is a unique (and therefore normal) subgroup C 6 ⊂ Aut K (E) of order 6. The map i : is injective except in the two cases 1) char(K) = 3 and G K/K acts trivially on C 6 ; 2) char(K) = 3 and the only elements in Aut K (E) fixed by G K/K are ±1.
Proof . In the case char(K) = 2, 3 we have for E as above that Aut K (E) is cyclic of order 6. So the result is trivial in this case. Now assume char(K) = 3. In this case # Aut K (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 Aut K (E). In the case that G K/K acts trivially on Aut K (E) we get once again that i is injective. Thus, we now assume that the Galois action is non-trivial.