Jacobians of Curves of Genus One

A thesis presented

by

Catherine Helen O’Neil

to

The Department of Mathematics in partial fulfillment of the requirements

for the degree of Doctor of Philosophy

in the subject of Mathematics

Harvard University Cambridge, Massachusetts

April, 1999

\begin{matrix}c \\ ◯ \end{matrix} 1999 by Catherine Helen O’Neil All rights reserved.

Abstract

In this thesis we consider the passage from curves of genus one to their Jacobian elliptic curves. More specifically, given the equations for a curve of genus one, our goal is to produce equations for its Jacobian. To achieve this goal we endow our curve with the structure of a torsion point on its Jacobian. We study this structure in generality and then apply these considerations to the three cases where our curve is embedded in ℙ^{2}, ℙ^{4}, and ℙ^{1} × ℙ^{1}, and is endowed with a 3-, 5-, and 2-torsion point on its Jacobian.

Acknowledgments

I would like to thank my advisor, Barry Mazur, for giving me a really great problem and also some really great advising.

I am deeply indebted to Brian Conrad for a careful reading of my thesis and many very helpful comments.

My understanding of the mathematics in this thesis has benefited hugely through conversations with the many wonderful people surrounding me. I will list only some of them here: Adam Logan, Jordan Ellenberg, Benedict Gross, Tomas Klenke, William McCallum, Michael Roth, Ravi Vakil, and Nicholas Katz.

Finally, I would like to thank my parents, my husband, and my friend Kenneth Ribet for their unbounded support.

Contents

1 Introduction 1

2 Conventions 2

3 General Results 4 3.1 n-prepared genus one curves . . . . . . . . . . . . . . . . . . . 4 3.2 Setting up over a field . . . . . . . . . . . . . . . . . . . . . . 6 3.3 Translation by T . . . . . . . . . . . . . . . . . . . . . . . . . 7 3.4 Embedding E . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.5 Finding x_{0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.6 Standardizing λ_{T} . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.7 The invariant b . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.8 The map det . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.9 The Model for E . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.10 Jacobians of n-prepared genus one curves over S . . . . . . . . 21

4 Curves of genus one in ℙ^{2} 22 4.1 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4.2 Setting up over a field . . . . . . . . . . . . . . . . . . . . . . 22 4.3 Standardizing F_{C} . . . . . . . . . . . . . . . . . . . . . . . . . 23 4.4 Standardizing F_{E} . . . . . . . . . . . . . . . . . . . . . . . . . 24 4.5 Finding x_{0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.6 Finding φ_{x_{0}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4.7 Main Theorem with ζ_{3} ∉ K . . . . . . . . . . . . . . . . . . . 32 4.8 Uniqueness of E’s Model . . . . . . . . . . . . . . . . . . . . . 36

5 Curves of Genus One in ℙ^{4} 37 5.1 Setting Up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 5.2 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . 38 5.3 Standardizing the model for C . . . . . . . . . . . . . . . . . . 39 5.4 Finding x_{0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 5.5 Standardizing the Model for E . . . . . . . . . . . . . . . . . . 43 5.6 Finding φ_{x_{0}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6 Curves of genus one in ℙ^{1} × ℙ^{1} 49 6.1 Setting up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 6.2 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . 50 6.3 Standardizing the Model for C . . . . . . . . . . . . . . . . . . 51 6.4 A Selmer-like Example . . . . . . . . . . . . . . . . . . . . . . 52 6.5 A Locally Trivial Sub-family . . . . . . . . . . . . . . . . . . . 54 6.6 Finding x_{0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 6.7 Standardizing the Model for E . . . . . . . . . . . . . . . . . . 61 6.8 Finding φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

1 Introduction

The purpose of my thesis is to explore the passage from smooth curves of genus one defined over a field K to their Jacobian curves. Specifically I consider curves of genus one in certain multiprojective spaces and provide explicit equations for their Jacobians, provided that their Jacobians have certain level structure. I do this in the three cases of ℙ^{2}, ℙ^{1} × ℙ^{1}, and ℙ^{4}. This problem has been and is currently being worked on by a number of people. This includes Salmon, who in 1873 published a book which contains formulas that (unbenownst to him) give a more general theorem than the one below, albeit quite complicated and lengthy; also, Shepherd-Barron and Fisher are looking at the problem in ℙ^{4}; An, Hammond, Kim, Kim, Marshall, McCallum, and Perlis are curresntly writing a survey article which explains what is and what is not known about the problem. In particular, the problem has been completely solved in ℙ^{3}.

For the case of ℙ^{2} we prove

Theorem 1.1 Let C be a smooth curve of genus one over a field K. Assume K does not have characteristic 3 and that K contains a primitive 3rd root of unity. Let f : C → ℙ_{K}^{2} be a closed immersion over K and let F_{C} (X, Y, Z) be the cubic defining C. Assume that the Jacobian E of C has a non-trivial rational three-torsion point T ∈ E[3](K). After modifying f by a K−linear automorphism of ℙ_{K}^{2} the cubic F_{C} is of the following form:

F_{C}(X, Y, Z) = α · (b^{2}X^{3} + bY^{3} + Z^{3}) + β · (bXY^{2} + bX^{2}Z + Y Z^{2})
+ γ · (bX^{2}Y + Z^{2}X + Y^{2}Z) + δ · (3XY Z),

for some elements α, β, γ, δ, b ∈ K. Then E = Jac(C) is given by the cubic

F_{E}(X, Y, Z) = X^{3} − Y^{3} + [(αb + δ)^{3} + β^{3}b^{2} + γ^{3}b − 3(αb + δ)βγb] · Z^{3}

+(2αb − 3δ) · XY Z,

with the origin of E given by O_{E} = ( 1 : 1 : 0 ).

The general idea is as follows. We consider C as a principal homogeneous space for its Jacobian elliptic curve. Then the assumption that there is a non-trivial rational three-torsion point on E gives a fixed-point free action of order three on the curve C, namely translation by that point. We show that this action can be extended to a linear automorphism of the entire projective plane. Moreover, we determine a standard form for the matrix representing this action which fixes a standard form for the cubic giving C. Putting this action into standard form amounts to composing the original embedding of C with a K-automorphism of ℙ_{K}^{2}. We then abstractly embed E in ℙ_{K}^{2} in such a way that, upon a choice of a special element x_{0} of C(L) for a suitable finite separable extension L of K, there exists a map φ (an isomorphism) from C to E over L which extends to a linear automorphism of ℙ_{L}^{2}. The x_{0} we have chosen is the L-rational point which maps to the identity of the group law on E. In much the same way as we did with C, we standardize the embedding of E, thereby rigidifying the map φ. We may think of the above theorem as giving us two families; the first is a family of genus one curves, parameterized by the coefficients α, β, γ, and δ, which contains every curve of genus one whose Jacobian has a non-trivial rational three-torsion point and which has an embedding into ℙ_{K}^{2} as a degree three curve. The second is the family of their Jacobians, and we find explicit formulas for a map between the two families over the field L = K(x_{0}).

In analogous ways we find families of genus one curves over K containing every curve of genus one whose Jacobian has a non-trivial rational two (resp. five) -torsion point and which has an embedding into ℙ_{K}^{1} × ℙ_{K}^{1} (resp. ℙ_{K}^{4}) as a homogeneous (2, 2)-form (resp. degree 5 curve), and we explicitly write down a map to the corresponding families of Jacobian curves after a suitable base extension.

2 Conventions

We will work over a fixed field K unless otherwise stated, with K_{s} a fixed separable closure of K and with G_{K} = Gal(K_{s}/K). A curve defined over K is an smooth, proper K-scheme of dimension 1, which is geometrically connected over K. All curves are projective over K.

Let L be a field extension of K. The automorphism group of ℙ_{L}^{n} defined over L is P GL_{n+1}(L), and a matrix class represented by M = (a_{i j})_{0≤i,j≤n} acts on the point (x_{0} : x_{1} : …: x_{n}) ∈ ℙ^{n}(L) by sending it to

(∑_{j=0}^{n} a_{0 j}x_{j} : ∑_{j=0}^{n} a_{1 j}x_{j} : …: ∑_{j=0}^{n} a_{n j}x_{j}) ∈ ℙ^{n}(L).

In other words, we represent points as column vectors and multiply by matrices on the left. The column vector whose topmost non-zero entry is 1 associated to a point x ∈ ℙ^{n}(L) will be denoted vec(x) : for example,

vec (2 : −2) = ( \begin{matrix}1 \\ −1 \end{matrix} ) .

Similarly, the point in projective space x corresponding to a vector v is denoted x = proj(v).

All fiber products are assumed to take place over Spec(K) when not otherwise indicated. If L is a field extension of K and X a K− scheme, we will write X ×_{K} L or X × L instead of X ×_{Spec(K)} Spec(L).

For integers n_{i} and closed points P_{i} of C, a divisor D = ∑ n_{i}P_{i} has degree ∑ n_{i} [K(P_{i}) : K]. Div_{K}^{n}(C) is the set of all divisors on C of degree n.

We will use the description of the Jacobian of a curve C as the abelian variety which coarsely represents the functor P_{C}^{0} which associates to a scheme

T →^{q} Spec(K) the set

P_{C}^{0}(T) = {L ∈ P ic (C × T) | deg(L_{t}) = 0 ∀ t ∈ T }/q^{∗}P ic (T).

Theorem 1.1 on page 168 of [6] states:

Theorem There is an abelian variety J over K and a morphism of functors ι : P_{C}^{0} ⟶ J such that ι : P_{C}^{0}(T) ⟶ J(T) is an isomorphism whenever C(T) is nonempty.

Indeed the Jacobian represents the functor T ↦ P_{C}^{0}(T_{L})^{G_{K}}, where L is any finite Galois extension of K such that C(L) is nonempty. In the case of T = Spec(K), we have an injective map ι : P_{C}^{0}(K) ⟶ J(K). We will sometimes write J = \underline{P ic}_{C/K}^{0}.

For n prime to the characteristic of K, ζ_{n} is understood to be a primitive nth root of unity in K_{s}. O is the origin of an elliptic curve, and O(D) is the sheaf associated to the Cartier divisor D (see page 144 of [4]).

3 General Results

3.1 n-prepared genus one curves

For an integer n > 2, we define an “n-prepared genus one curve (over S)” to be a triple

(C →^{π} S, L, T ),

where C →^{π} S is a projective flat morphism whose fibers are smooth genus one curves, L is a degree n line bundle on C (in particular L_{t} is a degree n line bundle for every geometric fiber C_{t} ), and T is an S-section of exact order n of the Jacobian group scheme J → S. Such a J exists by Theorem 8.1 of [6].

We define a “2-prepared genus one curve (over S)” to be a triple

(C →^{π} S, (L_{1}, L_{2}), T ),

where C →^{π} S is a projective flat morphism whose fibers are smooth genus one curves, L_{1} and L_{2} are line bundles of degree two on C which are linearly inequivalent on all fibers, and T ∈ J (S) has exact order 2.

Remark A. For n at least 3, an n-prepared genus one curve comes with a closed immersion of the curve C into ℙ(π_{∗}(L)) over S, using the notation of page 162 of [4]. In the case where S is the spectrum of a local ring or a field, ℙ(π_{∗}(L)) is isomorphic to ℙ_{S}^{n−1}, and the immersion is defined by a choice of basis of the module of global sections of the line bundle L. Hence this embedding is only defined up the action of Aut(ℙ_{S}^{n−1}) = PGL_{n}(S). Similarly, a 2-prepared genus one curve over the spectrum of a local ring or a field S comes with an embedding of the curve C into ℙ_{S}^{1} × ℙ_{S}^{1} defined up the action of Aut(ℙ_{S}^{1})×Aut(ℙ_{S}^{1}). Note that this map only uses the data of C →^{π} S and the line bundle L (resp. (L_{1}, L_{2})).

For n ≥ 3 (resp. n = 2) we define a morphism of two n-prepared curves over S

(C →^{π} S, L, T ) ⟶^{(f,α)} (C′ →^{π′} S, L′, T′)

(resp. (C →^{π} S, (L_{1}, L_{2}), T ) ⟶^{(f,(α_{1},α_{2}))} (C′ →^{π′} S, (L′_{1}, L′_{2}), T′))

to be a pair (f, α) (resp. (f, (α_{1}, α_{2}))) with:

1. f : C ⟶ C′ a morphism of schemes,

2. π = π′ · f,

3. α : L ≅ f^{∗}L′ (resp. α_{i} : L_{i} ≅ f^{∗}L′_{i}), and

4. the induced map of Jacobians f^{∗} : J′ → J sends T′ to T.

Remark B. A morphism between two n-prepared genus one curves with n ≥ 3 (resp. n = 2) induces a map between the spaces ℙ(π_{∗}(L)) and ℙ(π_{∗}(L′)) (resp. ℙ(π_{∗}(L_{i})) and ℙ(π_{∗}(L′_{i}))) which we also call f; in the case where S is the spectrum of a local ring and we have fixed choices of bases of global sections of L and L′, there is a unique map over S which extends f, which we also denote by f :

\begin{matrix}C & ⟶^{π} & ℙ_{S}^{n−1} \\ f ↓ & & ↓ f \\ C′ & ⟶^{π′} & ℙ_{S}^{n−1} \end{matrix} (resp. \begin{matrix}C & ⟶^{π} & ℙ_{S}^{1} × ℙ_{S}^{1} \\ f ↓ & & ↓ f \\ C′ & ⟶^{π′} & ℙ_{S}^{1} × ℙ_{S}^{1} \end{matrix} ) .

For a scheme S′ → S, an n-prepared genus one curve Λ over S naturally corresponds to one over S′, namely by base change. We will denote this Λ_{S′} = Λ ×_{S} S′. A “S′-morphism” of n-prepared genus one curves over S is defined as above by base changing the curves and replacing S by S′.

For n ≥ 3 (resp. n = 2), fix Λ = (C →^{π} S, L, T ) (resp.(C →^{π} S, (L_{1}, L_{2}), T )) an n-prepared curve of genus one. Let J be an elliptic curve over S. Fix a principal homogeneous space action of J on Λ, i.e. a morphism

λ : C ×_{S} J → C

which induces an isomorphism of J with the Jacobian of C. Any S′-valued point P of J gives us an S′-morphism

λ_{P} = λ(−, P) : C_{S′} ×_{S} S′ → C_{S′}.

We say that τ ∈Aut(Λ)(S′) “covers λ_{P} ” (or τ is a “λ_{P} -morphism”) if τ = (λ_{P} , α). Let Aut(Λ)(S′) be the set of S′-morphisms of Λ to itself covering λ_{P} for some P ∈ J (S′). Aut(Λ) is a contravariant functor. There is a natural map of functors

F : Aut(Λ) ⟶ J .

Claim 3.1 For any n-prepared genus one curve Λ, the image of

F : Aut(Λ) ⟶ J

is the functor associated to the finite group scheme J [n].

Proof. Given (λ_{P} , α) ∈ Aut(Λ)(S′), we have α : L_{S′} ≅ λ_{P}^{∗} L_{S′} ≅ P^{⊗ n} ⊗L_{S′}, so P ∈ J [n](S′). For P ∈ J [n](S′), there will clearly exist an isomorphism α : L_{S′} ≅ λ_{P}^{∗} L_{S′}. A similar proof works for n = 2.

For any S′ → S we have the exact sequence

1 ⟶ H^{0}(S′, O_{S}^{′∗}) ⟶ Aut(Λ)(S′) ⟶ J [n](S′) ⟶ 1.

Remark C. An n-prepared genus one curve Λ = (C →^{π} S, L, T ) (resp. (C →^{π} S, (L_{1}, L_{2}), T )) comes with a canonical subset F^{∗}(T )(S) of Aut(Λ)(S), namely pullback of the element corresponding to T (in the isomorphism of J with the Jacobian of C) by the functor F as in Claim 3.1. Moreover, by Remark B, if S is the spectrum of a local ring and we have fixed an embedding f : C ⟶ ℙ_{S}^{n−1}, there is a unique element of F^{∗}(T )(S) ⊂ Aut(Λ) which extends the automorphism of Λ to ℙ_{S}^{n−1}. We call this λ_{T} (f) = λ_{T} .

3.2 Setting up over a field

Let C be a smooth curve of genus one over a field K. All smooth curves of genus one defined over a finite field have a rational point, so they are isomorphic to their Jacobian. Thus we may assume K is infinite. Let n ∈ ℕ be an integer with n ≥ 3 (respectively n = 2), and suppose L ∈Pic_{K}^{n}(C).

Remark A tells us that the global sections of L gives us a closed immersion (resp. a map)

f : C ⟶ ℙ_{K}^{n−1},

defined over K, embedding C as a curve of degree n in ℙ_{K}^{n−1} (resp. exhibiting C as a double cover of ℙ_{K}^{1}) with f^{∗}(O(1)) = L. In fact there will be many such maps; fix one and call it f.

Let E be an elliptic curve over K and λ a principal homogeneous space action of E on C

λ : C ×_{K} E ⟶ C

which gives an isomorphism of E with the Jacobian E = \underline{P ic}_{C/K}^{0}.

3.3 Translation by T

From now on it will be a standing assumption that n is relatively prime to the characteristic of K. Let T ∈ E[n](K′), for some field extension K′ ⊂ K_{s} of K. Then (C_{K′}/K′, L_{K′}, T) is an n-prepared genus one curve. Restricting the second coordinate of λ from above to T we get the “translation by T map”

λ_{T} = λ(−, T) : C ×_{K} K′ ⟶ C ×_{K} K′,

defined over K′.

Claim 3.2 There exists a unique automorphism λ_{T} : ℙ_{K′}^{n−1} ⟶ ℙ_{K′}^{n−1} defined over K′ such that the following diagram commutes:

C ×_{K} K′ ⟶^{f} ℙ_{K′}^{n−1}
λ_{T} ↓  ↓ λ_{T}
C ×_{K} K′ ⟶^{f} ℙ_{K′}^{n−1} .

Proof. This follows from Remark C on page 6.□

From the above we get the following map:

χ : E[n](K_{s}) ⟶ P GL_{n}(K_{s}).
T ⟼ λ_{T} .

Claim 3.3 χ is an injective Galois-invariant homomorphism.

Proof. χ is injective because E[n] acts faithfully on C_{K} × K_{s}. χ is a homomorphism by uniqueness in Claim 3.2. Finally, χ is Galois-invariant because the maps λ and f are defined over K, so λ_{T}^{σ} = λ_{T^{σ}}. □

Let M_{1}, M_{2} ∈ GL_{n}(K_{s}) be any matrices that lift the images under χ of any two points T_{1}, T_{2} ∈ E[n](K_{s}). Then the commutator [M_{1}, M_{2}] does not depend on the lifts and is a scalar matrix of determinant 1, so is the identity matrix multiplied by some nth root of unity. Thus we have formed a pairing

e : E[n](K_{s}) × E[n](K_{s}) ⟶ μ_{n}(K_{s})

which is visibly bilinear and alternating.

By Galois-invariance, e is induced by a bilinear alternating pairing of group schemes over K :

χ : E[n] × E[n] ⟶ μ_{n}(K_{s}),

and this is clearly compatible with extensions of the base field.

Claim 3.4 The above e is the Weil pairing, defined on page 183 of [7].

Proof. We refine the section on the theta-group associated to a line bundle (pages 221-229 of [7]). A theta-group over K is a system of group schemes and homomorphisms

1 ⟶ 𝔾_{m} ⟶^{i} G ⟶^{π} K ⟶ 1

where

• K is commutative,

• ∃ an open covering {U_{i}} of K and section σ_{i} of π,

• i is a closed immersion, making 𝔾_{m} into the kernel of π, and

• 𝔾_{m} ⊂ center of G.

Let Λ be an n-prepared genus one curve over the field K, and let E an elliptic curve which is isomorphic to the Jacobian of C. From page 6 we have the following exact sequence for any K-scheme S :

ϵ(S) : 1 ⟶ H^{0}(S, O_{S}^{∗}) ⟶ Aut(Λ)(S) ⟶ E[n](S) ⟶ 1.

Theorem 1 on page 225 of [7] states that the functor S ↦ ϵ(S) is representable by the theta group sequence associated to the line bundle L :

1 ⟶ 𝔾_{m} ⟶ G(L) ⟶ K(L) ≅ E[n] ⟶ 1.

On page 222 of [7], Mumford defines a skew-symmetric bihomomorphism e : K ×_{K} K → 𝔾_{m} for any theta group, which he denotes by e^{L} in the case of the theta-group G(L). Using K(L) ≅ E[n], this recovers the pairing e in our theorem. Finally, on page 228 [7], Mumford proves crucial properties of e^{L}; here the symbols x, y, etc. are to be understood as R-valued points for any K−algebra R, and the Weil pairing is denoted \overline{e_{n}}.

• For x ∈ K(L) and y ∈ [n]^{−1}(K(L)), e^{L^{n}}(x, y) = e^{L}(x, ny)

• Define ϕ_{L} so that [n]^{−1}(K(L)) = ϕ_{L}^{−1}(X[n]). Then for x ∈ E[n] and y ∈ [n]^{−1}(K(L)) = ϕ_{L}^{−1}(X[n]), \overline{e_{n}}(x, ϕ_{L}(y)) = e^{L^{n}}(x, y).

In our situation K(L) = E[n] and ϕ_{L} = [n] since L has degree n, so combining the two properties proves the theorem.□

3.4 Embedding E

Let Λ = (C, L, T) (resp. Λ = (C, (L_{1}, L_{2}), T) be an n-prepared (resp. 2prepared) genus one curve over K, let E be the Jacobian of C, and let T′ = \frac{n (n−1)}{2} · T ∈ E[2](K). Using the K-action of E(K) = \underline{P ic}_{C/K}^{0}(K) on \underline{P ic}_{C/K}^{n}(K), L − T′ (resp. L_{1} − T′) ∈ \underline{P ic}_{C/K}^{n}(K). Let x_{0} ∈ C(K_{s}) be such that

L − T′ ≅ O(n · x_{0}) (resp.L_{1} − T′ ≅ O(n · x_{0})).

C × K_{s} is an elliptic curve and to find x_{0} we are asking for the inverse image of a specific point under the multiplication-by-n map.

We would like to have a map from the curve C to its Jacobian E which is a simple as possible. For example, we could ask for a linear automorphism of ℙ^{n−1} defined over a field extension K′ of K which takes C to E. By Remark B on page 5, this will happen whenever we have an n-prepared genus one curve ΛJac = (E, L_{E}, T) and a morphism from Λ to ΛJac defined over K′. For an x_{0} ∈ C(K_{s}) as above there is a K(x_{0})-morphism

φ_{x_{0}} : C ⟶ E

which takes a point x ∈ C(K_{s}) to the line bundle O(x − x_{0}) ∈ E(K_{s}). Then φ_{x_{0}}^{∗} takes a line bundle L of degree d on E to L∑ + d · x_{0}, a line bundle of degree d on C (we have identified E with its Jacobian by sending a line bundle L on E to the sum of its points L∑). In order for φ_{x_{0}} to extend to a morphism from Λ to ΛJac we need an isomorphism

α_{x_{0}} : L ≅ φ_{x_{0}}^{∗}(L_{E}) (resp.α_{i} : L_{i} ≅ f^{∗}L′_{i}).

Define

L_{E} = O((n − 1)O_{E} + T′)
(resp.(L_{E,1}, L_{E,2}) = (O((n − 1)O_{E} + T′), L_{2} ⊗ L_{1}^{−1} ⊗ O(T))).

Using Remark A, fix an embedding f_{E} of E in ℙ_{K}^{n−1} (resp. ℙ_{K}^{1} × ℙ_{K}^{1}) such that f_{E}^{∗}(O(1)) = L_{E} (resp. (pr_{i} · f_{E})^{∗}(O(1)) = L_{E,i}). Then

φ_{x_{0}}^{∗}L_{E} ≅ O(L_{E}^{∑} + n · x_{0}) ≅ O(n · x_{0} + T′),

so an α_{x_{0}} exists; by Remark B there is a unique extension of φ_{x_{0}} to ℙ_{K(x_{0})}^{n−1}. We have proved

Claim 3.5 With notation as above, let

Λ = (C, L, T) and ΛJac = (E, O((n − 1)O_{E} + T′), T)

be n-prepared genus one curves over K. Then there exists a unique element φ_{x_{0}} ∈ PGL_{n}(K(x_{0})) so that the following diagram commutes:

\begin{aligned}x & \underline{P ic}_{C/K}^{1} & = & C & ⟶^{f} & ℙ_{K(x_{0})}^{n−1} \\ \overline{↓} & ↓ & & φ_{x_{0}} ↓ & & ↓φ_{x_{0}} \\ (x − x_{0}) & \underline{P ic}_{C/K}^{0} & = & E & ⟶^{f_{E}} & ℙ_{K(x_{0})}^{n−1} \end{aligned} .

Remark. The same Claim can be made for 2-prepared curves of genus one where Λ = (C, (L_{1}, L_{2}), T) and ΛJac = (E, (L_{E,1}, L_{E,2}), T) are defined as above.

The above Claim motivates the following definition:

Definition 3.1 For n ≥ 3, given an n-prepared curve of genus one Λ = (C/S, L, T ), let

ΛJac = (E/S, O((n − 1)O_{E} + T′), T )

be the “Jacobian n-prepared genus one curve of Λ,” where E is the Jacobian of C and where T′ = \frac{n(n−1)}{2} · T . Given a 2-prepared genus one curve Λ_{2} = (C, (L_{1}, L_{2}), T ), let

Λ_{2,Jac} = (E/S, (O((n − 1)O_{E} + T′), L_{2} ⊗ L_{1}^{−1} ⊗ O(T )), T )

be the “Jacobian 2-prepared genus one curve of Λ_{2}.”

Remark. We are identifying T ∈ E[n](K) with its image under the canonical isomophism E ≅ Jac(E). Moreover, if we have an elliptic curve E ≅ E, we can define Λ_{E} = (E/K, O((n − 1)O_{E} + T′), T) using the isomorphsim E ≅_{λ} E ≅ Jac(E).

If n is odd, then T′ = O_{E}, and the condition in Claim 3.5 on x_{0} simplifies considerably:

L = O(n · x_{0}) ∈ P ic_{K}^{n}(C).

In other words, x_{0} is a flex point in ℙ^{2}, and in general if n = 2m + 1 in ℙ^{2m} is a hyperosculation point, i.e. a geometric point on C of maximal tangency. This implies that f_{E} is defined over K even when T is not. Of course, φ_{x_{0}} will still only be defined over K(x_{0}).

Recall that E is an elliptic curve over K and λ is a principal homogeneous space action of E on C

λ : C ×_{K} E ⟶ C

which gives an isomorphism of E with the Jacobian E. We can represent the cohomology class of the pair (Λ, λ) in the cohomology group H^{1}(G_{K}, E(K_{s}))[n] by the cocycle x_{0} − x_{0}^{σ} (see Theorem 2.2 on page 285 of [8]). The next Claim shows that it actually lifts to H^{1}(G_{K}, E[n](K_{s})).

Claim 3.6 With this choice of x_{0}, the 1-cocycle

ξ : G_{K} ⟶ E[\overline{K}]
ξ : σ ⟼ σϕ ∘ ϕ^{−1} = x_{0} − x_{0}^{σ}

lands in E [n](K_{s}).

Proof. n · (x_{0} − x_{0}^{σ}) ∼ n · x_{0} − n · x_{0}^{σ} ∼ (L − T′) − (L − T′)^{σ} = 0. Here we are using the rationality of L and T′. □

3.5 Finding x_{0}

We have two good reasons for choosing x_{0} as we did. First, It gives us a nice choice for an embedding of E, and second the resulting cocycle x_{0} − x_{0}^{σ} representing C in H^{1}(G, E(\overline{K})) comes from E[n](\overline{K}). A third reason is that we can actually find x_{0}. To do this we study the geometry arising from the ambient ℙ^{n−1} automorphisms and standardize the element λ_{T} ∈ PGL_{n}(K).

Lemma 3.2 λ_{T} is diagonalizable over K_{s}.

Proof. Lift λ_{T} to some matrix N ∈ GL_{n}(K_{s}) so that N^{n} −κ·I = 0 for some κ ∈ K_{s}^{∗}. Since x^{n}−κ ∈ K_{s}[x] is separable, N is semisimple and diagonalizable over K_{s}. □

Now let M be a lift of λ_{T} such that M ∈ GL_{n}(K). The eigenvectors of M correspond to fixed points of λ_{T} in ℙ^{n−1}(K_{s}). Fix n points {p_{0}, p_{1}, … p_{n−1}} in ℙ^{n−1}(K_{s}) fixed by M. We may assume the p_{i} are in general position, as λ_{T} is diagonalizable over K_{s}. Define ℍ_{i} to be the unique hyperplane in ℙ^{n−1}(K_{s}) containing all the p_{j}’s except p_{i}.

Claim 3.7 Let ℍ_{i}, (0 ≤ i ≤ n − 1) be as above. Then

• ℍ_{i} intersects C in n distince \overline{K}-rational points, and as we vary i we get n^{2} distinct \overline{K}-rational points.

• Call one such point x. Then x has the property that x_{0} has in claim

3.5, namely

L\overline{K} = O( n · x + T′ ) ∈ P ic_{C/K}^{n}(\overline{K}).

Proof. First we will show the second part. Without loss of generality, let i = 0. ℍ_{0} is parameterized by λ_{1}p_{1} + λ_{2}p_{2} + … + λ_{n−1}p_{n−1}, and as each p_{j} is fixed by M, so is ℍ_{0}. For each x in ℍ_{0} ∩ C, M · x = x + T is also on ℍ_{0}. Fixing some x ∈ ℍ_{0} ∩ C\overline{K}, x + i · T ∈ ℍ_{0} ∩ C\overline{K}, and so the degree n divisor ℍ_{0} ∩ C\overline{K} on C\overline{K} must be the sum of the n distinct points x + i · T so O(n · x + \frac{n(n−1)}{2} · T) = O(n · x + T′) = L.

To see why the n^{2} points described above are distinct we will need the following lemma:

Lemma 3.3 Any n distinct points of C\overline{K} do not lie in a codimension 2 hyperplane in ℙn−1 ; \overline{K}.

Proof of Lemma. Say n points of C(\overline{K}) lie on a linear subspace V of codimension 2 in ℙn−1 ; \overline{K}. Let p ∈ C\overline{K} \ V be a closed point. There exists a hyperplane ℍ containing p and V which intersects C at a divisor of degree ≥ n + 1, contradicting the fact that C is a degree n curve in ℙ_{K}^{n−1}.□

If x′ is on two hyperplanes ℍ_{i} and ℍ_{j}, then x′ + T is too, by the above argument, and so indeed the entire orbit of x′, consisting of n distinct points, lies on ℍ_{i} ∩ ℍ_{j}, contradicting Lemma 3.3.□

Corollary 3.4 The eigenvalues of M are distinct.

Proof. Assume not. Say p_{0} ≠ p_{1} are fixed by λ_{T} and correspond to eigenspaces of M with the same eigenvalue. For μ ∈ \overline{K}, let ℍ_{μ} be the hyperplane containing the n − 1 points μp_{0} + (1 − μ)p_{1}, p_{2}, p_{3}, …, p_{n−1}. As above, ℍ_{μ} is fixed by λ_{T} and the intersection of ℍ_{μ} with C gives n points x ∈ C(\overline{K}) with the property L\overline{K} = O( n · x + T′). Over the algebraic closure of K there are clearly only n^{2} such points (the above can be rewritten as L\overline{K} − O(T′) = O( n · x), and the map [n] is finite étale with degree n^{2}). As we vary μ through infinitely many values of \overline{K}, we produce a set of n such points which lie in more than one of the ℍ_{μ}’s, which contradicts Lemma 3.3. □

3.6 Standardizing λ_{T}

We say that an element of PGL_{n}(K) has distinct eigenvalues if a lift to GL_{n}(K) does. This is well-defined.

Claim 3.8 Assume K is infinite. If N ∈ P GL_{n}(K) has order n and if N has distinct eigenvalues, then there exists G ∈ P GL_{n}(K) such that

G^{−1}NG = [( \begin{matrix}0 & 1 & 0 & 0 & … & 0 \\ 0 & 0 & 1 & 0 & … & 0 \\ 0 & 0 & 0 & 1 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ \\ b & 0 & 0 & 0 & … & 0 \end{matrix} )] ,

where b is some lift to K^{∗} of det (N) · (−1)^{n+1}.

Proof. Lift N to Ñ ∈ GL_{n}(K). Define b = det (Ñ) · (−1)^{n+1}. Say we have a vector v in 𝔸_{K}^{n} such that the set of vectors

{ω_{1} = v, ω_{2} = Nv, ω_{3} = N^{2}v, …, ω_{n} = N^{n−1}v}

is a basis for 𝔸_{K}^{n}. Then Ñ sends ω_{i} to ω_{i+1}, for 0 ≤ i ≤ n − 1, and sends ω_{n} = N^{n−1}v to N · N^{n−1}v = b · v. Such a change of basis corresponds to conjugating Ñ as in the statement of the Claim. We define G to be the image of the change of basis matrix in PGL_{n}(K). It remains to prove such a v exists. A vector whose orbit under Ñ does not span all of 𝔸_{K}^{n} correspond to a point of ℙ^{n−1} which lies on one of the n hyperplanes ℍ_{i} (see Claim 3.7), fixed by N. Take v to be any vector corresponding to a point off of these hyperplanes. Such a v exists since K is infinite. □

Remarks. Using the above Claim we will now modify the embedding f of C so that

λ_{T} = [( \begin{matrix}0 & 1 & 0 & 0 & … & 0 \\ 0 & 0 & 1 & 0 & … & 0 \\ 0 & 0 & 0 & 1 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ \\ b & 0 & 0 & 0 & … & 0 \end{matrix} )] .

In particular we see the fixed points p_{i} of λ_{T} are the n points

p_{i} = (1 : ^{n} \sqrt{b}ζ_{n}^{i} : …: ^{n} \sqrt{b^{n−1}}ζ_{n}^{(n−1)i}) ∈ ℙ^{n−1}(\overline{K}),

where ζ_{n} is a primitive nth root of unity. The eigenvalues of the corresponding vectors are the n distinct roots of b = (−1)^{n−1} · det(λ_{T}).

3.7 The invariant b

Starting with an n-prepared genus one curve Λ/K = (C/K, L, T) we have the natural action λ_{T} ∈ PGL_{n}(K) on C as it is embedded in (multi-)projective space using the line bundle L. The invariant

b = (−1)^{n−1} · det(λ_{T})

arises from this action. We may write b = b(Λ). The next section entitled “The map det” will interpret b via Galois cohomology but will not completely explain it, for the following reason.

Given the equation(s) of a genus one curve in (multi-)projective space, we would like a formula for its Jacobian. This is like being given the information of C/K and L, the first two parts of Λ. If we were also told that its Jacobian has a rational n−torsion point, we would not know how to construct λ_{T} , so we would not know what b is. This is so even though, given the equation of C, there are only n^{2} possible order n fixed-point free actions on it. In other words, b is an invariant of the coefficients of the equations defining C but we have no formula for it. On the other hand b is vitally important and will appear in all of our formulas.

3.8 The map det

The following diagram illustrates the fact that elements of P GL_{n}(K) have determinants which are well defined only up to to nth powers in K.

0 ⟶ K^{∗} ⟶ GL_{n}(K) ⟶ P GL_{n}(K) ⟶ 0
↓ ↓ det ↓ det
0 ⟶ (K^{∗})^{n} ⟶ K^{∗}  ⟶ K^{∗}/(K^{∗})^{n} ⟶ 0

A different lifting c·Ñ of Ñ in the proof of claim 3.8 will have det (c·Ñ) = c^{n} · det (N).

We now give a Galois cohomological interpretation of the det map. Let Λ = (C/K, L, T) be an n-prepared genus one curve over the field K. From Λ we have the element λ_{T} ∈ PGL_{n}(K) (see page 7). Then define

det : {Λ} ⟶ K^{∗}/K^{∗n}
det : Λ ⟼ det(λ_{T}).

We make use of the following short exact sequence (page 197 of [8])

0 ⟶ E(K)/n E(K) ⟶^{δ} H^{1}(G, E[n](\overline{K})) ⟶ H^{1}(G, E(\overline{K}))[n] ⟶ 0.

Recall that λ (page 7) is a principal homogeneous space action of the elliptic curve E on C which induces an isomorphism of E with E, the Jacobian of C. Then the pair (Λ, λ) is represented by an element in the cohomolgy group H^{1}(G, E(\overline{K})). Moreover, (Λ, λ) has order dividing n in H^{1}(G, E(\overline{K})) since the order of C ≅ \underline{P ic}_{C/K}^{1} is the least d such that \underline{P ic}_{C/K}^{d}(K) ≠ ∅, and we have L ∈ \underline{P ic}_{C/K}^{n}(K). Thus the pair (Λ, λ) is represented by an element in H^{1}(G, E(\overline{K}))[n], the right-most group in the above exact sequence.

We can lift (Λ, λ) to the middle group, namely H^{1}(G, E[n](\overline{K})), by representing (Λ, λ) by the class of the cocycle x_{0} − x_{0}^{σ}, which takes values in E[n](\overline{K}) (see Claim 3.6). Recall that such an x_{0} ∈ C(\overline{K}) satisfies the relation O(n·x_{0}+T′) ≅ L. A different choice x′_{0} will differ (with respect to the action of E) from x_{0} by an n-torsion point, since n·x_{0}+T′ ∼ n·x′_{0}+T′ ⇒ n(x_{0}−x′_{0}) ∼ 0. Therefore the cocycles x_{0} − x_{0}^{σ} and x′_{0} − x_{0}^{′σ} differ by a coboundary. Thus we have a well-defined map

\{(Λ, λ)\} ⟶ H^{1}(G, E[n](\overline{K})).

Using T ∈ E[n](K) and the isomorphism E ≅ E we get a group scheme map ℤ/nℤ ⟶ E[n]; using the identification E[n] ≅ E[n]^{∧} by sending S ↦ e(−, S) we dualizes the above map to get E[n] ⟶ μ_{n}, giving the cohomological map

e^{∗} : H^{1}(G, E[n]) ⟶ H^{1}(G, μ_{n}) ≅ K^{∗}/K^{∗n}.

Claim 3.9 The composition map e^{∗} ∘ l is (−1)^{n−1} · det.

Remark. This is not a group homomorphism.

Proof. Define b = (−1)^{n−1}·det(λ_{T}). We need to show that (e^{∗}∘l)(C, D) = b. The map H^{1}(G, μ_{n}) ≅ K^{∗}/K^{∗n} is given by

[σ ↦ \frac{σ( ^{n} \sqrt{a})}{\sqrt[n]{a}}] ⟷ a.

Let S_{σ} be the n-torsion point x_{0} − x_{0}^{σ} ∈ E(\overline{K}). By definition, (e^{∗} ∘ l)(Λ) is the class of the cocycle

σ ↦ e(x_{0} − x_{0}^{σ}, T)

Using the version of the Weil pairing defined on page 8, we have

e(x_{0} − x_{0}^{σ}, T) · I = S_{σ} T S_{σ}^{−1} T^{−1}.

In other words, for any σ, the matrix S_{σ} T S_{σ}^{−1} T^{−1} is a multiple of the identity matrix; we need to prove that multiple is \frac{σ( ^{n} \sqrt{b})}{\sqrt[n]{b}}.

We can check this by seeing where one nonzero vector in 𝔸n ; \overline{K} is mapped by S_{σ} T S_{σ}^{−1} T^{−1}. We will choose this vector carefully. Let v_{i} be the vector associated to the fixed point p_{i} of λ_{T} (see Remark 3.6). Claim 3.7 implies that x_{0} is on a hyperplane ℍ containing n − 1 of the p_{i}’s. Relabeling if necessary, we may assume x_{0} lies on the hyperplane containing p_{1}, …, p_{n−1}. Define v = v_{n}. Then v is a vector whose projectivization is a fixed point of λ_{T} lying off of ℍ. We may assume that the eigenvalue of v is ^{n} \sqrt{b} (see Remark 3.6).

Lemma 3.5 For some ε ∈ \overline{K}^{∗}, S_{σ}^{−1}(v) = ε · v^{σ}.

Proof. The fixed points p_{i} of λ_{T} are permuted both by “translation by ntorsion” matrices, since all such matrices commute (in PGL_{n}(\overline{K})) with λ_{T}, and by elements of the Galois group G, since the set {p_{i}} is Galois invariant (as T is rational).

S_{σ}^{−1} maps the point x_{0} to x_{0}^{σ}, since S_{σ} is the n-torsion point x_{0} − x_{0}^{σ}. More generally, S_{σ}^{−1} maps the point x_{0} + i · T to x_{0}^{σ} + i · T = (x_{0} + i · T)^{σ}, by commutativity of n-torsion points. Since the n distinct points x_{0} + i · T generate the hyperplane ℍ, we know that the hyperplane ℍ is mapped by S_{σ}^{−1} to ℍ^{σ}. That means that the fixed points p_{1} through p_{n−1}, which are all on ℍ, are mapped by S_{σ}^{−1} to p_{1}^{σ} through p_{n−1}^{σ}. Then S_{σ}^{−1} must map p_{n} to p_{n}^{σ}, i.e. S_{σ}^{−1} maps v to ε · v^{σ} for some constant ε. □

Now we see how the matrix S_{σ} T S_{σ}^{−1} T^{−1} acts on the vector v. Since T^{−1}v = v/ ^{n} \sqrt{b}, we reduce to showing S_{σ} T S_{σ}^{−1} v = σ( ^{n} \sqrt{b}) v. By Lemma 3.5,

S_{σ} T S_{σ}^{−1} v = S_{σ} T ε · v^{σ} = ε · S_{σ} (T v)^{σ}· = ε · σ ( ^{n} \sqrt{b}) · S_{σ} v =

σ ( ^{n} \sqrt{b}) · ε · (\frac{1}{ε} · v) = σ ( ^{n} \sqrt{b}) v.

□

Identify T ∈ E[n](K) with the image of T ∈ E[n](K) underthe canonical isomorphism E ≅ E. Recall that T′ = \frac{n(n−1)}{2} · T.

Corollary 3.6 The map det satisfies the following properties:

1. The map e^{∗} ∘ δ : E(K)/n E(K) ⟶ K^{∗}/K^{∗n} given by the composition of the usual connecting map δ (page 16) in cohomology and the map e^{∗} (page 17), can be rewritten as follows: let P ∈ E(K), let P′ be the point P + T′, let L_{P} ∈ P ic_{E/K}^{n}(K) be the line bundle O((n − 1) · O_{E} + P′), and let Λ_{P} = (E/K, L_{P} , T_{E}). Then (e^{∗} ∘ δ)(P) = (−1)^{n−1}det(Λ_{P}).

2. If we fix C/K and T and vary L, then the image of

(−1)^{n−1} · det : {(Λ = (C/K, L, T), λ)} ⟶ K^{∗}/K^{∗n}

is a coset of the image of e^{∗} ∘ δ : E(K)/n E(K) ⟶ K^{∗}/K^{∗n}.

Remark. If n is odd, T′ = O_{E} so P′ = P. Also, for n odd we have the simplified formula (e^{∗} ∘ δ)(P) = det(Λ_{P}).

Proof. The connecting map δ brings P ∈ E(K) to the the class of the cocycle S − S^{σ}, where S ∈ E(\overline{K}) and n · S = P. A lift of the pair (Λ_{P} , λ_{E}) (where λ_{E} is the obvious action) is the class of the cocycle x_{0} − x_{0}^{σ} such that O(n · x_{0} + T′) ≅ L_{P} . We would like to see that S = x_{0}. This amounts to asking that L_{P} ≅ O(P + T′).

The difference of two lifts of (Λ, λ) ∈ H^{1}(G, E)[n] to the group H^{1}(G, E[n]) is the image of an element of E(K), because the short exact cohomology sequence (page 16) is exact.

3.9 The Model for E

We use the properties of the det map to further normalize the model for E.

Claim 3.10 Let E be an elliptic curve defined over K and let T ∈ E[n](K) be an element of exact order n. Let T′ = \frac{n(n−1)}{2} · T. Define the n-prepared curve

Λ_{E} = (E/K, O(n · O_{E} + T′), T).

Then det (Λ_{E}) = (−1)^{n−1}.

Proof. We will use Claim 3.9 to compute (−1)^{n−1} · det (Λ_{E}). We first lift the trivial element (Λ_{E}, λ_{E}) (where λ_{E} is the obvious E-action on itself, namely addition) from H^{1}(G, E(\overline{K}))[n] to H^{1}(G, E[n](\overline{K})) to the class of the cocycle x_{0} −x_{0}^{σ} where O(n·x_{0} +T′) ≅ O(n·O_{E} +T′). Since we can take x_{0} to be O_{E}, we are lifting (Λ_{E}, λ_{E}) to the trivial element in H^{1}(G, E[n](\overline{K})). The image of this in K^{∗}/K^{∗ n} is certainly 1. □

Claim 3.11 Let

Λ_{E} = (E/K, L_{E} = O(n · O_{E} + T′), T)

be as in Claim 3.10. Assume ζ_{n} ∈ K. We may choose an embedding f_{E} (see Remark A, page 4) associated to Λ so that O_{E} = (1 : 1 : 1 : …: 1 : 0), i · T = (1 : ζ_{n}^{i} : ζ_{n}^{2i} : …: ζ_{n}^{(n−2)i} : 0), and so that a lift of λ_{T} to GL_{n}(K) is

D = ( \begin{matrix}1 & 0 & 0 & … & 0 \\ 0 & ζ_{n} & 0 & … & 0 \\ 0 & 0 & ζ_{n}^{2} & … & 0 \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ 0 & 0 & 0 & … & ζ_{n}^{n−1} \end{matrix} ) .

Proof. Claims 3.10 and 3.8 show that there is a choice of f_{E} so that a lift to GL_{n}(K) of λ_{T} is

M = ( \begin{matrix}0 & 1 & 0 & 0 & … & 0 \\ 0 & 0 & 1 & 0 & … & 0 \\ 0 & 0 & 0 & 1 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ \\ 1 & 0 & 0 & 0 & … & 0 \end{matrix} ) .

We would like to show that there is an element G of GL_{n}(K) which conjugates M to D. Using the fact that (char(K), n) = 1, take G to be the matrix whose ith column is vec(1 : ζ_{n}^{i} : ζ_{n}^{2i} : …: ζ_{n}^{(n−1)i}). We now assume λ_{T} is the image in PGL_{n}(K) of D. If the coordinates of ℙ^{n−1} are given by x_{0}, …, x_{n}, the hyperplanes ℍ_{j} (see page 12) fixed by D are defined by x_{j} = 0. Since L_{E} = O(n·O_{E}+T′), by Claim 3.7 we know that O_{E} and its distinct translates by T are on one such hyperplane, say ℍ_{j}. We can modify our embedding of E by composing it with M, thereby conjugating λ_{T} by M. Since M commutes with D and cyclically permutes the coordinates of ℙ^{n−1}, this has the net effect of translating j by one. We can thus assume that j = n. Therefore we have i · T ∈ ℍ_{n} ∩ E. We need only put O_{E} = (1 : 1 : 1 : …: 1 : 0), since we have already fixed D. To do so, it is enough to show the existence of a invertible matrix N which commutes with D (so that upon composition we do not lose the form of λ_{T}), fixes ℍ_{n}, and sends (1 : 1 : 1 : …: 1 : 0) to O_{E}. Any matrix of the form N_{α} = α_{0} · I + α_{1} · D + … + α_{n−1} · D^{n−1} commutes with D and fixes ℍ_{n}. Moreover,

N_{α} · vec((1 : 1 : 1 : …: 1 : 0)) = G^{τ} · vec(( α_{0} : α_{1} : … α_{n−1} : 0)).

Since G^{τ} is invertible, we can choose the α_{i} ′s so that N_{α} sends (1 : 1 : 1 : …: 1 : 0) to O_{E}. □

3.10 Jacobians of n-prepared genus one curves over S

Given an n-prepared curve over a large base scheme S we would like to find its Jacobian. In some cases it is enough to find an elliptic curve over S whose generic fiber is the Jacobian of the generic fiber of the original curve. Indeed, Remark 1.10 on page 7 of [3] states that any morphism between abelian schemes defined over an open dense subset of a noetherian normal base scheme S extends to all of S.

Say we have an n-prepared curve (C →^{π} S, L, T ) over a base scheme S = Spec(K[x_{1}, …, x_{n}]_{P}), the localisation of a polynomial algebra over a field. Say we also have an (n-prepared) elliptic curve E over Spec(K[x_{1}, …, x_{n}]) whose generic fiber is isomorphic to the Jacobian of the generic fiber of C; moreover, assume there is a map from E to C which extends to a linear automorphism of ℙ_{(\overline{K(x_{1},...,x_{n}))}}^{n−1}; then E is actually smooth over S. The isomorphism between the generic fibers E and the Jacobian of C extends to an open set of S and we can apply the above Remark 1.10 to conclude that E/S is actually the Jacobian of C.

4 Curves of genus one in ℙ^{2}

4.1 Main Theorem

Let K be a field with char(K) not divisible by 3. Assume that ζ_{3} ∈ K. For variables α, β, γ, and δ, define

F_{C}(X, Y, Z) = α · (b^{2}X^{3} + bY^{3} + Z^{3}) + β · (bXY^{2} + bX^{2}Z + Y Z^{2})
+γ · (bX^{2}Y + Z^{2}X + Y^{2}Z) + δ(3XY Z),

over the base Spec(K(α, β, γ, δ)). Define S to be the largest open subscheme of Spec(K(α, β, γ, δ)) such that F_{C} defines a smooth flat genus one curve C over S, embedded in ℙ_{S}^{2} by f_{C}. Let L = f_{C}^{∗}(O(1)), and let E/S be the Jacobian of C. Note that the matrix

M_{T} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ b & 0 & 0 \end{matrix} )

acts on F_{C} and thus on C. Moreover, this action is of exact order n and has no fixed points on C; thus it is of the form λ_{T} for some T ∈ E[n](K) of exact order n. We have the n-prepared genus one curves Λ = (C/S, L, T ) and ΛJac = (E/S, O(3 · O_{E}), T ).

Theorem 4.1 With notation as above, E is given by the cubic

F_{E}(X, Y, Z) = X^{3}+Y^{3}+[(αb+δ)^{3}+β^{3}b^{2}+γ^{3}b−3(αb+δ)βγb]/(3(2αb−δ))^{3}·Z^{3}

+XY Z,

with O = (1 : −1 : 0).

4.2 Setting up over a field

Let K be a field with char(K) not divisible by 3 and with ζ_{3} ∈ K. Let Λ = (C/K, L, T) be an n-prepared curve of genus one over K. Let E be the Jacobian of C, and let Λ_{E} = (E/K, L_{E} = O(3 · O_{E}), T) be the n-prepared Jacobian curve of Λ (see Claim 3.5). By Remark A (page 4), Λ comes with an embedding f of C into ℙ_{K}^{2}. By Claim 3.8, we may assume f is such that λ_{T} ∈ PGL_{n}(K) lifts to the matrix

M_{T} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ b & 0 & 0 \end{matrix} ) .

Note that b is the determinant of λ_{T} ∈ P GL_{3}(K), and as such is only defined up to cubes in K^{∗}. From the embedding f we get a cubic F_{C}(X, Y, Z) which defines C in ℙ_{K}^{2}.

4.3 Standardizing F_{C}

Claim 4.1 With notation as in Section 4.2, there exist α, β, γ, and δ ∈ K such that

F_{C}(X, Y, Z) = α · (b^{2}X^{3} + bY^{3} + Z^{3})+

β · (bXY^{2} + bX^{2}Z + Y Z^{2}) + γ · (bX^{2}Y + Z^{2}X + Y^{2}Z) + δ · (3XY Z).

Proof. Consider the 10 dimensional vector space V over K generated by the monomials X^{3}, X^{2}Y, X^{2}Z, XY^{2}, XY Z, XZ^{2}, Y^{3}, Y^{2}Z, Y Z^{2}, and Z^{3}. M_{T} acts on this vector space (M_{T} sends X to Y, Y to Z, and Z to bX) as a linear operator whose third power acts as multiplying by b^{3}. Therefore the eigenvalues of M_{T} on V must be in the set {b, ζ_{3}b, ζ_{3}^{2}b}. On the other hand it is clear that we have the following linearly independent eigenvectors:

eigenvector  eigenvalue
XY Z  b
b^{2} ζ_{3}^{2i}X^{3} + b ζ_{3}^{i} Y^{3} + Z^{3}  b ζ_{3}^{i}
b ζ_{3}^{i} XY^{2} + b ζ_{3}^{2i} X^{2}Z + Y Z^{2}  b ζ_{3}^{i}
b ζ_{3}^{i}X^{2}Y + ζ_{3}^{2i}XZ^{2} + Y^{2}Z  b ζ_{3}^{i}

Since the equation F_{C} is fixed by the matrix M_{T} , F_{C} must be an element of one of the eigenspaces. Say that F_{C} lives in the eigenspace corresponding to the eigenvalue b ζ_{3}. Every eigenvector listed above with eigenvalue b ζ_{3} vanishes at the fixed points of M_{T}, namely (1; ^{3} \sqrt{b} ζ_{3}^{i}; ^{3} \sqrt{b^{2}} ζ_{3}^{2i}). Thus F_{C} must also vanish at these points. But we know that “translation by T ” has no fixed points on C. Thus F_{C} lives in the eigenspace corresponding to the eigenvalue b. □

Remark. We have shown that every n-prepared curve of genus one over the field K is in the family F_{C}(X, Y, Z) = α·(b^{2}X^{3}+bY^{3}+Z^{3})+β·(bXY^{2}+bX^{2}Z+ Y Z^{2}) + γ · (bX^{2}Y + Z^{2}X + Y^{2}Z) + δ · (3XY Z). Let K′ = K(α, β, γ, δ, b). By Section 3.10 we can prove Theorem 4.1 by proving that Λ ×_{K} K′ has Jacobian ΛJac ×_{K} K′. In other words, we will work over the field K′ and treat α, β, γ, δ, and b as variables.

4.4 Standardizing F_{E}

Let E be an elliptic curve with a principal homogeneous action on C

λ : C × E ⟶ C

which induces an isomorphism E ≅ E. Embed E in ℙ_{K′}^{2} by the line bundle O(3 · O_{E}). Call this embedding f_{E}. We now have the n-prepared genus one curve Λ_{E} = (E/K′, L_{E}, T) (see the Remark after Definition 3.1). By Claim 3.11, we may assume that the “translation by T on E” matrix class is represented by

D = ( \begin{matrix}1 & 0 & 0 \\ 0 & ζ_{3} & 0 \\ 0 & 0 & ζ^{2} \end{matrix} )

and that the origin O_{E} of E in ℙ^{2} is (1; −1; 0).

Claim 4.2 Let F_{E} be the cubic defining E in ℙ_{K′}^{2}. Then F_{E} is of the form

F_{E} = R_{1} · X^{3} + R_{2} · Y^{3} + R_{3} · Z^{3} + R_{4} · XY Z.

Proof. The proof is as in the argument as for Claim 4.1, where the eigenvectors of the linear operator D are the monomials X^{3}, Y^{3}, Z^{3}, and XY Z, and the eigenvalues are the cube roots of unity.

Remark. Going one step further, if we assume that we have another 3−torsion point S which is linearly independent of T and is K′−rational, we can standardize the matrix class of λ_{S} to be represented by the matrix

M_{S} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} )

(proof omitted). The argument of Claim 4.1 tells us E must also be of the form

F_{E} = S_{1} · (X^{3} + Y^{3} + Z^{3}) + S_{2} · (X^{2}Y + Y^{2}Z + Z^{2}X)+
S_{3} · (XY^{2} + Y Z^{2} + ZX^{2}) + S_{4} · XY Z.

Altogether we find that

F_{E} = A · (X^{3} + Y^{3} + Z^{3}) + B · XY Z.

This is a well-known model for the universal elliptic curve over Y (3), i.e. the moduli space of elliptic curves with full 3-torsion. In our case we only assume one K′−rational 3-torsion point T.

4.5 Finding x_{0}

Let Λ = (C/K′, L, T) and Λ_{E} = (E/K′, L_{E} ≅ O(3 · O_{E}), T) be as above. Let “translation by T on C” be represented by M_{T} as above, and recall that b ∈ K^{′∗}/(K^{′∗})^{3} is the determinant of λ_{T} ∈ P GL_{3}(K′). Let “translation by T on E” be represented by D as above, and let O_{E} = (1; −1; 0).

By Claim 3.5 we are assured of a linear automophism φ_{x_{0}} of ℙ_{K′(x_{0})}^{2} bringing C to E. To find φ_{x_{0}} we must first locate the point x_{0} ∈ C(\overline{K′}) in ℙ^{2}(\overline{K′}) which maps to O_{E} via φ_{x_{0}}. Claim 3.7 tells us two ways of searching: first, x_{0} is a flex point of the curve C. Unfortunately this seems hard to compute. Second, x_{0} is a \overline{K′}-point of the intersection of C with a hyperplane ℍ ⊂ ℙ^{2} defined by the property that it contains two of the fixed points p_{i} = (1 : ^{3} \sqrt{b} · ζ_{3}^{i} : ^{3} \sqrt{b^{2}} · ζ_{3}^{2i}) (for0 ≤ i ≤ 2) of M_{T} . An alternative definition of ℍ is that it is a hyperplane (a \overline{K′}− point of the dual of ℙ^{2}) which is fixed by M. There are 3 such ℍ’s. For 0 ≤ i ≤ 2, let

v_{i} = ( \begin{matrix}1 \\ ^{3} \sqrt{b}ζ_{3}^{i} \\ ^{3} \sqrt{b^{2}}ζ_{3}^{2i} \end{matrix} )

be vectors corresponding to the fixed points p_{i} of M, and let ℍ be the unique hyperplane containing the fixed points p_{1} and p_{2}. Then a \overline{K′}−point x of ℍ will be of the form proj(θ_{1} · v_{1} + θ_{2} · v_{2}).

Choose θ ∈ \overline{K′} so that the entries of the vector v_{1} − θ · v_{2} are all in K′(x_{0}) and so that x_{0} = proj(v_{1} − θ · v_{2}). Note that we may assume θ_{1} ≠ 0 since x_{0} ∈ C(\overline{K′}) is not fixed by λ_{T} . Then we have

x_{0} + T = λ_{T} · x_{0} = proj(M_{T} · (v_{1} − θ · v_{2})) =
proj(v_{1} ζ_{3} − θ · v_{2} ζ_{3}^{2}) = proj(v_{1} − θ · v_{2} ζ_{3})

and similarly we have x_{0} + 2T = proj(v_{1} − θ · v_{2} ζ_{3}^{2}). The fact that these points all lie on C means that

F (v_{1} − ζ_{3}^{i} · θ · v_{2} ) = 0, 0 ≤ i ≤ 2.

For any cubic form G, G(v + ζ_{3}^{i}w) = 0 ∀i ⇒ G(v) + G(w) = 0. This additive property can be checked on the elementary forms X^{3}, X^{2}Y, XY Z, …. This means that F (v_{1}) − θ^{3} · F (v_{2}) = 0, from which we conclude:

Claim 4.3

θ^{3} = \frac{F (v_{1})}{F (v_{2})}.

Claim 4.4

K′(x_{0}) = K′(θ, ^{3} \sqrt{b}) ⊃^{3} K′( ^{3} \sqrt{b}) ⊃^{3} K′.

Proof. Clearly x_{0} is defined over K′(θ, ^{3} \sqrt{b}). Moreover, both ^{3} \sqrt{b} and θ are in K′(x_{0}): x_{0} + T and x_{0} + 2T are defined over K′(x_{0}) (since T is K′-rational), so

∑_{i=0}^{2}(v_{1} − θ · v_{2} ζ_{3}^{i}) = −3 v_{1}

is defined over K′(x_{0}). This implies that ^{3} \sqrt{b} ∈ K′(x_{0}). Next, (v_{1}−θ·v_{2} )−v_{1} = −θ · v_{2} is defined over K′(x_{0}), so θ ∈ K′(x_{0}). □

4.6 Finding φ_{x_{0}}

Let the notation be as in Section 4.5. In particular we have

i · T = (1; −ζ_{3}^{i}; 0) ∈ E[3](K′).

We have determined that

x_{0} + i · T = proj(v_{1} − θ · v_{2} · ζ_{3}^{i}), where v_{i} = ( \begin{matrix}1 \\ ^{3} \sqrt{b}ζ_{3}^{i} \\ ^{3} \sqrt{b^{2}}ζ_{3}^{2i} \end{matrix} ) and θ^{3} = \frac{F (v_{1})}{F (v_{2})}.

Using Claim 3.5 we know

φ_{x_{0}} : C ⟶ E
φ_{x_{0}} : (x_{0} + i · T ) ↦ i · T.

Also by Claim 3.5, we know φ_{x_{0}} is representable by a matrix which we will denote for ease of computation by Φ_{x_{0}}^{−1} ∈ GL_{n}(K′(x_{0})). Then from the above we have

Φ_{x_{0}}^{−1}(v_{1} − θ · v_{2} · ζ_{3}^{i}) = ( \begin{matrix}1 \\ −ζ_{3}^{i} \\ 0 \end{matrix} ) ⟺ v_{1} − θ · v_{2} · ζ_{3}^{i} = Φ · ( \begin{matrix}1 \\ −ζ_{3}^{i} \\ 0 \end{matrix} ) .

We conclude that Φ is of the form

Φ = ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} ) ,

for h_{i} ∈ K′(x_{0}), 0 ≤ i ≤ 2.

Claim 4.5 Let L ⊂ K′_{s} be the Galois closure of K′(x_{0}) over K′. Let σ ∈ Gal(L/K′) be such that σ( ^{3} \sqrt{b}) = ^{3} \sqrt{b} and σ (θ) = θ · ζ_{3}^{2}. The automorphism φ^{σ}φ^{−1} of E is “translation by T.”

Proof of Claim. Such a σ exists since b and θ^{3} = \frac{F (v_{1})}{F (v_{2})} are transcendental over K. By construction the automorphism φ^{σ}φ^{−1} is translation by something, so we just need to see where the origin goes: φ^{σ}φ^{−1}O = φ^{σ} x_{0} = σ(Φ^{−1}) · x_{0} = σ (Φ^{−1}σ^{−1}x_{0}) = σ (Φ^{−1}(x_{0} + T)) = σT = T. □

The above Claim tells us that “translation by T on E” is given by the matrix class of σ(Φ^{−1})Φ; on the other hand, we have fixed “translation by T on E” to be represented by the matrix D. We have the following equality for some constant κ ∈ K′(x_{0}) :

κ · D = σ(Φ^{−1}) Φ ⟺ κ · σ(Φ) D = Φ,

so

κ · σ ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} ) · ( \begin{matrix}1 & 0 & 0 \\ 0 & ζ_{3} & 0 \\ 0 & 0 & ζ_{3}^{2} \end{matrix} ) = ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} ) ⟹

κ · ( v_{1} θ · v_{2} \begin{matrix}ζ_{3}^{2}σ(h_{1}) \\ ζ_{3}^{2}σ(h_{2}) \\ ζ_{3}^{2}σ(h_{3}) \end{matrix} ) = ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} ) ,

from which we conclude

σ(h_{i}) = ζ_{3} · h_{i}.

Claim 4.6 Let σ ∈ Gal(L/K′) be as in Claim 4.5, i.e. such that σ( ^{3} \sqrt{b}) = 3 \sqrt{b} and σ (θ) = θ · ζ_{3}^{2}. The automorphism φ_{x_{0}}^{−1}φ^{σ} of C is “translation by T.”

Proof. This is similar to the proof of Claim 4.5. □

Thus we have λ_{T} on C represented two different ways, namely by M_{T} and by Φ σ(Φ^{−1}); for some κ ∈ K′(x_{0}) we have

κ · M_{T} = Φ σ(Φ^{−1}) ⟺ κ · M_{T} σ(Φ) = Φ ⟺

κ · ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ b & 0 & 0 \end{matrix} ) · σ ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} ) = ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} ) ⟺

κ · ( ^{3} \sqrt{b} ζ_{3} · v_{1} θ · ^{3} \sqrt{b} ζ_{3}^{2} · v_{2} · ζ_{3}^{2} \begin{matrix}σ(h_{2}) \\ σ(h_{3}) \\ b · σ(h_{1}) \end{matrix} ) = ( v_{1} θ · v_{2} \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \end{matrix} )

We conclude from this that κ = \frac{1}{^{3} \sqrt{b} ζ_{3}} ,

h_{3} = ^{3} \sqrt{b^{2}} ζ_{3}^{2} · σ(h_{1}) = ^{3} \sqrt{b^{2}} · h_{1}, and
h_{2} = σ(h_{3})/( ^{3} \sqrt{b} ζ_{3}) = h_{3}/ ^{3} \sqrt{b} = ^{3} \sqrt{b} · h_{1}.

Replacing h_{2} and h_{3} by multiples of h_{1}, we see that the third column of Φ is the vector v_{0} multiplied by h_{1}. In fact we have

Claim 4.7

Φ = ( v_{1} θ · v_{2} θ^{2} · v_{0} · l ) ,

for some l ∈ K′( ^{3} \sqrt{b}).

Proof. We know that σ(h_{1}) = ζ_{3} · h_{1}. Since by definition σ(θ^{2}) = ζ_{3} · θ^{2}, the ratio h_{1}/θ^{2} ∈ K′(x_{0}) is fixed by all elements in Gal(K′_{s}/K′( ^{3} \sqrt{b})). Therefore h_{1}/θ^{2} ∈ K′( ^{3} \sqrt{b}). Write h_{1} = θ^{2} · l for some l ∈ K′( ^{3} \sqrt{b}). □

Claim 4.8 l / F (v_{2}) ∈ K′.

Proof. Fix a cubic form F_{C} defining C in ℙ_{K′}^{2} with coefficients in K′. Define the map F_{C} on vectors so that, for a point x ∈ ℙ^{2}(K′_{s}), F_{C}(vec(x)) = F_{C}(x). For vectors A, B, and C, define the trilinear form T as T (A, B, C) = F (A + B + C) − F (A + B) − F (A + C) − F (B + C) + F (A) + F (B) + F (C). We will use the equality

F (A + B + C) = T (A, B, C) + F (A + B) + F (A + C) + F (B + C)
− F (A) − F (B) − F (C)  (1)

We are going to use the fact that F_{E} has coefficients defined over K′ and that F_{E} is of a special form (see Claim 4.2). Indeed, we can find F_{E} since for a point (x : y : z) ∈ ℙ_{K′_{s}}^{2},

F_{E}(x : y : z) = 0 ⟺ F_{C}(φ^{−1}(x : y : z)) = 0 ⟺
F_{C}(Φ · vec(x : y : z)) = 0 ⟺ F_{C}(x · v_{1} + y · v_{2} θ + z · v_{0} θ^{2} l) = 0.

Using the trilinear form associated to F_{C}, we have

F_{C}(x · v_{1} + y · v_{2}θ + z · v_{0}θ^{2}l) =

T (x · v_{1}, y · v_{2}θ, z · v_{0}θ^{2}l) + F (x · v_{1}) + F (y · v_{2}θ) + F (z · v_{0}θ^{2}l) =

θ^{3} l · xyz · T (v_{1}, v_{2}, v_{0}) + x^{3} · F (v_{1}) + θ^{3} · y^{3} · F (v_{2}) + θ^{6} l^{3} · z^{3} · F (v_{0}).

The other terms drop out because of Claim 4.2. Dividing by F (v_{1}) we have proved

F_{E}(X, Y, Z) = X^{3} + B · Y^{3} + C · Z^{3} + D · XY Z,

with

• B = θ^{3} · F (v_{2})/F (v_{1}) = 1,

• C = θ^{6} l^{3} · F (v_{0})/F (v_{1}) = l^{3} · F (v_{0})F (v_{1})/F (v_{2})^{2}, and

• D = θ^{3} l · T (v_{1}, v_{2}, v_{0})/F (v_{1}) = l · T (v_{1}, v_{2}, v_{0})/F (v_{2}).

Note that

C = D^{3} · \frac{F (v_{0})F (v_{1})F (v_{2})}{T (v_{1}, v_{2}, v_{0})^{3}},

so we have

F_{E}(X, Y, Z) = X^{3} + Y^{3} + D^{3} · \frac{F (v_{0})F (v_{1})F (v_{2})}{T (v_{1}, v_{2}, v_{0})^{3}} · Z^{3} + D · XY Z;

D ∈ K′ since the coefficients of F_{E} are K′-rational, so if we change Z by a multiple of D we get

F_{E}(X, Y, Z) = X^{3} + Y^{3} + \frac{F (v_{0})F (v_{1})F (v_{2})}{T (v_{1}, v_{2}, v_{0})^{3}} · Z^{3} + XY Z.

Remark. Note that the Galois group Gal(L/K′) permutes the v_{i}’s. Thus both the denominator and numerator of \frac{F (v_{0})F (v_{1})F (v_{2})}{T (v_{1},v_{2},v_{0})^{3}} are rational. In fact this shows that T (v_{1}, v_{2}, v_{0}) ∈ K′.

We have proven

Theorem 4.2 Define C ⊂ ℙ_{K′}^{2} by the cubic

F_{C}(X, Y, Z) = α · (b^{2}X^{3} + bY^{3} + Z^{3}) + β · (bXY^{2} + bX^{2}Z + Y Z^{2})
+γ · (bX^{2}Y + Z^{2}X + Y^{2}Z) + δ(3XY Z).

If E is an elliptic curve over K′ isomorphic to the Jacobian of C, there is a model for E in ℙ_{K′}^{2} given by

F_{E}(X, Y, Z) = X^{3} + Y^{3} + \frac{F_{C}(v_{0})F_{C}(v_{1})F_{C}(v_{2})}{T (v_{1}, v_{2}, v_{0})^{3}} · Z^{3} + XY Z.

with

O_{E} = ( 1 −1 0 )

and a map

φ : C ⟶ E
φ : x ⟼ O(x − x_{0})

is given by the class of the matrix Φ^{−1}, where

Φ = ( v_{1} θ · v_{2} θ^{2} v_{0} · F_{C}(v_{2})/T (v_{1}, v_{2}, v_{0}) ) , v_{i} = ( \begin{matrix}1 \\ ^{3} \sqrt{b}ζ_{3}^{i} \\ ^{3} \sqrt{b^{2}}ζ_{3}^{2i} \end{matrix} ) ,

θ^{3} = \frac{F_{C}(v_{1})}{F_{C}(v_{2})}, and where T is the trilinear form associated to F (see page 29). We have

F_{E}(X, Y, Z) = X^{3} +Y^{3} + \frac{(αb + δ)^{3} + β^{3}b^{2} + γ^{3}b − 3(αb + δ)βγb}{(3(2αb − δ))^{3}} ·Z^{3} +XY Z.

So far, we have only fixed O and the image of T, namely the matrix D. There is still some freedom in choosing the model for E. A concrete way to see this ambiguity is by composing Φ with the matrix N_{c} = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & c′ \end{matrix} ) ,

for c ∈ K′. N_{c} conjugates the cocyle φ^{σ}φ^{−1}:

σ(Φ · N_{c})^{−1}(Φ · N_{c}) = N_{c}^{−1} · (σΦ^{−1}Φ) · N_{c}

A cocycle matrix will be fixed if it commutes with N_{c}. Since the matrix D commutes with N_{c} and O is fixed by N_{c}, a choice of c will fix the model of E and the image of χ.

4.7 Main Theorem with ζ_{3} ∉ K

Corollary 4.1 Let notation be as in Section 4.1 with the exception of having ζ_{3} ∈ K (and thus in K′). E is given by the cubic F_{E}(X, Y, Z) =

(3H +G)(X^{3}+Y^{3}+Z^{3})+(3G−3H)(X^{2}Y +XY^{2}+Y^{2}Z +Y Z^{2}+Z^{2}X +ZX^{2})

+(3H + 2G)(3XY Z),

• O = ( 1 −1 0 ) ,

• G = (αb + δ)^{3} + β^{3}b^{2} + γ^{3}b − 3(αb + δ)βγb, and

• H = (3(2αb − δ))^{3}.

Proof. Assume that ζ_{3} ∉ K. Let σ be the nontrivial Galois element of Gal(K′(ζ_{3})/K′). We work with an elliptic curve E which is K′-isomorphic to E. For a field extension L of K′ let E_{L} be the Jacobian of C ⊗_{K′} L over the field L. Then E_{K′(ζ_{3})} ≅ E_{K′} ⊗_{K′} K′(ζ_{3}); this isomorphism will be realized as a linear automorphism of ℙ_{K′(ζ_{3})}^{2}, so for some W ∈ PGL_{3}(K′(ζ_{3})), we have:

\begin{matrix}C ⊗ K′(ζ_{3}) & ⟶^{φ} & E_{K′(ζ_{3})} & ⟶^{W} & E_{K′} ⊗_{K′} K′(ζ_{3}) \\ ↓ & & ↓ & & ↓ \\ ℙ_{K′(ζ_{3})}^{2} & ⟶^{φ} & ℙ_{K′(ζ_{3})}^{2} & ⟶^{W} & ℙ_{K′(ζ_{3})}^{2} \end{matrix} .

Using Claim 3.10 we can assume the “translation by T on E” matrix is of the form

( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} ) ,

and so F_{E} has the following form (see the proof of Claim 4.1):

F_{E}(X, Y, Z) = j(X^{3} + Y^{3} + Z^{3}) + k(X^{2}Y + Y^{2}Z + Z^{2}X)+
l(XY^{2} + Y Z^{2} + ZX^{2}) + mXY Z.

Claim 4.9 Let (E/K′, O(3O_{E}), T) be a 3-prepared elliptic curve, and assume that we have embedded E (via Remark A) in ℙ_{K′}^{2} such that the “translation by T on E” matrix class is represented by

M_{E} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} ) .

We can modify the embedding f : E ⟶ ℙ_{K′}^{2} by a K′−linear automorphism of ℙ_{K′}^{2} so that M_{E} is as unaffected and O = (1 : −1 : 0).

Proof. O is one of the nine flex points on the embedded curve, since f^{∗}O(1) = O(3O_{E}). By the same argument as in Section 4.5, we can write O_{E} = proj(v_{1} − θv_{2}), where

v_{1} = ( \begin{matrix}1 \\ ζ_{3} \\ ζ_{3}^{2} \end{matrix} ) and v_{2} = ( \begin{matrix}1 \\ ζ_{3}^{2} \\ ζ_{3} \end{matrix} )

are the vectors associated to fixed points of M_{E} and

θ = ^{3} \sqrt{F_{E}(v_{1})/F_{E}(v_{2})}.

To finish the proof we need θ = ζ.

We have already fixed the shape of M_{E}; we’d like to see how much this has rigidified the value of δ_{E}. If we compose the embedding of E in ℙ^{2} with a linear automorphism V^{−1}, the automorphism M_{E} changes by its conjugation by V^{−1} : M_{E} ↦ V^{−1} · M_{E} · V. In order to leave M_{E} fixed, V should commute with M_{E} in P GL_{3}(K′). For example, we could let V = αM_{E} + βM_{E}^{2} + γI, for some constants α, β, γ ∈ K′. Then V^{−1} sends δ_{E} to

3 \sqrt{(V^{−1}F_{E})(v_{1})/(V^{−1}F_{E})(v_{2})} = ^{3} \sqrt{F_{E}(V · v_{1})/F_{E}(V · v_{2})} =
3 \sqrt{F_{E}((αζ_{3} + βζ_{3}^{2} + γ)v_{1})/F_{E}((αζ_{3}^{2} + βζ_{3} + γ)v_{2})} =

θ · \frac{αζ_{3} + βζ_{3}^{2} + γ}{αζ_{3}^{2} + βζ_{3} + γ} .

If we let r = αζ_{3} + βζ_{3}^{2} + γ and let σ ∈ Gal(K′(ζ_{3})/K′) be as above, we have

θ ⟼^{V} θ · \frac{r}{σ(r)}.

Since O_{E} is rational, σO_{E} = σ(v_{1} + θv_{2}) = v_{2} + σ(θ) · v_{1} = O_{E}, so σθ · θ = 1. In other words, θ is in the kernel of the norm map from K′(ζ_{3}) to K′. Since Gal(K′(ζ_{3})/K′) is cyclic, we know that the kernel of the norm map is the same as the set of elements of the from a/σ(a) for a ∈ K′(ζ_{3}) (see page 108 of [1]). Thus for some a ∈ K′(ζ_{3}), θ = a/σ(a). We have shown that we can change θ by multiplying it by r/σ(r) (for any r ∈ K′(ζ_{3})) while fixing M_{E}. Let r = 1/(ζ_{3} · a); then

θ = a/σ(a) ⟼^{V_{r}} θ · \frac{r}{σ(r)} = \frac{σ(ζ_{3})}{ζ_{3}} = ζ_{3}.

□

We have now shown that we may take θ = ^{3} \sqrt{F_{E}(v_{1})/F_{E}(v_{2})} = ζ_{3}; in terms of the coefficients of F_{E}, this means

3 \sqrt{(j + kζ_{3} + lζ_{3}^{2} + m)/(j + kζ_{3}^{2} + lζ_{3} + m)} = ζ_{3} ⟹ k = l.

Using Claim 3.11 we may assume that E_{K′(ζ_{3})} has O_{E_{K′(ζ_{3})}} = (1 : −1 :

0) and “translation by T on E_{K′(ζ_{3})}” is given by a diagonal matrix with entries the distinct cubes roots of unity. We are now searching for W ∈ PGL_{n}(K′(ζ_{3})) such that • W · O_{E_{K′(ζ_{3})}} = W · (1 : −1 : 0) = (1 : −1 : 0) = O_{E},

• W · T_{E_{K′(ζ_{3})}} = W · (1 : −ζ_{3} : 0) = (1 : 0 : −1) = T_{E},

• W · 2T_{E_{K′(ζ_{3})}} = W · (1 : −ζ_{3}^{2} : 0) = (0 : 1 : −1) = 2T_{E}, and

• W · M_{E_{K′(ζ_{3})}} · W^{−1} = W · ( \begin{matrix}1 & 0 & 0 \\ 0 & ζ_{3} & 0 \\ 0 & 0 & ζ_{3}^{2} \end{matrix} ) · W^{−1} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} ) = M_{E}.

Claim 4.10 We may take W = ( \begin{matrix}1 & ζ_{3} & ζ_{3}^{2} \\ 1 & ζ_{3}^{2} & ζ_{3} \\ 1 & 1 & 1 \end{matrix} ) .

Proof. First note that the above matrix satisfies the four properties above. Say we knew that W ∘ φ moves C to an elliptic curve E defined over K′. Then to see that E is actually the Jacobian of C, we need to check that the cocycle ξ : Gal(\overline{K′}/K′) ⟶ Aut(E), ξ : σ ⟼ σ(W ∘ φ) (W ∘ φ)^{−1} lands in the subgroup E ⊂ Aut(E). We will now simplify this problem. Rewrite σ(W ∘ φ) (W ∘ φ)^{−1} = σ(W · Φ^{−1}) (W · Φ^{−1})^{−1} = σW (σΦ^{−1} Φ )W^{−1}. For σ ∈ Gal(\overline{K′}/K′) such that σ(W) = W (⇔ σ(ζ_{3}) = ζ_{3}), ξ(σ) has been examined in the context of E_{K′(ζ_{3})} and was found be a translation map. That means we know ξ takes values in E for half of the relevant elements of Gal(\overline{K′}/K′).

To finish, we need only show that ξ(σ′) ∈ E for some σ′ such that σ′(ζ_{3}) = ζ_{3}^{2} : writing an element of Gal(\overline{K}′/K′) as the product ϵ · σ′ where ϵ fixes ζ_{3}, we have ξ(ϵ · σ′) = ξ(ϵ)^{σ′} · ξ(σ′), and the translation maps are clearly a Galois-invariant subgroup of Aut(E).

Let us choose σ′ to be such that σ′(ζ_{3}) = ζ_{3}^{2}, σ′( ^{3} \sqrt{b}) = ^{3} \sqrt{b} (⇒ σ′(v_{1}) = v_{2} and σ′(v_{0}) = v_{0}), and σ′(δ) = 1/δ. Then σ′ acts on the flex points of C in the following way:

• σ′(x_{0}) = σ′(v_{1} + δv_{2}) = v_{2} + 1/δv_{1} = x_{0}, and more generally

• σ′(x_{0} + i · T) = σ′(v_{1} + δ ζ_{3}^{i} v_{2}) = v_{2} + (1/δ) ζ_{3}^{2·i} v_{1} = x_{0} + i · T.

We have, for i ∈ ℤ,

ξ(σ′)(i T) = σ′(W Φ^{−1}) (Φ W^{−1}) (i T) = σ′(W Φ^{−1})(x_{0} + i T) =

σ′(W Φ^{−1}σ^{′−1}(x_{0} + i T)) = σ′(i T) = i T.

This means that ξ (σ′) lands in Aut^{0}(E) and has at least three fixed points in the three-torsion subgroup, so ξ (σ′) could only be trivial or ρ, of order three. Since ρ is an automorphism that acts only on curves with j-invariant 0, ξ(σ′) = ρ would mean that our family consisting of all curves of genus one whose Jacobian has a rational 3-torsion point all have j-invariant 0, which is clearly false. □

Finally, let’s see why W ∘ φ brings C to an elliptic curve defined over K′. This is the same thing as acting on E_{K′(ζ_{3})} by W.

F_{E}(X, Y, Z) = F_{E_{K′(ζ_{3})}}(X + ζ_{3}Y + ζ_{3}^{2}Z, X + ζ_{3}^{2}Y + ζ_{3}Z, X + Y + Z)

= (X + ζ_{3}Y + ζ_{3}^{2}Z)^{3} + (X + ζ_{3}^{2}Y + ζ_{3}Z)^{3} + \frac{G}{H} (X + Y + Z)^{3}+

(X + ζ_{3}Y + ζ_{3}^{2}Z)(X + ζ_{3}^{2}Y + ζ_{3}Z)(X + Y + Z) =

(3+ \frac{G}{H})(X^{3}+Y^{3}+Z^{3})+(−3+3\frac{G}{H})(X^{2}Y +XY^{2}+Y^{2}Z +Y Z^{2}+Z^{2}X +ZX^{2})

+(3 + 2\frac{G}{H})(3XY Z),

where G = (αb + δ)^{3} + β^{3}b^{2} + γ^{3}b − 3(αb + δ)βγb and H = (3(2αb − δ))^{3}. □

4.8 Uniqueness of E’s Model

F_{E} and all of the translation by three-torsion matrices for E are independent of the lift of b to K from K^{∗} /(K^{∗})^{3}. That is, a different choice bu^{3} instead of b, given by composing f by a simple matrix, changes many of our variables (for example, it sends α ↦ α/u^{6}, β ↦ β/u^{5}, etc.) but not the equation F_{E} and the matrices D and W. The equation F_{E} is defined up to a scalar multiple of F_{C}- that is, if we multiply α, β, γ, and δ by some number in K, we get the same Jacobian and the same maps.

5 Curves of Genus One in ℙ^{4}

5.1 Setting Up

Let K be a field with char(K) not divisible by 5. Assume that ζ_{5} ∈ K. Let

(C/S, L, T )

be a 5-prepared genus one curve. When S = Spec(K), we know that (Remark A on page 4) the line bundle L embeds C in ℙ_{K}^{4} as a degree 5 curve; in fact C will be given as the intersection of five quadrics.

Remark. A particularly nice way to get five such quadrics was explained to me by Prof. Shepherd-Baron: Give yourself a 5-by-5 skew-symmetric matrix (so the diagonal elements will be 0). There are then 5 4-by-4 minors along the diagonal; the determinants of these 5 matrices turn out to be perfect squares. The resulting 5 square roots are homogeneous degree 2 polynomials in the original entries; if the entries are linear in the 5 coordinates of ℙ^{4}, the resulting intersection of five quadrics is generically (over the space of the matrix entries) a smooth curve of genus one.

Let E be an elliptic curve with a principal homogeneous space action λ on C that induces an isomorphism of E with the Jacobian of C. We will sometimes identify the two elliptic curves by this isomorphism. We have the automorphism λ_{T} on C which by Remark C (page 6) extends to an automorphism of ℙ_{K}^{4}. Let M_{T} be a lift of λ_{T} to GL_{n}(K).

Fix five quadrics which define C, and let V be the vector space generated by those quadrics. Lemma 5.8.1 of [8] tells us that since V is a \overline{K}−vector space and since G = Gal(\overline{K}/K) acts continuously on V in a compatible manner with the \overline{K}− action, V has a basis of G−invariant vectors. M_{T} acts on V. Let Q ∈ V be a rational quadric (i.e. invariant by the G−action) such that < M_{T}^{i} · Q >= V. Such a Q exists by the same argument that proved Claim 3.8. Given such a Q we have, for any point x ∈ C and for 0 ≤ i ≤ 4,

M_{T}^{i} · Q(x) = 0 ⟺ Q(M_{T}^{−i} · (x)) = 0.

In essence, by finding such a Q, we have reduced the number of quadrics needed to define C from five to one. Let

Q(x) = ∑_{0≤i≤j≤4} a_{i,j} x_{i} x_{j}.

Let S be the largest subscheme of Spec(K[b, a_{i,j}]) over which the scheme defined by the formula

{x ∈ ℙ_{S}^{4} | M_{T}^{i} · Q(x) = 0 ∀i}

is a smooth curve over S. Let Λ_{S} = (C/S, L, T ) be the 5-prepared genus one curve defined by these quadrics M_{T}^{i} · Q(x).

5.2 The Main Theorem

Theorem 5.1 Let S, Λ_{S} = (C/S, L, T ) be as above. Let

Λ_{Jac,S} = (E/S, O(5 · O_{E}), T )

be the Jacobian of Λ_{S}. Then E is given by quadrics D_{T}^{i} · Q_{E}(x), 0 ≤ i ≤ 4, where

• D_{T} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & ζ_{5} & 0 & 0 & 0 \\ 0 & 0 & ζ_{5}^{2} & 0 & 0 \\ 0 & 0 & 0 & ζ_{5}^{3} & 0 \\ 0 & 0 & 0 & 0 & ζ_{5}^{4} \end{matrix} ) ,

• Q_{E}(x) = Q(Φ(x)), where Φ ∈ PGL_{3}(K(b, a_{i j})_{s}) is given by the matrix

Φ = ( v_{0} ^{5} \sqrt{d} · v_{1} ^{5} \sqrt{d^{2}} · α · v_{2} ^{5} \sqrt{d^{3}} · β · v_{3} ^{5} \sqrt{d^{4}} · γ · v_{4} · k ) , with

• v_{i}, 0 ≤ i ≤ 4, are the eigenvectors of M_{T} ,

• B(X, Y) is the bilinear form Q(X + Y) − Q(X) − Q(Y)

• d = \frac{Q(v_{0})·B(v_{0},v_{3})·B(v_{0},v_{2})^{2}}{B(v_{2},v_{3})·B(v_{1},v_{2})·Q(v_{1})^{2}} , α = −\frac{Q(v_{2})}{B(v_{1},v_{3})}, β = −\frac{Q(v_{2})·Q(v_{1})^{2}}{B(v_{1},v_{3})·B(v_{0},v_{2})^{2}} ,

γ = \frac{Q(v_{0})}{d·B(v_{1},v_{4})}, k ∈ K, • vec(x_{0}) = v_{0} + ^{5} \sqrt{d} · v_{1} + ^{5} \sqrt{d^{2}} · α · v_{2} + ^{5} \sqrt{d^{3}} · β · v_{3} is the pre-image of

O under φ, and

• O_{E} = (1 : 1 : 1 : 1 : 0).

To prove Theorem 5.1 we would like to use the comments in Section 3.10 to reduce to proving it over the generic point of the base S. The snag is that we would have to prove that S as defined above is normal. I believe it is, but I do not have an elegant proof. Instead we will prove our theorem first over the generic point and then deduce it for all other smooth fibers of our family of curves. Let K′ = K(b, a_{i j}). We need to find the Jacobian of the 5-prepared genus one curve Λ = (C/K′, L, T) where C ∈ ℙ_{K′}^{4} is given by the formula

{x ∈ ℙ_{S}^{4} | M_{T}^{i} · Q(x) = 0 ∀i}.

5.3 Standardizing the model for C

We already know that C is given in ℙ_{K′}^{4} by the five quadrics M_{T}^{i} · Q. By Claim 3.8 we may assume

M_{T} = [( \begin{matrix}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ b & 0 & 0 & 0 & 0 \end{matrix} )] ,

where b is some lift to K^{∗} of det (M_{T}).

By standardizing M_{T} we get also change Q but its salient properties remain- it is still K′-rational and its M_{T} -orbit still generates V.

5.4 Finding x_{0}

Let Λ_{E} = (E/K′, O(5 · O_{E}), T) be isomorphic to the Jacobian of Λ. From Claim 3.7, page 12 we know that x_{0}, the point of C which maps to O_{E} by φ_{x_{0}} (Claim 3.5), lives on the hyperplane in ℙ^{4} which contains four of the five fixed \overline{K}-points of M_{T}: p_{i} = (1 : ^{5} \sqrt{b}ζ_{5}^{i} : …: ^{5} \sqrt{b^{4}}ζ_{5}^{4i}) ∈ ℙ^{4}(\overline{K}). The vectors v_{i}

associated to the p_{i} are the eigenvectors of M, where v_{i} has eigenvalue ^{5} \sqrt{b}·ζ_{5}^{i}. Then x_{0} = proj(λ_{0} · v_{0} + λ_{1} · v_{1} + λ_{2} · v_{2} + λ_{3} · v_{3}) for some λ_{i} ∈ K′_{s}, 0 ≤ i ≤ 3. Since x_{0} is also a point on C, we know Q(M^{i} · x_{0}) = 0, 0 ≤ i ≤ 4. We have

M^{i}(x_{0}) = M^{i}(proj(λ_{0} · v_{0} + λ_{1} · v_{1} + λ_{2} · v_{2} + λ_{3} · v_{3}))
= proj(λ_{0} · M^{i}(v_{0}) + λ_{1} · M^{i}(v_{1}) + λ_{2} · M^{i}(v_{2}) + λ_{3} · M^{i}(v_{3}))
= proj(λ_{0} · v_{0} + λ_{1} · ζ_{5}^{i} · v_{1} + λ_{2} · ζ_{5}^{2}i · v_{2} + λ_{3} · ζ_{5}^{3}i · v_{3}).

We have:

Q(M^{i} · x_{0}) = 0 for all i, 0 ≤ i ≤ 4.
⇕

Q(λ_{0} · v_{0} + λ_{1} · ζ_{5}^{i} · v_{1} + λ_{2} · ζ_{5}^{2i} · v_{2} + λ_{3} · ζ_{5}^{3i} · v_{3}) = 0 for all i, 0 ≤ i ≤ 4.

Claim 5.1 Write x_{0} = proj(λ_{0} · v_{0} + λ_{1} · v_{1} + λ_{2} · v_{2} + λ_{3} · v_{3}). Then

• \frac{λ_{1}}{λ_{0}} = ^{5} \sqrt{d} for some d ∈ K′( ^{5} \sqrt{b})

• \frac{λ_{2}}{λ_{0}} = ^{5} \sqrt{d^{2}} · α for some α ∈ K′( ^{5} \sqrt{b})

• \frac{λ_{3}}{λ_{0}} = ^{5} \sqrt{d^{3}} · β for some β ∈ K′( ^{5} \sqrt{b})

Proof. We will be studying the following diagram of fields:

K′(x_{0}) | K′( ^{5} \sqrt{b}) | K′ There are five points in the intersection of C with the hyperplane

ℍ = λ_{0} · v_{0} + λ_{1} · v_{1} + λ_{2} · v_{2} + λ_{3} · v_{3}.

Once we rationalize one point, all of them are rational, since once we have x_{0} we also have x_{0}+i·T, and toegether they make up the intersection (see Claim 3.7). Any field over which all of the x_{0} + i · T ’s are defined contains ^{5} \sqrt{b}. Thus to rationalize one point of intersection will require an extension of degree five of K′( ^{5} \sqrt{b}); that is, [K′(x_{0}) : K′( ^{5} \sqrt{b})] = 5. Moreover, K′(x_{0})/K′( ^{5} \sqrt{b}) is a Galois extension, since the points in the intersection are sent only to each other by elements of Gal(\overline{K′( ^{5} \sqrt{b})}/K′( ^{5} \sqrt{b})) (since ℍ is fixed). Let σ generate the group Gal(K′(x_{0})/K′( ^{5} \sqrt{b})) ≅ ℤ/5ℤ. We may assume that σ(x_{0}) = x_{0}+T. We now have:

σ(x_{0}) = λ_{0} · v_{0} + λ_{1} · ζ_{5} · v_{1} + λ_{2} · ζ_{5}^{2} · v_{2} + λ_{3} · ζ_{5}^{3} · v_{3}.

We can also write

σ(x_{0}) = σ(λ_{0}) · v_{0} + σ(λ_{1}) · v_{1} + σ(λ_{2}) · v_{2} + σ(λ_{3}) · v_{3},

since the v_{j}’s are defined over K( ^{5} \sqrt{b}). Over K( ^{5} \sqrt{b}), ℍ ≅ ℙ_{K( ^{5} \sqrt{b})}^{3}, with coordinates λ_{0}, λ_{1}, λ_{2}, and λ_{3}. We have

(λ_{0} : λ_{1} : λ_{2} : λ_{3}) ↦^{σ} (λ_{0} : λ_{1} · ζ_{5} : λ_{2} · ζ_{5}^{2} : λ_{3} · ζ_{5}^{3}) .

We conclude

σ (\frac{λ_{1}}{λ_{0}} ) = ζ_{5} · \frac{λ_{1}}{λ_{0}} ,  σ (\frac{λ_{2}}{λ_{0}} ) = ζ_{5}^{2} · \frac{λ_{2}}{λ_{0}} ,  and σ (\frac{λ_{3}}{λ_{0}} ) = ζ_{5}^{3} · \frac{λ_{3}}{λ_{0}} .

Define d = (\frac{λ_{1}}{λ_{0}} )^{5} . The norm of (\frac{λ_{1}}{λ_{0}} ) is d :

ℕ_{K(x_{0})/K( ^{5} \sqrt{b})} (\frac{λ_{1}}{λ_{0}} ) = ∏_{i=0}^{4} σ^{i} (\frac{λ_{1}}{λ_{0}} ) = ∏_{i=0}^{4} ζ_{5}^{i} · \frac{λ_{1}}{λ_{0}} = \frac{λ_{1}}{λ_{0}}^{5} = d ∈ K( ^{5} \sqrt{b}).

This implies K(x_{0}) = K( ^{5} \sqrt{b}, ^{5} \sqrt{d}). Since σ(\frac{λ_{2}}{λ_{0}} ) = ζ_{5}^{2} · \frac{λ_{2}}{λ_{0}} , we see that \frac{λ_{2}}{λ_{0}} / ^{5} \sqrt{d^{2}} is fixed by σ, so define α = \frac{λ_{2}}{λ_{0}} / ^{5} \sqrt{d^{2}} ∈ K( ^{5} \sqrt{b}). Similarly, define β = \frac{λ_{3}}{λ_{0}} / ^{5} \sqrt{d^{3}} ∈ K( ^{5} \sqrt{b}). □

It remains to describe d, α, β, and γ. Let B(X, Y) be the bilinear form associated to Q, i.e.

B(X, Y) = Q(X + Y) − Q(X) − Q(Y).

We may write

x_{0} = λ_{0} · v_{0} + λ_{1} · v_{1} + λ_{2} · v_{2} + λ_{3} · v_{3},

where \frac{λ_{1}}{λ_{0}} = ^{5} \sqrt{d}, \frac{λ_{2}}{λ_{0}} = ^{5} \sqrt{d^{2}} · α, and \frac{λ_{3}}{λ_{0}} = ^{5} \sqrt{d^{3}} · β. Fixing k,

M^{−k} · Q(x_{0}) = Q(M^{k}(x_{0})) = Q(λ_{0} · v_{0} + λ_{1} ζ_{5}^{k} · v_{1} + λ_{2} ζ_{5}^{2k} · v_{2} + λ_{3} ζ_{5}^{3k} · v_{3}) =

∑_{i=0}^{4} λ_{i}^{2} ζ_{5}^{2 i k} · Q(v_{i}) + ∑_{i≠j} λ_{i} λ_{j} ζ_{5}^{k (i+j)} · B(v_{i}, v_{j}) = 0  (2)

The above equation depends on k, and as we vary k we get five different equations. If we add up all those equations, we get:

5 (λ_{0}^{2} · Q(v_{0}) + λ_{2} λ_{3} · B(v_{2}, v_{3})) = 0,

because ∑_{i=0}^{4} ζ_{5}^{i} = 0. However if we divide each equation by the appropriate power of ζ_{5} and then add, we get:

5 (λ_{1}^{2} · Q(v_{1}) + λ_{0} λ_{2} · B(v_{0}, v_{2})) = 0,
5 (λ_{2}^{2} · Q(v_{2}) + λ_{1} λ_{3} · B(v_{1}, v_{3})) = 0,
5 (λ_{3}^{2} · Q(v_{3}) + λ_{0} λ_{1} · B(v_{0}, v_{1})) = 0,

and finally

5 (λ_{0} λ_{3} · B(v_{0}, v_{3}) + λ_{1} λ_{2} · B(v_{1}, v_{2})) = 0.

Since char(K′) ≠ 5, we can divide and rearrange to get:

d · α · β = \frac{λ_{2} λ_{3}}{λ_{0}^{2}} = −\frac{Q(v_{0})}{B(v_{2}, v_{3})}

α = \frac{λ_{0} λ_{2}}{λ_{1}^{2}} = −\frac{Q(v_{1})}{B(v_{0}, v_{2})}

\frac{β}{α^{2}} = \frac{λ_{1} λ_{3}}{λ_{2}^{2}} = −\frac{Q(v_{2})}{B(v_{1}, v_{3})}

\frac{1}{β^{2} · d} = \frac{λ_{0} λ_{1}}{λ_{3}^{2}} = −\frac{Q(v_{3})}{B(v_{0}, v_{1})}

\frac{α}{β} = \frac{λ_{1} λ_{2}}{λ_{0} λ_{3}} = −\frac{B(v_{0}, v_{3})}{B(v_{1}, v_{2})}.

From the above equalities we deduce the value of d, α, and β in terms of the values of B and Q on the fixed points v_{i} of M.

5.5 Standardizing the Model for E

By Claim 3.11 we may assume that “translation by T” on E is given by the matrix D, where

D = ( \begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & ζ_{5} & 0 & 0 & 0 \\ 0 & 0 & ζ_{5}^{2} & 0 & 0 \\ 0 & 0 & 0 & ζ_{5}^{3} & 0 \\ 0 & 0 & 0 & 0 & ζ_{5}^{4} \end{matrix} ) ,

and that

O_{E} = (1 : 1 : 1 : 1 : 0).

The quadrics defining E are the quadrics that Q and its orbit under M_{T} are sent to under φ_{x_{0}}. Define Q_{E} to be the quadric that Q is sent to by φ_{x_{0}}.

Claim 5.2 E is defined by the five quadrics Q_{E}, D · Q_{E}, D^{2} · Q_{E}, D^{3} · Q_{E}, and D^{4} · Q_{E}.

Proof. Let σ ∈ Gal(K′_{s}/K′) be an element such that σ( ^{5} \sqrt{b}) = ^{5} \sqrt{b} and σ( ^{5} \sqrt{d}) = ζ_{5}^{4} ^{5} \sqrt{d}. The “translation by T map on E” is given by the map D = φ_{x_{0}}^{σ} φ_{x_{0}}^{−1}, and the “translation by T map on C” is given by the map D = φ_{x_{0}}^{−1} φ_{x_{0}}^{σ} (see the proof of Claim 4.5). But φ_{x_{0}}^{σ} φ_{x_{0}}^{−1} = (φ_{x_{0}}φ_{x_{0}}^{−1})φ_{x_{0}}^{σ} φ_{x_{0}}^{−1} = φ(φ_{x_{0}}^{−1}φ_{x_{0}}^{σ})φ_{x_{0}}^{−1} = φ_{x_{0}} [M]φ_{x_{0}}^{−1}. The map φ_{x_{0}}^{−1} takes the quadrics defining E to the quadrics defining C, M permutes them, and φ_{x_{0}} brings them back. □

5.6 Finding φ_{x_{0}}

We hope to find φ_{x_{0}}, the map from C to E which exists by Claim 3.5. Lift φ_{x_{0}} ∈ PGL_{n}(K′(x_{0})) to the matrix Φ^{−1} ∈ GL_{5}(K(x_{0})). We have already fixed

φ_{x_{0}} (x_{0} + i · T) = (1 ; ζ_{5}^{i}; ζ_{5}^{2i}; ζ_{5}^{3i}; 0).

Then

Φ^{−1} · M^{i}(x_{0}) = (1 ; ζ_{5}^{i}; ζ_{5}^{2i}; ζ_{5}^{3i}; 0) ⟺ M^{i}(x_{0}) = Φ · (1 ; ζ_{5}^{i}; ζ_{5}^{2i}; ζ_{5}^{3i}; 0).

Since we already know the form of each M^{i}(x_{0}) (see page 40) we deduce that there exist h_{i} ∈ K′(x_{0}) such that

Φ = ( v_{0} v_{1} ^{5} \sqrt{d} v_{2} ^{5} \sqrt{d^{2}}α v_{3} ^{5} \sqrt{d^{3}}β h ) , for h = ( \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \\ h_{4} \\ h_{5} \end{matrix} ) .

Claim 5.3 Φ is of the form

Φ = ( v_{0} v_{1} ^{5} \sqrt{d} v_{2} ^{5} \sqrt{d^{2}}α v_{3} ^{5} \sqrt{d^{3}}β v_{4} ^{5} \sqrt{d^{4}}γ ) ,

for γ ∈ K′( ^{5} \sqrt{b}).

Proof. Let σ^{−1} be as in the proof of Claim 4.5, i.e. such that σ^{−1}( ^{5} \sqrt{b}) = ^{5} \sqrt{b}, σ^{−1}( ^{5} \sqrt{d}) = ζ_{5}^{4} ^{5} \sqrt{d}, and the map φ_{x_{0}}^{σ^{−1}}φ_{x_{0}}^{−1} is “translation by T on E.” In terms of matrices this means that for some κ ∈ K′(x_{0}), σ^{−1}(Φ^{−1} ) · Φ = κ · D. We rewrite this as σΦ = κ · Φ · D :

σΦ = σ ( v_{0} v_{1} ^{5} \sqrt{d} v_{2} ^{5} \sqrt{d^{2}}α v_{3} ^{5} \sqrt{d^{3}}β h )
= ( v_{0} v_{1} ^{5} \sqrt{d}ζ_{5} v_{2} ^{5} \sqrt{d^{2}}ζ_{5}^{2}α v_{3} ^{5} \sqrt{d^{3}}ζ_{5}^{3}β σ(h) )
= κ · Φ · D = κ · ( v_{0} v_{1} ^{5} \sqrt{d}ζ_{5} v_{2} ^{5} \sqrt{d^{2}}ζ_{5}^{2}α v_{3} ^{5} \sqrt{d^{3}}ζ_{5}^{3}β ζ_{5}^{4} · h ) .

Similarly we know the map φ_{x_{0}}^{−1} φ_{x_{0}}^{σ^{−1}} is “translation by T on C.” This means that for some κ′ ∈ K′(x_{0}), Φ · σ^{−1}Φ^{−1} = κ′ · M, or in other words σΦ = κ′ · M · Φ :

σΦ = ( v_{0} v_{1} ^{5} \sqrt{d}ζ_{5} v_{2} ^{5} \sqrt{d^{2}}ζ_{5}^{2}α v_{3} ^{5} \sqrt{d^{3}}ζ_{5}^{3}β σ(h) ) = κ′ · M · Φ

= κ′ · ( \begin{matrix}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ b & 0 & 0 & 0 & 0 \end{matrix} ) · ( v_{0} v_{1} ^{5} \sqrt{d} v_{2} ^{5} \sqrt{d^{2}}α v_{3} ^{5} \sqrt{d^{3}}β \begin{matrix}h_{1} \\ h_{2} \\ h_{3} \\ h_{4} \\ h_{5} \end{matrix} )

= κ′ · ^{5} \sqrt{b} · ( v_{0} v_{1}ζ_{5} ^{5} \sqrt{d} v_{2}ζ_{5}^{2} ^{5} \sqrt{d^{2}}α v_{3}ζ_{5}^{3} ^{5} \sqrt{d^{3}}β \begin{matrix}h_{2} \\ h_{3} \\ h_{4} \\ h_{5} \\ bh_{1} \end{matrix} ) .

We know that σ(h_{1}) = h_{2} and σ(h_{1}) = ζ_{5}^{4}h_{1}, from which we conclude h_{2} = h_{1}ζ_{5}^{4}. We continue this way and conclude that the right-most column of Φ is h_{1} · v_{4}. Moreover, we know h_{1} ∈ K′(x_{0}) and we know how σ from above acts on h_{1}; namely, σ(h_{1} ) = ζ_{5}^{4}h_{1}. This means that \frac{h_{1}}{\sqrt[5]{d^{4}}} ∈ K′( ^{5} \sqrt{b}), or in other words h_{1} = ^{5} \sqrt{d^{4}} · γ for some γ ∈ K′( ^{5} \sqrt{b}). □

It remains to find γ. We already know γ ∈ K( ^{5} \sqrt{b}), so if we understand how γ moves under a Galois element moving ^{5} \sqrt{b} to ^{5} \sqrt{b}·ζ_{5}, we will understand γ up to a multiple of something in K. To find γ we will investigate the model for E :

Q_{E}(y) = 0 ⟺ Q(φ_{x_{0}}^{−1}(y)) = Q(Φy) = 0.

Let y = (V : W : X : Y : Z); then up to scalar multiples,

Q_{E} (y) = Q (Φ · y)
= Q(V v_{0} + W v_{1}ζ_{5} ^{5} \sqrt{d} + X ζ_{5}^{2} ^{5} \sqrt{d^{2}}α + Y ζ_{5}^{3} ^{5} \sqrt{d^{3}}β + Z ζ_{5}^{4} ^{5} \sqrt{d^{4}}γ).

We expand to get

Q_{E}(y) = V^{2} · Q(v_{0}) + W^{2} · ζ_{5}^{2} ^{5} \sqrt{d^{2}} Q(v_{1}) + … + Y Z · ζ_{5}^{7} ^{5} \sqrt{d^{7}} β γ B(v_{3}, v_{4}).

We now find a new representation of E; we will find five quadrics which still define E but are fixed by a multiple of D. The reason we do so is that a quadratic form fixed by D has many fewer terms than the general quadratic form. Define

R_{k} = ∑_{i=0}^{4}(D/ζ_{5}^{k})^{i} Q_{E}.

Claim 5.4 Each R_{k} is fixed by D/ζ_{5}^{k}.

Proof.

D/ζ_{5}^{k} (R_{k}) = D/ζ_{5}^{k} (∑_{i=0}^{4}(D/ζ_{5}^{k})^{i} Q_{E}) = ∑_{i=0}^{4}(D/ζ_{5}^{k})^{i+1} Q_{E} = R_{k}.□

If we write

R_{k} = ∑_{0≤i ≤ j≤4} r_{i,j} x_{i} x_{j},

then all r_{i j} = 0 unless i + j − 2 · k ≡ 0 (mod 5 ), since R_{k} is fixed by D/ζ_{5}^{k}. For each k, we have three terms in R_{k} (for example, R_{0} = r_{0 0} x_{0}^{2} +r_{1 4} x_{1} x_{4} + r_{2 3} x_{2} x_{3}).

Claim 5.5 Let τ ∈ Gal (\overline{K}/K) be an element such that σ( ^{5} \sqrt{b}) = ζ_{5} ^{5} \sqrt{b}. In particular, this means τv_{i} = v_{i+1}. Then τφ φ^{−1} is represented by the matrix

S_{τ} = [( \begin{matrix}0 & \sqrt[5]{d} & 0 & 0 & 0 \\ 0 & 0 & α ^{5} \sqrt{\frac{d^{2}}{τd}} & 0 & 0 \\ 0 & 0 & 0 & \frac{β}{τα} ^{5} \sqrt{\frac{d^{3}}{(τd)^{2}}} & 0 \\ 0 & 0 & 0 & 0 & \frac{γ}{τβ} ^{5} \sqrt{\frac{d^{4}}{(τd)^{3}}} \\ \frac{1}{τγ} ^{5} \sqrt{\frac{1}{(τd)^{4}}} & 0 & 0 & 0 & 0 \end{matrix} )] .

Proof. We need to see why τΦ^{−1} Φ = S_{τ}, or in other words why κ·Φ = τΦ S_{τ} for some κ ∈ K′(x_{0}) :

τΦ S_{τ} = τ ( v_{0} ^{5} \sqrt{d} · v_{1} ^{5} \sqrt{d^{2}} · α · v_{2} ^{5} \sqrt{d^{3}} · β · v_{3} ^{5} \sqrt{d^{4}} · γ · v_{4} ) · S_{τ} =

( v_{1} ^{5} \sqrt{τd} · v_{2} ^{5} \sqrt{(τd)^{2}} · τα · v_{3} ^{5} \sqrt{(τd)^{3}} · τβ · v_{4} ^{5} \sqrt{(τd)^{4}} · τγ · v_{0} )·S_{τ} =

( v_{0} ^{5} \sqrt{d} · v_{1} ^{5} \sqrt{d^{2}} · α · v_{2} ^{5} \sqrt{d^{3}} · β · v_{3} ^{5} \sqrt{d^{4}} · γ · v_{4} ) = Φ.

Corollary 5.1 S_{τ} permutes the quadrics R_{k}.

Proof. The matrix S_{τ} moves x_{i} to some multiple of x_{i+1} (i ∈ 𝔽_{5}), so S_{τ} moves the term r_{i j} x_{i} x_{j} of R_{k} to b_{i+1 j+1} x_{i+1} x_{j+1}, with i + j − 2 · k ≡ 0 (mod 5 ) ⇔ (i + 1) + (j + 1) − 2 · (k + 1) ≡ 0 (mod 5 ). In other words S_{τ} moves R_{0} to a quadric which has exactly the same non-zero coefficients as R_{1}. By a dimension count we conclude that S_{τ} · R_{0} is a multiple of R_{1}. We can rescale the R_{k}’s so that S_{τ} actually permutes them; rescaling will keep them fixed by D/ζ_{5}^{k} and won’t change E.

Claim 5.6 γ can be taken to be \frac{Q(v_{0})}{d·B(v_{1},v_{4})}.

Proof. First note that

( \begin{matrix}V \\ W \\ X \\ Y \\ Z \end{matrix} ) ↦^{S_{τ}} ( \begin{matrix}W ^{5} \sqrt{d} \\ X \frac{\sqrt[5]{d^{2}}}{τd} α \\ Y \frac{\sqrt[5]{d^{3}}}{(τd)^{2}} \frac{β}{τα} \\ Z \frac{\sqrt[5]{d^{4}}}{(τd)^{3}} \frac{γ}{τβ} \\ V \frac{1}{(τd)^{4}} \frac{1}{τγ} \end{matrix} ) ,

so

S(R_{0}) = S(V^{2} · Q(v_{0}) + W Z · B(v_{1}, v_{4}) d γ + X Y · B(v_{2}, v_{3}) d α β) =

W^{2}·Q(v_{0}) ^{5} \sqrt{d^{2}}+X V ·B(v_{1}, v_{4}) \frac{d γα}{τγ} ^{5} \sqrt{\frac{d^{2}}{(τd)^{5}}} +Y Z·B(v_{3}, v_{3}) \frac{d α β^{2} γ}{τα τβ} ^{5} \sqrt{\frac{d^{7}}{(τd)^{5}}} .

Since

R_{1} = W^{2} · Q(v_{1}) ^{5} \sqrt{d^{2}} + V X · B(v_{0}, v_{2}) ^{5} \sqrt{d^{2}} α + Y Z · B(v_{3}, v_{4}) ^{5} \sqrt{d^{7}} β γ,

we have

\frac{B(v_{0}, v_{2}) ^{5} \sqrt{d^{2}} α}{Q(v_{1}) ^{5} \sqrt{d^{2}}} = \frac{B(v_{1}, v_{4}) \frac{d γα}{τγ} ^{5} \sqrt{\frac{d^{2}}{(τd)^{5}}}}{Q(v_{0}) ^{5} \sqrt{d^{2}}} .

This simplifies to

\frac{γ}{τγ} = \frac{τd · Q(v_{0}) · B(v_{0}, v_{2})}{d · Q(v_{1}) · B(v_{1}, v_{4})} = \frac{τd · Q(v_{0}) · τB(v_{4}, v_{1})}{d · τQ(v_{0}) · B(v_{1}, v_{4})},

so we have

\frac{γ}{τγ} = \frac{Q(v_{0})}{d · B(v_{1}, v_{4})} · τ (\frac{d · B(v_{1}, v_{4})}{Q(v_{0})}) .

In other words we have an element γ · \frac{d·B(v_{1}, v_{4})}{Q(v_{0})} of K′( ^{5} \sqrt{b}) which is fixed by

τ - this means that it is in K′. We conclude that γ = k · \frac{Q(v_{0})}{d·B(v_{1}, v_{4})} for some k ∈ K′.

We can choose k as we wish; once we do we will have completely rigidified the model for E.

We have found the Jacobian of the generic point of our family C/S from Theorem 5.1. Now say C_{a_{i j},b} is a smooth curve over K with a_{i j}, b ∈ K. Note that φ_{x_{0}} brings C_{a_{i j},b} to some elliptic curve over K. We want to show this is (isomorphic to) the Jacobian of C_{a_{i j},b}. We will do this directly, i.e. by looking at the cocycle created by φ_{x_{0}}. Let L be the Galois closure of K(x_{0}). If ^{5} \sqrt{b}, ^{5} \sqrt{d} ∉ K, then our previous Galois theory computations have already found the Jacobian of C_{a_{i j},b}. If ^{5} \sqrt{b} ∈ K, then we see that

Φ = ( v_{0} ^{5} \sqrt{d} · v_{1} ^{5} \sqrt{d^{2}} · α · v_{2} ^{5} \sqrt{d^{3}} · β · v_{3} ^{5} \sqrt{d^{4}} · γ · v_{4} · k ) =

( \begin{matrix}1 & 1 & 1 & 1 & 1 \\ 1 & ζ_{5} & ζ_{5}^{2} & ζ_{5}^{3} & ζ_{5}^{4} \\ 1 & ζ_{5}^{2} & ζ_{5}^{4} & ζ_{5} & ζ_{5}^{3} \\ 1 & ζ_{5}^{3} & ζ_{5}^{3} & ζ_{5}^{4} & ζ_{5}^{2} \\ 1 & ζ_{5}^{4} & ζ_{5} & ζ_{5}^{2} & ζ_{5} \end{matrix} ) · ( \begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & \sqrt[5]{d} & 0 & 0 & 0 \\ 0 & 0 & ^{5} \sqrt{d^{2}} · α & 0 & 0 \\ 0 & 0 & 0 & ^{5} \sqrt{d^{3}} · β & 0 \\ 0 & 0 & 0 & 0 & ^{5} \sqrt{d^{4}} · γ \end{matrix} ) .

If we modify our embedding of C_{a_{i j},b} by the left matrix, then we get a “Selmer-like” curve where “Translation by T ” is represented by D both on C_{a_{i j},b} and on E. Moreover, the map φ_{x_{0}} is now a diagonal matrix. The proof that this matrix (the one on the right above) actually maps C_{a_{i j},b} to its Jacobian is very similar to the proof of Theorem 6.2. If ^{5} \sqrt{d} ∈ K, we have L = K( ^{5} \sqrt{b}). Note if both ^{5} \sqrt{d} ∈ K, and ^{5} \sqrt{b} ∈ K, there is a rational point on C and Φ is an K-isomorphism of C with its Jacobian. So assume that K( ^{5} \sqrt{b})/K is a degree five extension, and let σ generate the Galois group. We just need to show that the map φ_{x_{0}}^{σ} · φ_{x_{0}}^{−1} on E is a translation map. It is enough to show that φ_{x_{0}}^{σ} · φ_{x_{0}}^{−1}(x) = x + T′ for five values of x ∈ E and some T′ ∈ E[5](\overline{K}), since composing φ_{x_{0}}^{σ} · φ_{x_{0}}^{−1} with λ_{−T′} gives an automorphism of E fixing the origin and four other points, so it must be the identity. By construction we know φ_{x_{0}}(x_{0} + i · T) = i · T. This means

(φ_{x_{0}} · λ_{T,C}^{i}) · x_{0} = (λ_{T,E}^{i} · φ_{x_{0}}) · x_{0}.

The two maps φ_{x_{0}} ·λ_{T,C}^{i} and λ_{T,E}^{i} ·φ_{x_{0}} act the same on the five points x_{0}+i·T, which by the above argument means φ_{x_{0}} · λ_{T,C}^{i} = λ_{T,E}^{i} · φ_{x_{0}}. Then because T is rational, λ_{T,E} commutes with the cocycle φ_{x_{0}}^{σ} · φ_{x_{0}}^{−1}. we can take for x the five points of any orbit of x ∈ E by Λ_{T,E}.

6 Curves of genus one in ℙ^{1} × ℙ^{1}

6.1 Setting up

Let K be a field with char(K) not divisible by 2. Let

(C/S, (L_{1}, L_{2}), T )

be a 2-prepared genus one curve over some base S. When S = Spec(K), we know that (Remark A on page 4) the pair (L_{1}, L_{2}) embeds (see Remark A) C in ℙ_{K}^{1} ×_{K} ℙ_{K}^{1} as a double cover of each ℙ_{K}^{1}.; in fact C will be given by a (2, 2)-form. There are two involutions κ_{1} and κ_{2} of C given an embedding C in ℙ_{K}^{1} ×_{K} ℙ_{K}^{1}; namely, since each map C ⟶ ℙ_{K}^{1} is a double cover, let for i = 1, 2, x ∈ C,

κ_{i}(x) = y ∈ C such that O(y + x) ≅ L_{i}.

Let E be an elliptic curve with a principal homogeneous space action λ on C that induces an isomorphism of E with the Jacobian of C. We will sometimes identify the two elliptic curves by this isomorphism.

Assume that S = Spec(K). We have the automorphism λ_{T} on C which by Remark C (page 6) extends to an automorphism of ℙ_{K}^{1} ×_{K} ℙ_{K}^{1}. Let (M_{1}, M_{2}) be a lift of λ_{T} to GL_{n}(K)×GL_{n}(K). By Claim 3.8 we may modify the embedding of C in ℙ_{K}^{1} ×_{K} ℙ_{K}^{1} by a linear automorphism so that

(M_{1}, M_{2}) = (( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) , ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} )) .

Fix a (2, 2)-form F_{C} which defines C. Then (M_{1}, M_{2}) acts on (2, 2)-forms, and if we represent

F_{C} = s^{2}(f_{1}x^{2} + f_{2}xy + f_{3}y^{2}) + st (g_{1}x^{2} + g_{2}xy + g_{3}y^{2}) + t^{2}(h_{1}x^{2} + h_{2}xy + h_{3}y^{2})

by the matrix

M_{F_{C}} = ( \begin{matrix}f_{1} & f_{2} & f_{3} \\ g_{1} & g_{2} & g_{3} \\ h_{1} & h_{2} & h_{3} \end{matrix} ) ,

the condition that (M_{1}, M_{2}) fixes F_{C} means that M_{F_{C}} actually has the form (see Section 6.3)

M_{F_{C}} = ( \begin{matrix}bf_{3} & f_{2} & f_{1} \\ bg_{3} & g_{2} & g_{3} \\ abf_{1} & af_{2} & af_{3} \end{matrix} ) .

Let S be the largest subscheme of Spec(K[a, b, f_{1}, f_{2}, f_{3}, g_{2}, g_{3}]) over which the scheme defined by the formula

{x ∈ ℙ_{S}^{1} ×_{S} ℙ_{S}^{1} | F_{C}(x) = 0}

is a smooth curve over S. Let Λ_{S} = (C/S, (L_{1}, L_{1}), T ) be the 2-prepared genus one curve defined by F_{C}.

6.2 The Main Theorem

Theorem 6.1 Let S, Λ_{S} be as above. Then

Λ_{Jac,S} = (E/S, (O(O_{E} + T ), L_{2} ⊗ L_{1}^{−1} ⊗ O(T )), T )

is the Jacobian of Λ_{S}. The elliptic curve E is given by the (2, 2)-form

F_{E} = −\frac{(a(f_{1} + f_{3})^{2} − g_{3}^{2}) S_{1}^{2}}{(f_{3} − f_{1})^{2}} s^{2}x^{2} + st (\frac{b S_{1}^{2}}{S_{2}} x^{2} − S_{2} y^{2})

+t^{2} (b \frac{S_{1}^{2}}{S_{2}} x^{2} + S_{2} xy + S_{2} y^{2}) ,

where

S_{1} = 2\sqrt{a}\frac{2f_{2}g_{3} − f_{1}g_{2} − f_{3}g_{2}}{a(f_{1} + f_{3})^{2} − g_{3}^{2}} , S_{2} = \frac{4b(a(f_{1} + f_{3})^{2} − g_{3}^{2}) − 4af_{2}^{2} − g_{2}^{2}}{a(f_{1} + f_{3})^{2} − g_{3}^{2}} ,

and with O = ((1 : 0), (1 : 0)).

To prove Theorem 6.1 we will use the comments in Section 3.10 to reduce to proving it over the generic point of the base S. Let

K′ = K(a, b, f_{1}, f_{2}, f_{3}, g_{2}, g_{3}).

Then we need to find the Jacobian of the 2-prepared genus one curve

Λ = (C/K′, (L_{1}, L_{2}), T)

where

C ⟶^{f_{C}} ℙ_{K′}^{1} ×_{K′} ℙ_{K′}^{1}

is given by the equation F_{C}(x) = 0, for

F_{C} : s^{2}(bf_{3}x^{2}+f_{2}xy+f_{1}y^{2})+st (bg_{3}x^{2}+g_{2}xy+g_{3}y^{2})+t^{2}(abf_{1}x^{2}+af_{2}xy+af_{3}y^{2}),

T is a K′-rational point of exact order two of the Jacobian of C, and

L_{i} = (pr_{i} ∘ f_{C})^{∗}O(1).

6.3 Standardizing the Model for C

Claim 6.1 With the above representation (M_{1}, M_{2}) of λ_{T} acting on C, the

equation for C is of the following form: M_{F} = ( \begin{matrix}bf_{3} & f_{2} & f_{1} \\ bg_{3} & g_{2} & g_{3} \\ abf_{1} & af_{2} & af_{3} \end{matrix} ) .

Proof. (M_{1}, M_{2}) is a linear operator on V =< s^{2}x^{2}, s^{2}xy, s^{2}y^{2}, stx^{2}, … >, a nine dimensional K-vector space. Moreover, (M_{1}, M_{2})^{2} = (M_{1}^{2}, M_{2}^{2}) acts by multiplcation by a^{2}b^{2} on every basis vector s^{2}x^{2}, s^{2}xy, … of V. This means that λ_{T} has eigenspaces in V corresponding to the eigenvalues ±ab. We have the following table which lists these eigenspaces:

eigenvector eigenvalue
bs^{2}x^{2} ± at^{2}y^{2}  ±ab
s^{2}xy ± at^{2}xy  ±ab
s^{2}y^{2} ± abt^{2}x^{2}  ±ab
bstx^{2} ± sty^{2}  ±ab
stxy  ab

The claim follows from the simple observation that every eigenvector with

eigenvalue −ab vanishes at the point ( \sqrt{a} 1 ) , ( \begin{matrix}1 \\ √ & \overline{b} \end{matrix} ) . That means every

(2, 2)-form coming form the eigenspace corresponding to −ab has a fixed point under λ. Since F has no such fixed point, it lives in the eigenspace corresponding to ab.□

Remark.We have not demonstrated an algorithm which produces the f_{i}’s, g_{j}’s, a and b from the coefficients of a (2, 2)-form; however, we have demonstrated that every 2-prepared genus one curve over K is in the family of (2, 2)-forms above.

6.4 A Selmer-like Example

Assume for now that a = b = 1, that is the determinants of M_{1} and M_{2} are both 1 (mod squares). Then by Claim 3.11 we may assume

(M_{1}, M_{2}) = (( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )) .

Since (M_{1}, M_{2}) fixes F_{C} we have

(M_{1}, M_{2}) · M_{F_{C}} = ( \begin{matrix}f_{1} & −f_{2} & f_{3} \\ −g_{1} & g_{2} & −g_{3} \\ h_{1} & −h_{2} & h_{3} \end{matrix} ) = M_{F_{C}} = ( \begin{matrix}f_{1} & f_{2} & f_{3} \\ g_{1} & g_{2} & g_{3} \\ h_{1} & h_{2} & h_{3} \end{matrix} )

⟹ M_{F_{C}} = ( \begin{matrix}f_{1} & 0 & f_{3} \\ 0 & g_{2} & 0 \\ h_{1} & 0 & h_{3} \end{matrix} ) .

Theorem 6.2 Let Λ = (C, (L_{1}, L_{2}), T) be a 2-prepared curve of genus one over K whose embedding in ℙ_{K}^{1} × ℙ_{K}^{1} is given by

M_{F} = ( \begin{matrix}f_{1} & 0 & f_{3} \\ 0 & g_{2} & 0 \\ h_{1} & 0 & h_{3} \end{matrix} ) .

Then the Jacobian Jac(C) = E also embeds in ℙ^{1}×ℙ^{1} with an equation given by

M_{E} = ( \begin{matrix}f_{1} & 0 & −f_{1} \\ 0 & g_{2} & 0 \\ −h_{1}f_{3}/f_{1} & 0 & h_{3} \end{matrix} )

The map from C to E is given by an L−linear automorphism of ℙ^{1} × ℙ^{1}, where L is the field K(\sqrt{−\frac{f_{3}}{f_{1}}} ).

Proof. We can assume that f_{1} ≠ 0 since if f_{1} = 0, C is not smooth. I claim that the map ϕ_{x_{0}} : C ⟶ J(C) = E is the restriction of a linear automorphism of ℙ^{1} × ℙ^{1} defined over K(\sqrt{−\frac{f_{3}}{f_{1}}} ), namely the map

ϕ : ( s t ) , ( \begin{matrix}x \\ y \end{matrix} ) ↦ ( s t ) ( \begin{matrix}1 & 0 \\ 0 & \sqrt{−\frac{f_{3}}{f_{1}}} \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 0 & \sqrt{−\frac{f_{1}}{f_{3}}} \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} )

= ( s \sqrt{−\frac{f_{3}}{f_{1}}} t ) , ( \begin{matrix}x \\ \sqrt{−\frac{f_{1}}{f_{3}}} y \end{matrix} )

This is clearly a map from C to E. Also, note that

E : s^{2} (x^{2} − y^{2}) + \frac{g_{2}}{f_{1}} s t x y + t^{2} (−\frac{h_{1}f_{3}}{f_{1}^{2}} x^{2} + \frac{h_{3}}{f_{1}} y^{2}) = 0

has the rational point (1 : 0), (1 : 1) so we can assign a group structure to E by letting this point be the origin.

To prove that E is actually isomorphic to the Jacobian of C, we form the cocycle ξ ∈ H^{1}(Gal(K̄)/K), Aut(E)(K_{s})) defined by ξ(σ) = ϕ^{σ}ϕ^{−1} for σ ∈ Gal(\overline{K}/K). We’d like to show that ξ takes values in the group E(K_{s}), thinking of E(K_{s}) sitting inside Aut(E)(K_{s}) as the translation maps. The image of ξ naturally falls into two cases; in the first, σ(\sqrt{−a}) = \sqrt{−a}, which clearly means that ξ = ϕ^{σ}ϕ^{−1} = Id_{E}. In the second, σ(\sqrt{−a}) = −\sqrt{−a}, and we calculate

ξ ((s : t) , (x : y)) = (−s : t) , (−x : y) .

Notice we can think of ξ as acting on the whole of ℙ^{1} × ℙ^{1}, since ϕ does. But this action is fixed-point free when we restrict to E, and so must be translation by some point of E. Moreover, the origin (1 : 0), (1 : 1), is clearly mapped to (1 : 0), (−1 : 1), and since the map is clearly an involution, we also conclude that (1 : 0), (−1 : 1) is a two-torsion point of E. □^{1}

6.5 A Locally Trivial Sub-family

A main motivation for this thesis is to understand elements of Sha better. In that light we have the following example coming from ℙ^{1} × ℙ^{1} :

Let a and c be relatively prime integers, with a even. Define b = c^{2} + 4a. Giving the coordinates (s, t) to the first ℙ^{1} and the coordinates (x, y) to the second, define the following family of curves:

(1) C_{a,c} : s^{2}(x^{2} − ay^{2}) + c(a − b) stxy − ab t^{2} (x^{2} − ay^{2}) = 0.

From Theorem 6.2, the Jacobian of this curve is the following:

(2) E_{a,c} : s^{2} (x^{2} − a^{2}y^{2}) + c(a − b) stxy − b t^{2}(x^{2} − a^{2}y^{2}) = 0.

The origin of the Jacobian curve can be taken to be O = (1 : 0), (a : 1); then there is full rational 2-torsion, given by

T = (1 : 0), (−a : 1), T′ = (0 : 1), (a : 1), and T + T′ = (0 : 1), (−a : 1).

The action of the non-trivial rational two-torsion point T is given on both curves by

( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) ,

1This proof is similar to the proof in the “Construction of the Jacobian” chapter of Cassel’s book [2].

sending t to −t and y to −y and fixing s and x. This is because the above curve is a “Selmer-like” curve. The action of “translation by T’ on C” is given by

( \begin{matrix}0 & −1 \\ ab & 0 \end{matrix} ) , ( \begin{matrix}0 & a \\ 1 & 0 \end{matrix} ) ,

and on E is given by

( \begin{matrix}0 & −1 \\ b & 0 \end{matrix} ) , ( \begin{matrix}0 & a^{2} \\ 1 & 0 \end{matrix} ) .

Claim 6.2 The above family C_{a,c} is locally trivial.

Proof. Assume that a, b, and c are in ℤ. Let’s see why there is a rational point over ℚ_{p}. If we let t = 0, s = 1, we are cutting the curve in a line and we get two point which are rational as a pair so either they are rational or they are conjugates defined over a quadratic field. This is determined by whether there is a solution to the equation x^{2} − ay^{2} = 0, i.e. whether a is a square in ℚ_{p}. If instead we let x = 1, y = 0, we are asking whether ab is a square in ℚ_{p}. Finally, if we let s = a, t = 1, we have

x^{2}(a^{2} − ab) + xyac(a − b) + y^{2} (a^{2}b − a^{3}) = 0,

Which has a solution in ℚ_{p} whenever the discriminant (c(a − b)a)^{2} + 4(a^{2} − ab)a(a^{2} −ab) is a square in ℚ_{p}. This is the same as asking whether c^{2} +4a = b is a square in ℚ_{p}. Now for any p |̸2ab, one of a, b, and ab will be a square mod p, and by Hensel’s Lemma we can lift to a solution in ℚ_{p}. For p|2ab, we let x = y = 1, and we have s^{2}(1 − a) + c(a − b) st − ab t^{2} (1 − a) = 0, Which has a solution in ℚ_{p} whenever the discriminant c^{2}(a−b)^{2} +4ab(1−a)^{2} is a square in ℚ_{p}. If p|ab, p ≠ 2, this is congruent to a non-zero square mod p. To see that it is non-zero, assume p | c(a − b); if p | c, since p | ab and since (a, c) = 1 we have p | b = c^{2} + 4a ⇒ p | 4a. If p | (a − b), since p | ab we have p | a and p | b ⇒ p | c. If p = 2, we need to check that c^{2}(a − b)^{2} + 4ab(1 − a)^{2} is 1 mod 8. Since c^{2}(a − b)^{2} is an odd square and thus 1 mod 8, we want 4ab(1 − a)^{2} 0 mod 8. Since a is even, we are done. □

Claim 6.3 For specific choices a_{0} and c_{0} of a and c, The curve C_{a_{0},c_{0}} is nontrivial in the Sha group of its Jacobian E_{a_{0},c_{0}} whenever E_{a_{0},c_{0}} has Mordell-Weil rank 0.

Proof. We will make use of Claim 3.9 and Corollary 3.6. Let E = E_{a_{0},c_{0}} and C = C_{a_{0},c_{0}}. We will again use the short exact sequence (page 197 of [8])

0 ⟶ E(Q)/n E(Q) ⟶^{δ} H^{1}(G, E[n](\overline{ℚ})) ⟶ H^{1}(G, E(\overline{ℚ}))[n] ⟶ 0

in conjunction with the map e_{2}(−, T′)^{∗} : H^{1}(G, E[n](\overline{ℚ})) ⟶ ℚ^{∗}/(ℚ^{∗})^{2} which comes from pairing a cocycle with the 2-torsion point T′. First let’s look at the composition of δ with e(−, T′)^{∗}. Certainly O is mapped to 1, since e(−, T′)^{∗} ∘ δ is a homomorphism. Next, by Corollary 3.6, to find the image of T +T′ we need to map E to ℙ^{1} by the divisor (T +T′)+T′ ∼ O +T, define the action “translation by T′” on this embedding to get an element λ_{T′}, and finally take the negative of its determinant. We already have a map like this, namely the projecion onto the first ℙ^{1} of the above embedding

E ↪ ℙ^{1} × ℙ^{1}. Since λ_{T′} on this ℙ^{1} is ( \begin{matrix}0 & −1 \\ b & 0 \end{matrix} ), the image of T + T′ in

ℚ^{∗}/(ℚ^{∗})^{2} is −b. So we know the image of O and T + T′ so far.

Lift C ∈ H^{1}(G, E(\overline{ℚ}))[n] to the pair (C, 2 · x_{0} + T′) ∈ H^{1}(G, E[n](\overline{ℚ})); by Claim 3.9 its image in ℚ^{∗}/(ℚ^{∗})^{2} is the negative of the determinant of λ_{T′} acting on the second projection of the embedding C ↪ ℙ^{1} × ℙ^{1}, i.e. the map to ℙ^{1} corresponding to the divisor 2 · x_{0} + T′. We have seen above

this λ_{T′} is ( \begin{matrix}0 & a \\ 1 & 0 \end{matrix} ) , so the image of C is a. To finish we want to see two

things, firstly that the image of C in ℚ^{∗}/(ℚ^{∗})^{2} is disjoint from the image of E[2](\overline{ℚ}), and secondly that the image of E[2](\overline{ℚ}) is non-trivial. The latter will let us conclude that there are no rational four-torsion points, which means the contribution of the left-hand group E(Q)/n E(Q) in the above exact sequanece comes only from 2-torsion and possibly rank; the former will let us conclude that C does not come from the 2-torsion, so must either be nontrivial in Sha or correspond to a point of infinite order in the Mordell-Weil group of E(K). To finish it is sufficient to show that the image of T′ is not in general −ab. we will study the map e(−, T′)^{∗} ∘δ directly. δ maps T′ to the cocycle class of S −S^{σ}, where 2·S = T′. This means S is a point of strict order 4. Using properties of the Weil pairing from [8], e(−, T′)^{∗}(S − S^{σ}) = e_{2}(S − S^{σ}, T) = e_{2}(S − S^{σ}, 2 · S) = e_{2}(2 · S, S − S^{σ})^{−1} = e_{4}(S, S − S^{σ})^{−1} = e_{4}(S − S^{σ}, S) = e_{4}(S, S)/e_{4}(S^{σ}, S) = e_{4}(S, S^{σ}). Given an explicit formula for S, we can ascertain this cocycle. Since E ↪ ℙ^{1} × ℙ^{1} ⟶^{pr_{2}} ℙ^{1} is given by the divisor O + T′, such an S will be a ramification point of this double cover of ℙ^{1}. In other words S will be fixed by one of the two geometric involutions. We see that S is one of the four points that has fixed (x : y) coordinates under the second involution. We can find S from the formula for E by looking for x/y in terms of s and t and seeing where we get a double root. E : s^{2} (x^{2} − a^{2}y^{2}) + c(a − b) stxy − b t^{2}(x^{2} − a^{2}y^{2}) = 0. We rewrite this as E : x^{2} · α + xy · β + y^{2} · γ = 0, with α = s^{2} − b · t^{2}, β = s · t · c(a − b), and γ = −a^{2} · α. The solution

x/y = \frac{−β ± \sqrt{β^{2} − 4αγ}}{2α}

will have a double root when the discriminant β^{2} − 4α · γ is zero, i.e. when β^{2} = 4α·γ = −4a^{2}·α^{2}. This happens when β = ±2ai·α. Choosing β = 2ai·α and solving for s and t we find

s/t = \frac{c(a − b) ± \sqrt{(c(a − b))^{2} − 4(2ai)(−2abi)}}{2ai} =

\frac{c(a − b) ± \sqrt{(c(a − b))^{2} − 16a^{2}b}}{2ai}

Also we see that x/y = −β/(2α) = ai. So the point S is given by

(s : t), (x : y) = (\frac{c(a − b) + \sqrt{(c(a − b))^{2} − 16a^{2}b}}{2ai} : 1) , (ai : 1).

There are two obvious Galois elements that act on S; first, complex conjugation ^{−} which takes i to −i, and second τ with τ(\sqrt{(c(a − b))^{2} − 16a^{2}b}) = −\sqrt{(c(a − b))^{2} − 16a^{2}b}. The Galois conjugates of S are the four points of exact order 4 whose double is T′, namely S, S + T, S + T′, and S + T + T′. Only S + T and S + T + T′ pair with S non-trivially. We can act on S by λ_{T}

to find S + T, and we get S + T = (−^{\frac{c(a−b)±\sqrt{(c(a−b))^{2}−16a^{2}b}}{2ai}} : 1) , (−ai : 1)

Also, (ai : 1) is mapped to (−ai : 1) by λ_{T′}. Thus λ_{T +T′} must fix (ai : 1), so

S + T + T′ = (\frac{c(a − b) − \sqrt{(c(a − b))^{2} − 16a^{2}b}}{2ai} : 1) , (ai : 1)

and we find that τ(S) = S + T + T′ so τ ∈ ker(e^{∗}(S − S^{σ}, T)). This means that the image of T′ is (c(a − b))^{2} − 16a^{2}b. Whenever a is not equivalent mod squares to −b, (c(a − b))^{2} − 16a^{2}b, and b · ((c(a − b))^{2} − 16a^{2}b), and when (c(a−b))^{2}−16a^{2}b is not a square itself, C does not come from the two-torsion group. □

Corollary 6.1 There is no global section in the family C_{a,c}

Proof. The curve E_{a = −4, c = 3} has a conductor which is not divisible by 9 (so E is modular) and an L−series which does not vanish at s = 1 (both were computed by gp with the help of Adam Logan). We need only check that (c(a − b))^{2} − 16a^{2}b = (3 · 3)^{2} − 16 · 4^{2} · (−7) = 1873 is not a square and that is not congruent modulo sqaures to -4 or 7. It is prime. □

6.6 Finding x_{0}

By Claim 3.7, we are searching for a point x_{0} such that O(x_{0} + (x_{0} + T)) ≅ L_{1}. The cocycle x_{0} − x_{0}^{σ} will then take values in E[2](K_{s}) (see Claim 3.6). Moreover as in the proof of Claim 3.5 we will be able to extend a map φ_{x_{0}} : C ⟶ E to the entire multi-projective space ℙ_{K′}^{1} × ℙ_{K′}^{1} We can use geometry to locate the potential x_{0}’s as follows: We want (see page 49)

O(x_{0} + (x_{0} + T)) ≅ L_{1} ⟺ κ(x_{0}) = x_{0} + T = (M_{1}, M_{2}) · x_{0}.

We observe that κ_{1} fixes the first coordinate of points, whereas

M_{1} = ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) .

Thus the first coordinate of x_{0} is a fixed point under that matrix, i.e. (\sqrt{a} : 1) or (−\sqrt{a} : 1). We will denote these v_{0} and v_{1}, respectively, to maintain consistent notation with the other chapters. There are four possible choices for x_{0}, so it makes sense that we have two possible choices for the first coordinate; let’s choose the first coordinate of x_{0} to be v_{0} = ( \sqrt{a} 1 ) . Then the fact that x_{0} is in C means that the second coordinate of x_{0} must satisfy the equation

F_{C}(x_{0}) = F_{C}((\sqrt{a} : 1), (X : Y)) = 0,

or in other words

( a \sqrt{a} 1 ) ( \begin{matrix}bf_{3} & f_{2} & f_{1} \\ bg_{3} & g_{2} & g_{3} \\ abf_{1} & af_{2} & af_{3} \end{matrix} ) ( \begin{matrix}X^{2} \\ XY \\ Y^{2} \end{matrix} ) = 0.

We simplify to get the quadratic equation

bX^{2} + \frac{2af_{2} + \sqrt{a}g_{2}}{a(f_{1} + f_{3}) + \sqrt{a}g_{3}} XY + Y^{2} = 0.

For ease of notation, define

B = \frac{2af_{2} + \sqrt{a}g_{2}}{a(f_{1} + f_{3}) + \sqrt{a}g_{3}} and B′ = \frac{2af_{2} − \sqrt{a}g_{2}}{a(f_{1} + f_{3}) − \sqrt{a}g_{3}} .

Then we have the equation bX^{2} + BXY + Y^{2} = 0, which we solve for Y/X and get two solutions δ_{0} = (−B+\sqrt{B^{2} − 4b})/2 and δ_{1} = (−B−\sqrt{B^{2} − 4b})/2, with the properties

• δ_{0} + δ_{1} = −B

• δ_{0} · δ_{0} = b

• δ_{0} − δ_{1} = \sqrt{B^{2} − 4b}

• δ_{0} and δ_{1} are defined over the field K′ (\sqrt{a}, \sqrt{B^{2} − 4b}) .

Let Δ_{0} = ( \begin{matrix}1 \\ δ_{0} \end{matrix} ) and Δ_{1} = ( \begin{matrix}1 \\ δ_{1} \end{matrix} ); we now have x_{0} = (v_{0}, Δ_{0}) and x_{0}+T =

(v_{0}, Δ_{1}). The map φ from C to E is defined over K′(x_{0}) = K′(\sqrt{a}, \sqrt{B^{2} − 4b}).

Remark. Work for the duration of this Remark over the field K, where we consider f_{i}’s, g_{j}’s, a and b as elements of K. It doesn’t look as if the field of definition of φ is well-defined, since a and b are only defined “up to squares” in K. If we change a by u^{2} we see that K(\sqrt{a}) hasn’t changed, but it doesn’t look like \sqrt{B^{2} − 4b} behaves well with such a change. We will see that it does, however. M_{C}, the matrix of coefficients, has the property that it is fixed by the map

λ_{T} = (α, β) = (( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) , ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} )) .

That is, we have Ψ(α) · M_{C} · Φ(β) = M_{C}. Changing b by a square v^{2} must change M_{C} = M_{C}(a, b) to M_{C}(a, bv^{2}) such that it is fixed by the map λ_{T} (a, bv^{2}), i.e. such that Ψ(α) · M_{C}(a, bv^{2}) · Φ(β(a, bv^{2})) = M_{C}. Since

( \begin{matrix}v & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ bv^{2} & 0 \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & v \end{matrix} ) = ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} ) ,

the equality Ψ(α) · M_{C} · Φ(β) = M_{C} is equivalent to

Ψ (α) (M_{C} · Φ ( \begin{matrix}v & 0 \\ 0 & 1 \end{matrix} )) Φ ( \begin{matrix}0 & 1 \\ bv^{2} & 0 \end{matrix} ) = (M_{C} · Φ ( \begin{matrix}v & 0 \\ 0 & 1 \end{matrix} )) ,

and hence we have

M_{C}(a, bv^{2}) = M_{C} · Φ ( \begin{matrix}v & 0 \\ 0 & 1 \end{matrix} ) = M_{C} · ( \begin{matrix}v^{2} & 0 & 0 \\ 0 & v & 0 \\ 0 & 0 & 1 \end{matrix} )

= ( \begin{matrix}bv^{2}f_{3} & f_{2}v & f_{1} \\ bv^{2}g_{3} & g_{2}v & g_{3} \\ abv^{2}f_{1} & af_{2}v & af_{3} \end{matrix} ) .

We see that the change b ↦ bv^{2} transforms f_{1} ↦ f_{1}, f_{3} ↦ f_{3}, g_{3} ↦ g_{3}, f_{2} ↦ vf_{2}, and g_{2} ↦ vg_{2}. Thus B ↦ Bv, and the field K(\sqrt{a}, \sqrt{B^{2} − 4b}) is indeed well-defined.

Claim 6.4 The normal closure L of K′(x_{0}) is a degree eight extension of K′ with a dihedral Galois group; it is the field of definition of x_{0} and its conjugates.

Proof. Neither \sqrt{a} is in K′ nor \sqrt{B^{2} − 4b} is in K′(\sqrt{a}). Notice that the set of points x such that O(2 · x + T) ≅ L_{1} is a Galois invariant set, contains x_{0} by construction, and is in fact the orbit of x_{0} under the translation by two-torsion point maps, since for T′ ∈ E[2](K′_{s}),

O(2 · (x + T′) + T) ≅ L_{1} ⇔ O(2 · x + T) ≅ L_{1}.

In the Galois group G of the normal closure L of K′(x_{0}) there will be some element σ such that σ(\sqrt{a}) = −\sqrt{a}; Then σ brings x_{0} to x_{0} + T′, where T′ ∈ E[2](\overline{K}). We know T ≠ T′ because x_{0} + T has the same first coordinate as x_{0}, namely v_{0}. The second coordinate of x_{0} + T′ must satisfy the equation

(Y/X)^{2} + (σB) Y/X + b = 0,

so x_{0} + T′ is defined over the field K(−\sqrt{a}, \sqrt{(σB)^{2} − 4b}). Note that σB is B′ from above. We have exhibited that L is at most a degree 8 extension of K, as all of the possible conjugates of x_{0} are defined over the field K(\sqrt{a}, \sqrt{(B′)^{2} − 4b}, \sqrt{B^{2} − 4b}). Moreover, L is the splitting field of the following polynomial defined over K:

[Z^{2}+BZ+b]·[Z^{2}+B′Z+b] = Z^{4}+(B+B′)Z^{3}+(BB′+2b)Z^{2}+b(B+B′)Z+b^{2}.

The Galois group is a non-commutative group, since the field extension K(\sqrt{a}, \sqrt{(B′)^{2} − 4b}) is not normal. Thus it must be the dihedral group of eight elements. □

6.7 Standardizing the Model for E

Claim 6.5 We may assume that E is embedded in ℙ^{1} ×ℙ^{1} with the following properties:

• O = (1 : 0) , (1 : 0)

• T = (1 : 0) , (0 : 1)

• Let T′ and T′ + T be the other two-torsion points in E(\overline{K}). They both have first coordinate (0 : 1) .

• “Translation by T ” = λ_{T,E} on E is given by

(( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , ( \begin{matrix}0 & 1 \\ ab & 0 \end{matrix} )) .

Proof. The first representative matrix of λ_{T,E} can be put in the above form by Claim 3.10. The map proj_{2} ∘ f_{E} from E to the second ℙ^{1} is given by the divisor L_{2} ⊗ L_{1}^{−1} ⊗ O(T) (see the discussion at the beginning of Section 6.6), and by the properties of the det map (see Claim 3.6) we have

det(E, Q_{2}) = \frac{det(C, D_{2}) · det(E, T)}{det(C, D_{1})} = −a/b.

Since det takes values in K^{∗}/K^{∗2}, we may lift

det(E, Q_{2}) = −ab,

and we can compose f_{E} with a linear map to put λ_{T} in the above form. Since proj_{1} ∘ f_{E} is given by the divisor O + T, we know the points O and T lie on a (1,0)-form which is stable under the action of λ_{T} , and the same can be said about the points T′ and T′ + T. Then the first coordinate of O is either (1 : 0) or (0 : 1), and if it’s (0 : 1) we compose f_{E} with the map

(( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , I);

this permutes the two coordinates and commutes with λ_{T} .

Finally, to see that we can fix the second coordinates of O and T as above, note that we need only fix the second coordinate of O since we can act by λ_{T} to get T; to do so, we note that O is not fixed by λ_{T} , so since the first coordinate is fixed, the second coordinate isn’t. We can compose f_{E} by the linear map

(I, αI + β ( \begin{matrix}0 & 1 \\ ab & 0 \end{matrix} )) ,

and for appropriate α and β we can move the second coordinate of O to (1 : 0).□

6.8 Finding φ

Let L be the Galois closure of K′(x_{0}) as in Claim 6.4. Fix σ ∈ Gal(L/K′) such that σ(\sqrt{a}) = −\sqrt{a}. There will be four such elements, since the natural map

Gal(L/K′) ⟶ Gal(K′(\sqrt{a})/K′)

is surjective with kernel Gal(L/K′(\sqrt{a})), of size four. Let τ ∈ Gal(L/K′) be an element such that τ(\sqrt{a}) = \sqrt{a} and τ(\sqrt{B^{2} − 4b}) = −\sqrt{B^{2} − 4b}. Then τ(δ_{0}) = δ_{1}, τ(δ_{1}) = δ_{0}, τ(Δ_{0}) = δ_{1}, τ(Δ_{1}) = Δ_{0}, and τ^{2} = 1.

Claim 6.6 Let σ and τ be as above. Let φ be the linear automorphism carrying C to Jac(C) = E. Then if φ is represented by the automorphism (α^{−1}, β^{−1}, ϵ = 0) ∈ Aut(ℙ_{L}^{1} ×_{L} ℙ_{L}^{1}), we have

τα^{−1}α = ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) and βτβ^{−1} = ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} ) .

Moreover, σα^{−1}α and βσβ^{−1}, which correspond to translation by another two-torsion point on C, are both non-trivial elements of P GL_{2}(L) of order two and have trace zero.

Proof. By the choice of x_{0}, we know that the cocycle created by φ on C, namely ξ_{C}(γ) = φ^{−1}φ^{γ} for γ ∈ Gal(L/K′), will be the translation map by some two torsion point; which will be determined by seeing where x_{0} goes.

φ^{γ}(x_{0}) = (α^{−1}, β^{−1})^{γ} ∘ (v_{0}, Δ_{0})
= ( \sqrt{a} 1 ) (α^{−1})^{γ}, (β^{−1})^{γ} ( \begin{matrix}1 \\ δ_{0} \end{matrix} )
= (( γ^{−1}(\sqrt{a}) 1 ) α^{−1}, β^{−1} ( \begin{matrix}1 \\ γ^{−1}δ_{0} \end{matrix} ))^{γ} .

Let’s replace γ by τ above:

φ^{τ} (x_{0}) = (( \sqrt{a} 1 ) α^{−1}, β^{−1} ( \begin{matrix}1 \\ δ_{1} \end{matrix} ))^{τ} = (φ (x_{0} + T ))^{τ} = τ (T ) = T.

By Claim 3.4 we know “translation by torsion point matrices” (defined over \overline{K}′) have distinct eigenvalues. Since any such element of P GL_{2}(\overline{K}′) of order two can be conjugated over \overline{K}′ to a diagonal matrix with eigenvalues ±1, its trace is zero. Note that the trace function is invariant under conjugation and although it is not invariant under scaling, the vanishing of the trace function is well-defined for elements of P GL_{n}(K′).

Let’s see what the previous conditions impose on the cocycle on E: for τ as in Claim 6.6,

ταα^{−1} = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) & β^{−1}τβ = ( \begin{matrix}0 & 1 \\ ab & 0 \end{matrix} )

⟹ τα = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) · α & τβ = β · ( \begin{matrix}0 & 1 \\ ab & 0 \end{matrix} ) .

Since φ sends x_{0} to O, we have

v_{0} · α^{−1} = ( ∗ 0 ) .

The first coordinate of T′ is also known up to a scalar:

v_{1} · α^{−1} = ( 0 ∗′ ) .

We conclude that α is of the form

α = ( \begin{matrix}v_{0} \\ c · v_{1} \end{matrix} ) ,

for some c ∈ K′(x_{0}). A similar analysis of β reveals that

β = ( d · Δ_{0} Δ_{1} )

for some d ∈ K′(x_{0}).

The above conditions on α and β give

τα = ( \begin{matrix}τ(v_{0}) \\ τ(c · v_{1}) \end{matrix} ) = ( \begin{matrix}v_{0} \\ τ(c) · v_{1} \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) · α = ( \begin{matrix}v_{0} \\ −c · v_{1} \end{matrix} ) .

Then τ(c) = −c. We know α has coefficients in K′(x_{0}), and since c/\sqrt{B^{2} − 4b} is fixed by the non-trivial Galois element in Gal(K′(x_{0})/K′(\sqrt{a})),

c = c′ · \sqrt{B^{2} − 4b} = c′(δ_{0} − δ_{1})

for some c′ ∈ K′(\sqrt{a}) :

α = ( \begin{matrix}v_{0} \\ c′ · (δ_{0} − δ_{1}) · v_{1} \end{matrix} ) ,

Similarly for β, we write τβ =

( τ(d) · Δ_{1} Δ_{0} ) = β ( \begin{matrix}0 & 1 \\ ab & 0 \end{matrix} ) = ( ab · Δ_{1} d · Δ_{0} ) = ( \frac{ab}{d} · Δ_{1} Δ_{0} ) ,

from which we conclude that τ(d) = ab/d.

From Claim 6.6 we have the “Translation by T on C” matrices:

τα^{−1}α = ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) , βτβ^{−1} = ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} ) .

This doesn’t give us any new information about α, but for β we have:

β = ( d · Δ_{0} Δ_{1} ) = ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} ) · τβ = ( \begin{matrix}0 & 1 \\ b & 0 \end{matrix} ) · ( τ(d) · Δ_{1} Δ_{0} ) =

( \begin{matrix}τ(d) · δ_{1} & δ_{0} \\ τ(d) · b & b \end{matrix} ) = ( \begin{matrix}τ(d) · δ_{1}/δ_{0} & 1 \\ τ(d) · b/δ_{0} & b/δ_{0} \end{matrix} ) = ( (τ(d)δ_{1}/δ_{0}) · Δ_{0} Δ_{1} ) ,

since δ_{0} δ_{1} = b. We conclude that d/τ(d) = δ_{1}/δ_{0} = δ_{1}/τδ_{1}. That is, d/δ_{1} is fixed by τ, which is a nontrivial element of Gal(K′(x_{0})/K′(\sqrt{a})); therefore d = d′ · δ_{1}, for some d′ ∈ K′(\sqrt{a}). We have

α = ( \begin{matrix}v_{0} \\ c′(δ_{0} − δ_{1}) · v_{1} \end{matrix} ) , β = ( d′δ_{1} · Δ_{0} Δ_{1} ) .

We are left to find c′, d′ ∈ K(\sqrt{a}). Let F_{E} be (2,2)-form which gives E. Since the coefficients of F_{E} are rational, we will be able to deduce c′ and d′.

The linear operators α and β also act linearly on the nine dimensional vector space V of (2, 2)-forms in the following way:

γ ∘ M_{F_{C}} = Ψ(α) · M_{F_{C}} · Φ(β),

where Ψ and Φ are the “Sym^{2}” homomorphisms

Ψ, Φ : P GL_{2}(\overline{K}) ⟶ P GL_{3}(\overline{K}),

with Ψ ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}a^{2} & ab & b^{2} \\ 2ac & ad + bc & 2bd \\ c^{2} & cd & d^{2} \end{matrix} ) and Φ(α^{τ}) = Ψ(α)^{τ}. If M_{F_{C}} is

the matrix of coefficients of C, then M_{F_{E}} = Ψ(α)·M_{F_{C}} ·Φ(β) is the matrix of coefficients of E. The top left and right corners of M_{F_{E}} the product vanish, since O = ((1 : 0) , (1 : 0)) and T = ((1 : 0) , (0 : 1)) are on the curve E.

The middle entry in the matrix is also zero, since:

( 2ac′(δ_{0} − δ_{1}) 0 −2c′(δ_{0} − δ_{1}) ) · ( \begin{matrix}bf_{3} & f_{2} & f_{1} \\ bg_{3} & g_{2} & g_{3} \\ abf_{1} & af_{2} & af_{3} \end{matrix} ) · ( \begin{matrix}2d′δ_{1} \\ d′(δ_{1}^{2} + b) \\ 2d′bδ_{1} \end{matrix} ) =

(−2c′(δ_{0} − δ_{1}) · d′δ_{1}) ( −a 0 1 ) · ( \begin{matrix}bf_{3} & f_{2} & f_{1} \\ bg_{3} & g_{2} & g_{3} \\ abf_{1} & af_{2} & af_{3} \end{matrix} ) · ( \begin{matrix}2 \\ δ_{1} + δ_{0} \\ 2b \end{matrix} ) =

(−2c′(δ_{0} − δ_{1}) · d′δ_{1}) ( ab(f_{1} − f_{3}) 0 −a(f_{1} − f_{3}) ) · ( \begin{matrix}2 \\ δ_{1} + δ_{0} \\ 2b \end{matrix} ) = 0.

So far we have computed the following coefficients of M_{F_{E}}: M_{F_{E}}(1, 1) = M_{F_{E}}(1, 3) = M_{F_{E}}(2, 2) = 0. In order to compute the other entries, we will define two bilinear forms that will be useful. Since F_{C} is a (2,2)-form, if we fix one of the coordinates it becomes a quadratic function of the other coordinates. Then we can define the following:

B_{L}(v, v′, w) = F_{C}(v + v′, w) − F_{C}(v, w) − F_{C}(v′, w)
B_{R}(v, w, w′) = F_{C}(v, w + w′) − F_{C}(v, w) − F_{C}(v, w′)

Then:

M_{F_{E}}(1, 2) = B_{R}(v_{0}, d′δ_{1}Δ_{0}, Δ_{1}) = B_{R}(v_{0}, Δ_{0}, Δ_{1}) · d′δ_{1} M_{F_{E}}(2, 1) = B_{L}(v_{0}, c′(δ_{0} − δ_{1})v_{1}, d′δ_{1}Δ_{0}) = B_{L}(v_{0}, v_{1}, Δ_{0}) · c′(δ_{0} − δ_{1})(d′δ_{1})^{2} M_{F_{E}}(2, 3) = B_{L}(v_{0}, c′(δ_{0} − δ_{1})v_{1}, Δ_{1}) = B_{L}(v_{0}, v_{1}, Δ_{1}) · c′(δ_{0} − δ_{1}) M_{F_{E}}(3, 1) = F_{C}(c′(δ_{0} − δ_{1})v_{1}, d′δ_{1}Δ_{0}) = F_{C}(v_{1}, Δ_{0}) · (c′(δ_{0} − δ_{1})d′δ_{1})^{2} M_{F_{E}}(3, 2) = B_{R}(c′(δ_{0} − δ_{1})v_{1}, d′δ_{1}Δ_{0}, Δ_{1}) = B_{R}(v_{1}, Δ_{0}, Δ_{1}) · (c′(δ_{0} − δ_{1}))^{2}d′δ_{1} M_{F_{E}}(3, 3) = F_{C}(c′(δ_{0} − δ_{1})v_{1}, Δ_{1}) = F_{C}(v_{1}, Δ_{1}) · (c′(δ_{0} − δ_{1}))^{2}

We compute

B_{R}(v_{0}, Δ_{0}, Δ_{1}) = (a(f_{1} + f_{3}) + \sqrt{a}g_{3})(4b − B^{2}) B_{L}(v_{0}, v_{1}, Δ_{0}) = 2aδ_{0}(δ_{1} − δ_{0})(f_{1} − f_{3}) B_{L}(v_{0}, v_{1}, Δ_{1}) = −2aδ_{1}(δ_{1} − δ_{0})(f_{1} − f_{3}) F_{C}(v_{1}, Δ_{0}) = δ_{0}(B′ − B)(a(f_{1} + f_{3}) − \sqrt{a}g_{3}) B_{R}(v_{1}, Δ_{0}, Δ_{1}) = (4b − BB′)(a(f_{1} + f_{3}) − \sqrt{a}g_{3}) F_{C}(v_{1}, Δ_{1}) = δ_{1}(B′ − B)(a(f_{1} + f_{3}) − \sqrt{a}g_{3})

From the above we can determine c′ and d′ up to multiplication by an element of K′. Since E is defined over K′, the ratios of the above coefficients are in K′ : we can divide out by any non-zero coefficient to get rational coefficients. In general the above coefficients are non-zero. Then

M_{F_{E}}(3, 2)/M_{F_{E}}(3, 3) = \frac{B_{R}(v_{1}, Δ_{0}, Δ_{1}) · (c′(δ_{0} − δ_{1}))^{2}d′δ_{1}}{F_{C}(v_{1}, Δ_{1}) · (c′(δ_{0} − δ_{1}))^{2}} = d′′ ∈ K′,

so we have

d′ = d′′ \frac{F_{C}(v_{1}, Δ_{1})}{B_{R}(v_{1}, Δ_{0}, Δ_{1}) · δ_{1}} .

We simplify and expand this to get

d′ = d′′ \frac{(B′ − B)}{(4b − BB′)}

If σ is the nontrivial element of Gal(K′(\sqrt{a})/K′), then σ(d′) = −d′, since σ(B) = B′ and σ(B′) = B.

Next we find c′ up to an element of K′.

M_{F_{E}}(3, 1)/M_{F_{E}}(2, 1) = \frac{F_{C}(v_{1}, Δ_{0}) · (c′(δ_{0} − δ_{1})d′δ_{1})^{2}}{B_{L}(v_{0}, v_{1}, Δ_{0}) · c′(δ_{0} − δ_{1})(d′δ_{1})^{2}} = c′′ ∈ K′,

so

c′ = c′′ \frac{B_{L}(v_{0}, v_{1}, Δ_{0})}{F_{C}(v_{1}, Δ_{0}) · (δ_{0} − δ_{1})}.

This simplifies to

c′ = c′′ \frac{2a(f_{3} − f_{1})}{(B′ − B)(a(f_{1} + f_{3}) − \sqrt{a}g_{3})}, c′′ ∈ K′.

Now we have almost finished defining the map φ; the ambiguities remaining are the constants c′′ and d′′ in K′. As in the cases of ℙ^{2} and ℙ^{4}, we will have more rigidity if we fixed the other translation by two torsion matrices on the Jacobian curve. However, unlike those cases, we will not be completely rigid even when we do that- the reason is that the centralizer of the image of χ is no longer itself. Indeed since the image of χ lands in the product space P GL_{2}(\overline{K}) × P GL_{2}(\overline{K}), any element of \{I\} × Im(pr_{2} ∘ χ) is in the centralizer of χ. Altogether we have

α = ( \begin{matrix}v_{0} \\ c′′ \frac{a(f_{3}−f_{1})(δ_{0}−δ_{1})}{(B′−B)(a(f_{1}+f_{3})−\sqrt{a}g_{3})} · v_{1} \end{matrix} ) , β = ( d′′ \frac{(B′−B)}{(4b−BB′)}δ_{1} · Δ_{0} Δ_{1} ) .

Let S = (c′(δ_{0} − δ_{1}))^{2}(d′δ_{1})(a(f_{1} + f_{3}) − \sqrt{a}g_{3}). Then upon dividing M_{F_{E}}(3, 2) by S, we get

M_{F_{E}}(3, 2)/S = \frac{B_{R}(v_{1}, Δ_{0}, Δ_{1})}{(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} = 4b − BB′ ∈ K′.

Similarly, we get

M_{F_{E}}(3, 3)/S = \frac{F_{C}(v_{1}, Δ_{1})}{(d′δ_{1})(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} = \frac{B′ − B}{d′} = \frac{(4b − BB′)}{d′′},

M_{F_{E}}(3, 1)/S = \frac{d′δ_{1}F_{C}(v_{1}, Δ_{0})}{a(f_{1} + f_{3}) − \sqrt{a}g_{3}} = d′δ_{1}δ_{0}(B′ − B) = d′′b\frac{(B′ − B)^{2}}{(4b − BB′)},

M_{F_{E}}(1, 2)/S = \frac{B_{R}(v_{0}, Δ_{0}, Δ_{1})}{(c′(δ_{0} − δ_{1}))^{2}(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} =

\frac{(a(f_{1} + f_{3}) + \sqrt{a}g_{3})(4b − B^{2})}{(c′(δ_{0} − δ_{1}))^{2}(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} = \frac{−(a(f_{1} + f_{3})^{2} − g_{3}^{2})(B′ − B)^{2}}{(c′′(f_{3} − f_{1}))^{2}} ,

M_{F_{E}}(2, 1)/S = \frac{B_{L}(v_{0}, v_{1}, Δ_{0})d′δ_{1}}{c′(δ_{0} − δ_{1})(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} = \frac{−2aδ_{0}(f_{1} − f_{3})d′δ_{1}}{c′(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} =

\frac{−2aδ_{0}(f_{1} − f_{3})d′′ \frac{(B′−B)}{(4b−BB′)}δ_{1}}{c′′ \frac{2a(f_{3}−f_{1})}{(B′−B)(a(f_{1}+f_{3})−\sqrt{a}g_{3})}(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} = \frac{bd′′(B′ − B)^{2}}{c′′(4b − BB′)},

M_{F_{E}}(2, 3)/S = \frac{B_{L}(v_{0}, v_{1}, Δ_{1})}{c′d′(δ_{0} − δ_{1})δ_{1}(a(f_{1} + f_{3}) − \sqrt{a}g_{3})} = \frac{2a(f_{1} − f_{3})}{c′d′(a(f_{1} + f_{3}) − \sqrt{a}g_{3})}

= \frac{−(4b − BB′)}{c′′d′′}

We have now shown:

Claim 6.7 The resulting matrix for E is given by:

M_{F_{E}} = ( \begin{matrix}0 & \frac{−(a(f_{1}+f_{3})^{2}−g_{3}^{2})(B′−B)^{2}}{(c′′(f_{3}−f_{1}))^{2}} & 0 \\ \frac{bd′′(B′−B)^{2}}{c′′(4b−BB′)} & 0 & \frac{−(4b−BB′)}{c′′d′′} \\ d′′b\frac{(B′−B)^{2}}{(4b−BB′)} & 4b − BB′ & \frac{(4b−BB′)}{d′′} \end{matrix} )

Now our earlier observation that τdd = ab tells us (d′′)^{2} = 1.

Note that if we multiplied the (2,2)-form F_{C} by a constant, we would not have a different matrix for E; this is because each coefficient has “weight zero” in terms of the coefficients of F. A simple calculation shows that

B′ − B = 2\sqrt{a}\frac{2f_{2}g_{3} − f_{1}g_{2} − f_{3}g_{2}}{a(f_{1} + f_{3})^{2} − g_{3}^{2}}

and

4b − BB′ = \frac{4b(a(f_{1} + f_{3})^{2} − g_{3}^{2}) − 4af_{2}^{2} − g_{2}^{2}}{a(f_{1} + f_{3})^{2} − g_{3}^{2}} .

To finish the proof of Theorem 6.1 let c′′ = d′′ = 1.

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