ON SOME ALGEBRAS ASSOCIATED TO GENUS ONE CURVES

TOM FISHER

Abstract. Haile, Han and Kuo have studied certain non-commutative algebras associated to a binary quartic or ternary cubic form. We extend their construction to pairs of quadratic forms in four variables, and conjecture a further generalisation to genus one curves of arbitrary degree. These constructions give an explicit realisation of an isomorphism relating the Weil-Châtelet and Brauer groups of an elliptic curve.

1. Introduction

Let C be a smooth curve of genus one, written as either a double cover of ℙ^{1} (case n = 2), or as a plane cubic in ℙ^{2} (case n = 3), or as an intersection of two quadrics in ℙ^{3} (case n = 4). We write C = C_{f} where f is the binary quartic form, ternary cubic form, or pair of quadratic forms defining the curve. In this paper we investigate a certain non-commutative algebra A_{f} determined by f.

The algebra A_{f} was defined in the case n = 2 by Haile and Han [10], and in the case n = 3 by Kuo [12]. We simplify some of their proofs, and extend to the case n = 4. We also conjecture a generalisation to genus one curves of arbitrary degree n. The following theorem was already established in [10, 12] in the cases n = 2, 3. We work throughout over a field K of characteristic not 2 or 3.

Theorem 1.1. If n ∈ {2, 3, 4} then A_{f} is an Azumaya algebra, free of rank n^{2} over its centre. Moreover the centre of A_{f} is isomorphic to the co-ordinate ring of E \ {0_{E}} where E is the Jacobian elliptic curve of C_{f}.

Let E/K be an elliptic curve. A standard argument (see Section 6.1) shows that the Weil-Châtelet group of E is canonically isomorphic to the quotient of Brauer groups Br(E)/ Br(K). For our purposes it is more convenient to write this isomorphism as

(1)  H^{1}(K, E) ≅ ker (Br(E) ⟶^{ev_{0}} Br(K)) .

where ev_{0} is the map that evaluates a Brauer class at 0 ∈ E(K). The algebras we study explicitly realise this isomorphism.

Theorem 1.2. If n ∈ {2, 3, 4} then the isomorphism (1) sends the class of C_{f} to the class of A_{f}.

Date: 25th July 2017.

The following two corollaries were proved in [10, 12] in the cases n = 2, 3.

Corollary 1.3. Let n ∈ {2, 3, 4}. The genus one curve C_{f} has a K-rational point if and only if the Azumaya algebra A_{f} splits over K.

Proof. This is the statement that the class of C_{f} in H^{1}(K, E) is trivial if and only if the class of A_{f} in Br(E) is trivial. □

For 0 ≠ P ∈ E(K) we write A_{f,P} for the specialisation of A_{f} at P. This is a central simple algebra over K of dimension n^{2}.

Corollary 1.4. Let n ∈ {2, 3, 4}. The map E(K) → Br(K) that sends P to the class of A_{f,P} is a group homomorphism.

Proof. By Theorem 1.2 the Tate pairing E(K) × H^{1}(K, E) → Br(K) is given by (P, [C_{f}]) ↦ [A_{f,P}]. This corollary is the statement that the Tate pairing is linear in the first argument. □

The algebras A_{f} are interesting for several reasons. They have been used to study the relative Brauer groups of curves (see [5, 8, 11, 13]) and to compute the Cassels-Tate pairing (see [9]). We hope they might also be used to construct explicit Brauer classes on surfaces with an elliptic fibration. This could have important arithmetic applications, extending for example [17].

In Sections 2 and 3 we define the algebras A_{f} and describe their centres. In Section 4 we show that these constructions behave well under changes of co-ordinates. The proofs of Theorems 1.1 and 1.2 are given in Sections 5 and 6.

The hyperplane section H on C_{f} is a K-rational divisor of degree n. Let P ∈ E(K) where E is the Jacobian of C_{f}. In Section 7 we explain how finding an isomorphism A_{f,P} ≅ Mat_{n}(K) enables us to find a K-rational divisor H′ on C_{f} such that [H − H′] ↦ P under the isomorphism Pic^{0}(C_{f}) ≅ E. In the cases n = 2, 3 our construction involves some of the representations studied in [3].

Nearly all our proofs are computational in nature, and for this we rely on the support in Magma [4] for finitely presented algebras. We have prepared a Magma file checking all our calculations, and this is available online. It would of course be interesting to find more conceptual proofs of Theorems 1.1 and 1.2.

2. The algebra A_{f}

In this section we define the algebras A_{f} for n = 2, 3, 4, and suggest how the definition might be generalised to genus one curves of arbitrary degree. The prototype for these constructions is the Clifford algebra of a quadratic form. We therefore start by recalling the latter, which will in any case be needed for our treatment of the case n = 2. We write [x, y] for the commutator xy − yx.

2.1. Clifford algebras. Let Q ∈ K[x_{1}, …, x_{n}] be a quadratic form. The Clifford algebra of Q is the associative K-algebra A generated by u_{1}, …, u_{n} subject to the relations deriving from the formal identity in α_{1}, …, α_{n},

(α_{1}u_{1} + … + α_{n}u_{n})^{2} = Q(α_{1}, …, α_{n}).

The involution u_{i} ↦ −u_{i} resolves A into eigenspaces A = A_{+} ⊕ A_{−}. By diagonalising Q, it may be shown that A and A_{+} are K-algebras of dimensions 2^{n} and 2^{n−1}. Moreover, rescaling Q does not change the isomorphism class of A_{+}.

In the case n = 3 we let

η = u_{1}u_{2}u_{3} − u_{3}u_{2}u_{1} = u_{2}u_{3}u_{1} − u_{1}u_{3}u_{2} = u_{3}u_{1}u_{2} − u_{2}u_{1}u_{3}.

Then η belongs to the centre Z(A), and η^{2} = disc Q, where if Q(x) = x^{T} Mx then disc Q = −4 det M. Moreover, if disc Q ≠ 0 then A_{+} is a quaternion algebra and A = A_{+} ⊗ K[η]. Although not needed below, it is interesting to remark that the well known map

H^{1}(K, PGL_{2}) → Br(K)

is realised by sending the smooth conic {Q = 0} ⊂ ℙ^{2} (which as a twist of ℙ^{1} corresponds to a class in H^{1}(K, PGL_{2})) to the class of A_{+}.

2.2. Binary quartics. Let f ∈ K[x, z] be a binary quartic, say

f(x, z) = ax^{4} + bx^{3}z + cx^{2}z^{2} + dxz^{3} + ez^{4}.

Haile and Han [10] define the algebra A_{f} to be the associative K-algebra generated by r, s, t subject to the relations deriving from the formal identity in α and β,

(α^{2}r + αβs + β^{2}t)^{2} = f(α, β).

Thus A_{f} = K{r, s, t}/I where I is the ideal generated by the elements

r^{2} − a,
rs + sr − b,
rt + tr + s^{2} − c,
st + ts − d,
t^{2} − e.

We have [r, s^{2}] = [r, rs + sr] = [r, b] = 0, and likewise [s^{2}, t] = 0. Therefore ξ = s^{2} − c belongs to the centre Z(A_{f}). By working over the polynomial ring K[ξ], instead of the field K, we may describe A_{f} as the Clifford algebra of the quadratic form

Q_{ξ}(x, y, z) = ax^{2} + bxy + cy^{2} + dyz + ez^{2} + ξ(y^{2} − xz).

This quadratic form naturally arises as follows. Let C ⊂ ℙ^{3} be the image of the curve Y^{2} = f(X, Z) embedded via (x_{1} : x_{2} : x_{3} : x_{4}) = (X^{2} : XZ : Z^{2} : Y). Then C is defined by a pencil of quadrics with generic member x_{4}^{2} = Q_{ξ}(x_{1}, x_{2}, x_{3}).

2.3. Ternary cubics. Let f ∈ K[x, y, z] be a ternary cubic, say

f(x, y, z) = ax^{3} + by^{3} + cz^{3} + a_{2}x^{2}y + a_{3}x^{2}z
+ b_{1}xy^{2} + b_{3}y^{2}z + c_{1}xz^{2} + c_{2}yz^{2} + mxyz.

In the special case c = 1 and a_{3} = b_{3} = c_{1} = c_{2} = 0, Kuo [12] defines the algebra A_{f} to be the associative K-algebra generated by x and y subject to the relations deriving from the formal identity in α and β,

f(α, β, αx + βy) = 0.

We make the same definition for any ternary cubic f with c ≠ 0. Thus A_{f} = K{x, y}/I where I is the ideal generated by the elements

cx^{3} + c_{1}x^{2} + a_{3}x + a,
c(x^{2}y + xyx + yx^{2}) + c_{1}(xy + yx) + c_{2}x^{2} + mx + a_{3}y + a_{2},
c(xy^{2} + yxy + y^{2}x) + c_{2}(xy + yx) + c_{1}y^{2} + my + b_{3}x + b_{1},
cy^{3} + c_{2}y^{2} + b_{3}y + b.

2.4. Quadric intersections. Let f = (f_{1}, f_{2}) be a pair of quadratic forms in four variables, say x_{1}, …, x_{4}. Assuming C_{f} = {f_{1} = f_{2} = 0} ⊂ ℙ^{3} does meet the line {x_{3} = x_{4} = 0}, we define the algebra A_{f} to be the associative K-algebra generated by p, q, r, s subject to the relations deriving from the formal identities in α and β,

(2)  f_{i}(αp + βr, αq + βs, α, β) = 0, i = 1, 2
(3)  [αp + βr, αq + βs] = 0.

Explicitly if f_{1} = ∑_{i≤j} a_{ij}x_{i}x_{j} and f_{2} = ∑_{i≤j} b_{ij}x_{i}x_{j} then A_{f} = K{p, q, r, s}/I where I is the ideal generated by the elements

a_{11}p^{2} + a_{12}pq + a_{22}q^{2} + a_{13}p + a_{23}q + a_{33},

a_{11}(pr + rp) + a_{12}(ps + rq) + a_{22}(qs + sq) + a_{14}p + a_{24}q + a_{13}r + a_{23}s + a_{34},

a_{11}r^{2} + a_{12}rs + a_{22}s^{2} + a_{14}r + a_{24}s + a_{44},

b_{11}p^{2} + b_{12}pq + b_{22}q^{2} + b_{13}p + b_{23}q + b_{33},

b_{11}(pr + rp) + b_{12}(ps + rq) + b_{22}(qs + sq) + b_{14}p + b_{24}q + b_{13}r + b_{23}s + b_{34},

b_{11}r^{2} + b_{12}rs + b_{22}s^{2} + b_{14}r + b_{24}s + b_{44},

pq − qp,

ps + rq − qr − sp,

rs − sr.

One motivation for including the commutator relation (3) is that without it, the relations (2) would be ambiguous.

2.5. Genus one curves of higher degree. Let C be a smooth curve of genus one. If D is a K-rational divisor on C of degree n ≥ 3 then the complete linear system |D| defines an embedding C → ℙ^{n−1}. We identify C with its image, which is a curve of degree n. If n = 3 then C is a plane cubic, whereas if n ≥ 4 then the homogeneous ideal of C is generated by quadrics.

Let A be the associative K-algebra generated by u_{1}, u_{2}, …, u_{n−2}, v_{1}, v_{2}, …, v_{n−2}, subject to the relations deriving from the formal identities in α and β,

f(αu_{1} + βv_{1}, αu_{2} + βv_{2}, …, αu_{n−2} + βv_{n−2}, α, β) = 0 for all f ∈ I(C),
[αu_{i} + βv_{i}, αu_{j} + βv_{j}] = 0 for all 1 ≤ i, j ≤ n − 2.

This definition may be thought of as writing down the conditions for C to contain a line. The fact that C does not contain a line then tells us that there are no non-zero K-algebra homomorphisms A → K.

We conjecture that the analogues of Theorems 1.1 and 1.2 hold for these algebras. In support of this conjecture, we have checked that Theorem 1.1 holds in some numerical examples with n = 5.

3. The centre of A_{f}

In this section we exhibit some elements ξ and η in the centre of A_{f}. In each case ξ and η generate the centre, and satisfy a relation in the form of a Weierstrass equation for the Jacobian elliptic curve.

3.1. Binary quartics. Let C_{f} be a smooth curve of genus one defined as a double cover of ℙ^{1} by y^{2} = f(x, z), where f is a binary quartic. It already follows from the results in Sections 2.1 and 2.2 that the centre of A_{f} is generated by ξ = s^{2} − c and η = rst − tsr. Alternatively, this was proved by Haile and Han [10] for quartics with b = 0, and the general case follows by making a change of co-ordinates (see Section 4). The elements ξ and η satisfy η^{2} = F (ξ) where

(4)  F (x) = x^{3} + cx^{2} − (4ae − bd)x − 4ace + b^{2}e + ad^{2}.

This is a Weierstrass equation for the Jacobian of C_{f}.

There is a derivation D : A_{f} → A_{f} defined on the generators r, s, t by Dr = [s, r], Ds = [t, r] and Dt = 0. To see this is well defined, we checked that the derivation acts on the ideal of relations defining A_{f}. It is easy to see that D must act on the centre of A_{f}. We find that Dξ = 2η and Dη = 3ξ^{2} + 2cξ − (4ae − bd).

3.2. Ternary cubics. Let C_{f} ⊂ ℙ^{2} be a smooth curve of genus one defined by a ternary cubic f. With notation as in Section 2.3, the centre of A_{f} contains

ξ = c^{2}(xy)^{2} − (cy^{2} + c_{2}y + b_{3})(cx^{2} + c_{1}x + a_{3}) + (cm − c_{1}c_{2})xy + a_{3}b_{3}.

There is a derivation D : A_{f} → A_{f} defined on the generators x, y by Dx = c[xy, x] and Dy = c[y, yx]. Let a′_{1}, a′_{2}, a′_{3}, a′_{4}, a′_{6} ∈ ℤ[a, b, c, …, m] be the coefficients of a Weierstrass equation for the Jacobian of C_{f}, as specified in [7, Section 2], i.e. a′_{1} = m, a′_{2} = −(a_{2}c_{2} + a_{3}b_{3} + b_{1}c_{1}), a′_{3} = 9abc − (ab_{3}c_{2} + ba_{3}c_{1} + ca_{2}b_{1}) − (a_{2}b_{3}c_{1} + a_{3}b_{1}c_{2}), a′_{4} = . . .. (These formulae were originally given in [2].) Then η = \frac{1}{2}(Dξ − a′_{1}ξ − a′_{3}) is also in the centre of A_{f}, and these elements satisfy

η^{2} + a′_{1}ξη + a′_{3}η = ξ^{3} + a′_{2}ξ^{2} + a′_{4}ξ + a′_{6}.

In fact ξ and η generate the centre of A_{f}. This was proved by Kuo [12] in the case c = 1 and a_{3} = b_{3} = c_{1} = c_{2} = 0. The general case follows by making a change of co-ordinates (see Section 4).

3.3. Quadric intersections. Let C_{f} ⊂ ℙ^{3} be a smooth curve of genus one defined by a pair of quadratic forms f = (f_{1}, f_{2}). Let a_{1}, …, a_{10} and b_{1}, …, b_{10} be the coefficients of f_{1} and f_{2}, where we take the monomials in the order

x_{1}^{2}, x_{1}x_{2}, x_{1}x_{3}, x_{1}x_{4}, x_{2}^{2}, x_{2}x_{3}, x_{2}x_{4}, x_{3}^{2}, x_{3}x_{4}, x_{4}^{2}.

Let d_{ij} = a_{i}b_{j} − a_{j}b_{i}. With notation as in Section 2.4 we put

p_{i} = d_{1i}p + d_{2i}q + d_{3i},  r_{i} = d_{1i}r + d_{2i}s + d_{4i},
q_{i} = d_{2i}p + d_{5i}q + d_{6i},  s_{i} = d_{2i}r + d_{5i}s + d_{7i}

and t = qr − ps = rq − sp. Then

ξ = (p_{5}s)^{2} + (s_{1}p)^{2}

+ (d_{56}p_{4} + d_{29}p_{5} + d_{37}p_{5} − d_{27}p_{6})s − d_{56}(d_{13}r + d_{23}s − d_{17}q + d_{12}t − d_{19})s + (d_{14}s_{6} + d_{29}s_{1} − d_{37}s_{1} − d_{23}s_{4})p − d_{14}(d_{27}p + d_{57}q + d_{35}r − d_{25}t − d_{59})p

belongs to the centre of A_{f}. We give a slightly simpler expression for ξ in Section 4.3, but this alternative expression is only valid when t is invertible.

There is a derivation D : A_{f} → A_{f} defined on the generators p, q, r, s by Dp = \frac{1}{2}[p, ε], Dq = \frac{1}{2}[q, ε] and Dr = Ds = 0 where

ε = d_{12}(pr + rp) + d_{15}(ps + qr + sp + rq) + d_{25}(qs + sq).

Then η = \frac{1}{2}Dξ is also in the centre of A_{f}. We show in Section 5 that ξ and η generate the centre, and that they satisfy a Weierstrass equation for the Jacobian of C_{f}.

4. Changes of co-ordinates

In this section we show that making a change of coordinates does not change the isomorphism class of the algebra A_{f}. We also describe the effect this has on the central elements ξ and η, and on the derivation D.

Let G_{2}(K) = K^{×} × GL_{2}(K) act on the space of binary quartics via

(λ, M) : f(x, z) ↦ λ^{2}f(m_{11}x + m_{21}z, m_{12}x + m_{22}z).

Let G_{3}(K) = K^{×} × GL_{3}(K) act on the space of ternary cubics via

(λ, M) : f(x, y, z) ↦ λf(m_{11}x + m_{21}y + m_{31}z, …, m_{13}x + m_{23}y + m_{33}z).

Let G_{4}(K) = GL_{2}(K) × GL_{4}(K) act on the space of quadric intersections via

(Λ, I_{4}) : (f_{1}, f_{2}) ↦ (λ_{11}f_{1} + λ_{12}f_{2}, λ_{21}f_{1} + λ_{22}f_{2}),
(I_{2}, M) : (f_{1}, f_{2}) ↦ (f_{1}(∑_{i=1}^{4} m_{i1}x_{i}, …), f_{2}(∑_{i=1}^{4} m_{i1}x_{i}, …)).

We write det(λ, M) = λ det M in the cases n = 2, 3, and det(Λ, M) = det Λ det M in the case n = 4. A genus one model is a binary quartic, ternary cubic, or pair of quadratic forms, according as n = 2, 3 or 4.

Theorem 4.1. Let f and f′ be genus one models of degree n ∈ {2, 3, 4}. In the case n = 3 we suppose that f(0, 0, 1) ≠ 0 and f′(0, 0, 1) ≠ 0. In the case n = 4 we suppose that C_{f} and C_{f′} do not meet the line {x_{3} = x_{4} = 0}. If f′ = γf for some γ ∈ G_{n}(K) then there is an isomorphism ψ : A_{f′} → A_{f} with

(5)  ξ ↦ (det γ)^{2}ξ + ρ
(6)  η ↦ (det γ)^{3}η + σξ + τ

for some ρ, σ, τ ∈ K, with σ = τ = 0 if n ∈ {2, 4}. Moreover there exists κ ∈ A_{f} such that

(7)  ψD(x) = (det γ)Dψ(x) + [κ, ψ(x)]

for all x ∈ A_{f′}.

Proof: We prove the theorem for γ running over a set of generators for G_{n}(K). The set of generators will be large enough that the extra conditions in the cases n = 3, 4 (avoiding a certain point or line) do not require special consideration.

Writing η in terms of the Dξ we see that (6) is a formal consequence of (5) and (7). It therefore suffices to check (5) and (7). We may paraphrase (7) as saying that ψDψ^{−1} and (det γ)D are equal up to inner derivations. In particular we only need to check this statement for x running over a set of generators for A_{f}.

We now split into the cases n = 2, 3, 4.

4.1. Binary quartics. Let γ = (λ, M). There is an isomorphism ψ : A_{f′} → A_{f} given by

r ↦ λ(m_{11}^{2}r + m_{11}m_{12}s + m_{12}^{2}t),
s ↦ λ(2m_{11}m_{21}r + (m_{11}m_{22} + m_{12}m_{21})s + 2m_{12}m_{22}t),
t ↦ λ(m_{21}^{2}r + m_{21}m_{22}s + m_{22}^{2}t).

We find that (5) and (7) are satisfied with

ρ = −λ^{2}(2m_{11}^{2}m_{21}^{2}a + m_{11}m_{21}(m_{11}m_{22} + m_{12}m_{21})b

+ 2m_{11}m_{12}m_{21}m_{22}c + m_{12}m_{22}(m_{11}m_{22} + m_{12}m_{21})d + 2m_{12}^{2}m_{22}^{2}e)

and κ = λ(m_{11}m_{21}r + m_{12}m_{21}s + m_{12}m_{22}t).

4.2. Ternary cubics. The result is clear for γ = (λ, I_{3}). We take γ = (1, M). If this change of co-ordinates fixes the point (0 : 0 : 1), equivalently m_{31} = m_{32} = 0, then there is an isomorphism ψ : A_{f′} → A_{f} given by

x ↦ m_{33}^{−1}(m_{11}x + m_{12}y − m_{13}),
y ↦ m_{33}^{−1}(m_{21}x + m_{22}y − m_{23}).

We checked (5) by a generic calculation (leading to a lengthy expression for ρ which we do not record here), and find that (7) is satisfied with

κ = cm_{33}(m_{23}(m_{11}x + m_{12}y) − m_{13}(m_{21}x + m_{22}y)).

It remains to consider a transformation that moves the point (0 : 0 : 1). Let f′(x, y, z) = f(z, x, y). By hypothesis a = f′(0, 0, 1) ≠ 0. From the first relation defining A_{f} it follows that x is invertible, i.e. x^{−1} = −(cx^{2} + c_{1}x + a_{3})/a. There is an isomorphism ψ : A_{f′} → A_{f} given by x ↦ −yx^{−1} and y ↦ x^{−1}. We find that (5) and (7) are satisfied with ρ = 0 and κ = cyx + c_{1}y.

4.3. Quadric intersections. The result for γ = (Λ, I_{4}) follows easily from the fact our expressions for ε and ξ are linear and quadratic in the d_{ij}. We take γ = (I_{2}, M). If

M = (\begin{matrix}U^{−1} & 0 \\ 0 & I_{2} \end{matrix}) or (\begin{matrix}I_{2} & 0 \\ V & I_{2} \end{matrix})

then an isomorphism ψ : A_{f′} → A_{f} is given by

\{ \begin{matrix}p & ↦ u_{11}p + u_{21}q \\ q & ↦ u_{12}p + u_{22}q \\ r & ↦ u_{11}r + u_{21}s \\ s & ↦ u_{12}r + u_{22}s \end{matrix}\} or \{ \begin{matrix}p & ↦ p − v_{11} \\ q & ↦ q − v_{12} \\ r & ↦ r − v_{21} \\ s & ↦ s − v_{22} \end{matrix}\} .

We checked (5) by a generic calculation, and find that (7) is satisfied with κ = 0 or κ = v_{11}(d_{12}r + d_{15}s) + v_{12}(d_{15}r + d_{25}s).

It remains to consider a transformation that moves the line {x_{3} = x_{4} = 0}. Let f′_{i}(x_{1}, x_{2}, x_{3}, x_{4}) = f_{i}(x_{3}, x_{4}, x_{1}, x_{2}) for i = 1, 2. By hypothesis C_{f} does not meet the line {x_{1} = x_{2} = 0} and so t = qr − ps is invertible, i.e.

t^{−1} = −(d_{89}(s_{1}r + s_{4}) + d_{8,10}(r_{5}q + r_{6} + p_{5}s + p_{7} + d_{29}) + d_{9,10}(q_{1}p + q_{3}))/Δ

where Δ = d_{8,10}^{2} − d_{89}d_{9,10}. There is an isomorphism ψ : A_{f′} → A_{f} given by p ↦ −st^{−1}, q ↦ qt^{−1}, r ↦ rt^{−1}, s ↦ −pt^{−1} and t ↦ t^{−1}. Under our assumption that t is invertible, we have ξ = ξ_{1} + c_{1} where

ξ_{1} =(d_{15}^{2} − d_{12}d_{25})t^{2} + (d_{15}d_{37} − d_{12}d_{67} − d_{15}d_{46} − d_{25}d_{34})t
+ (d_{37}d_{8,10} − d_{36}d_{9,10} + d_{46}d_{8,10} − d_{47}d_{89})t^{−1} + (d_{8,10}^{2} − d_{89}d_{9,10})t^{−2},

and c_{1} ∈ K is a constant (depending on f). Working with ξ_{1} in place of ξ makes it easy to check (5). Finally (7) is satisfied with

κ = λ(p(s_{1}r + s_{4}) + q_{10}) + μ(r(q_{1}p + q_{3}) + s_{8}) + r(d_{12}p + d_{15}q) − \frac{1}{2}(d_{23}r + d_{26}s)

for certain λ, μ ∈ K. In fact we may take λ = (2d_{48}d_{8,10} − d_{38}d_{9,10} − d_{89}d_{49} + d_{89}d_{3,10})/(2Δ) and μ = (2d_{3,10}d_{8,10} − d_{4,10}d_{89} − d_{9,10}d_{39} + d_{9,10}d_{48})/(2Δ). □

5. Proof of Theorem 1.1

In this section we prove the following refined version of Theorem 1.1. The first two parts of the theorem show that A_{f} is an Azumaya algebra.

Theorem 5.1. Let C_{f} be a smooth curve of genus one, defined by a genus one model f of degree n ∈ {2, 3, 4}. Then

(i) The algebra A = A_{f} is free of rank n^{2} over its centre Z (say).

(ii) The map A ⊗_{Z} A^{op} → End_{Z}(A); a ⊗ b ↦ (x ↦ axb) is an isomorphism. (iii) The centre Z is generated by the elements ξ and η specified in Section 3,

subject only to these satisfying a Weierstrass equation. (iv) The Weierstrass equation in (iii) defines the Jacobian of C_{f}.

For the proof of the first three parts of Theorem 5.1 we are free to extend our field K. However working over an algebraically closed field, it is well known that smooth curves of genus one C_{f} and C_{f′} are isomorphic as curves (i.e. have the same j-invariant) if and only if the genus one models f and f′ are in the same orbit for the group action defined at the start of Section 4. We now split into the cases n = 2, 3, 4 and verify the theorem by direct computation for a family of curves covering the j-line. The general case then follows by Theorem 4.1.

The generic calculations in Sections 3.1 and 3.2 already prove Theorem 5.1(iv) in the cases n = 2, 3. The case n = 4 will be treated in Section 5.3.

5.1. Binary quartics. Let K[x_{0}, y_{0}] = K[x, y]/(F) where

F (x, y) = y^{2} − (x^{3} + a_{2}x^{2} + a_{4}x + a_{6}).

We consider the binary quartic f(x, z) = a_{6}x^{4} + a_{4}x^{3}z + a_{2}x^{2}z^{2} + xz^{3}. Specialising the formulae in Section 3.1 we see that ξ, η ∈ A_{f} satisfy F (ξ, η) = 0.

Lemma 5.2. There is an isomorphism of K-algebras θ : A_{f} → Mat_{2}(K[x_{0}, y_{0}]) given by

r ↦ (\begin{matrix}−y_{0} & x_{0}^{2} + a_{2}x_{0} + a_{4} \\ −x_{0} & y_{0} \end{matrix}) , s ↦ (\begin{matrix}0 & x_{0} + a_{2} \\ 1 & 0 \end{matrix}) , t ↦ (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) .

Moreover we have θ(ξ) = x_{0}I_{2} and θ(η) = y_{0}I_{2}.

Proof: We write E_{ij} for the 2 by 2 matrix with a 1 in the (i, j) position and zeros elsewhere. Then Mat_{2}(K[x_{0}, y_{0}]) is generated as a K[x_{0}, y_{0}]-algebra by E_{12} and E_{21} subject to the relations E_{12}^{2} = E_{21}^{2} = 0 and E_{12}E_{21} + E_{21}E_{12} = 1. We define a K-algebra homomorphism ϕ : Mat_{2}(K[x_{0}, y_{0}]) → A_{f} via

x_{0} ↦ ξ, y_{0} ↦ η, E_{12} ↦ t, E_{21} ↦ s − s^{2}t.

We checked by direct calculation that θ and ϕ are well defined (i.e. they send all relations to zero), and that they are inverse to each other. □

5.2. Ternary cubics. Let K[x_{0}, y_{0}] = K[x, y]/(F) where

F (x, y) = y^{2} + a_{1}xy + a_{3}y − (x^{3} + a_{2}x^{2} + a_{4}x + a_{6}).

We consider the ternary cubic f(x, y, z) = x^{3}F (z/x, y/x). Specialising the formulae in Section 3.2 we see that ξ, η ∈ A_{f} satisfy F (ξ, η) = 0.

Lemma 5.3. There is an isomorphism of K-algebras θ : A_{f} → Mat_{3}(K[x_{0}, y_{0}]) given by

x ↦ (\begin{matrix}−x_{0} − a_{2} & −1 & 0 \\ x_{0}^{2} + a_{2}x_{0} + a_{4} & 0 & y_{0} \\ y_{0} + a_{1}x_{0} + a_{3} & 0 & x_{0} \end{matrix}) , y ↦ (\begin{matrix}0 & 0 & 0 \\ −a_{1} & 0 & −1 \\ 1 & 0 & 0 \end{matrix}) .

Moreover we have θ(ξ) = x_{0}I_{3} and θ(η) = y_{0}I_{3}.

Proof: We write E_{ij} for the 3 by 3 matrix with a 1 in the (i, j) position and zeros elsewhere. Then Mat_{3}(K[x_{0}, y_{0}]) is generated as a K[x_{0}, y_{0}]-algebra by E_{12}, E_{23} and E_{31} subject to the relations

E_{12}^{2} = E_{23}^{2} = E_{31}^{2} = E_{12}E_{31} = E_{23}E_{12} = E_{31}E_{23} = 0,

and

E_{12}E_{23}E_{31} + E_{23}E_{31}E_{12} + E_{31}E_{12}E_{23} = 1.

We define a K-algebra homomorphism ϕ : Mat_{3}(K[x_{0}, y_{0}]) → A_{f} via x_{0} ↦ ξ, y_{0} ↦ η and

E_{12} ↦ −xy^{2}(x + ξ + a_{2}), E_{23} ↦ −y^{2}(xy − a_{1}), E_{31} ↦ (yx − a_{1})y^{2}.

We checked by direct calculation that θ and ϕ are well defined, and that they are inverse to each other. □

5.3. Quadric intersections. Let f′ ∈ K[x, z] be a binary quartic, say

f′(x, z) = ax^{4} + bx^{3}z + cx^{2}z^{2} + dxz^{3} + ez^{4}.

The morphism C_{f′} → ℙ^{3} given by (x_{1} : x_{2} : x_{3} : x_{4}) = (xz : y : x^{2} : z^{2}) has image C_{f} where f = (f_{1}, f_{2}) and

(8)  \begin{aligned}f_{1}(x_{1}, x_{2}, x_{3}, x_{4}) & = x_{1}^{2} − x_{3}x_{4}, \\ f_{2}(x_{1}, x_{2}, x_{3}, x_{4}) & = x_{2}^{2} − (ax_{3}^{2} + bx_{1}x_{3} + cx_{3}x_{4} + dx_{1}x_{4} + ex_{4}^{2}). \end{aligned}

We write r′, s′, t′ and ξ′, η′ for the generators and central elements of A_{f′}.

Lemma 5.4. There is an isomorphism of K-algebras θ : A_{f} → Mat_{2}(A_{f′}) given by

p ↦ (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) , q ↦ (\begin{matrix}r′ & s′ \\ 0 & r′ \end{matrix}) , r ↦ (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}) , s ↦ (\begin{matrix}t′ & 0 \\ s′ & t′ \end{matrix}) .

Moreover we have θ(ξ) = (ξ′ + c)I_{2} and θ(η) = −η′I_{2}.

Proof. Again the proof is by direct calculation, the K-algebra homomorphism inverse to θ being given by E_{12} ↦ p, E_{21} ↦ r and

r′I_{2} ↦ pqr + rqp, s′I_{2} ↦ prqr + rqrp, t′I_{2} ↦ psr + rsp.  □

To complete the proof of Theorem 5.1, and hence of Theorem 1.1, it remains to show that in the case n = 4 the Weierstrass equation satisfied by ξ and η is in fact an equation for the Jacobian of C_{f}.

Let A and B be the 4 by 4 matrices of partial derivatives of f_{1} and f_{2}. We define a, b, c, d, e by writing \frac{1}{4} det(Ax + B) = ax^{4} + bx^{3} + cx^{2} + dx + e. As shown in [1], the Jacobian of C_{f} has Weierstrass equation y^{2} = F (x) where F is the monic cubic polynomial defined in (4).

We claim that η^{2} = F (ξ + c_{0}) for some constant c_{0} ∈ K (depending on f). In verifying this claim we are free to extend our field. We are also free to make changes of coordinates. Indeed if f′ = γf for some γ = (Λ, M) ∈ G_{4}(K) then by Theorem 4.1 there is an isomorphism ψ : A_{f′} → A_{f} with ξ ↦ (det γ)^{2}ξ + ρ and η ↦ (det γ)^{3}η. On the other hand the monic cubic polynomials F and F′ (associated to f and f′) are related by F′(x − \frac{1}{3}c′) = (det γ)^{6}F ((det γ)^{−2}x − \frac{1}{3}c). Finally we checked that for f as specified in (8), the claim is satisfied with c_{0} = 0.

6. Proof of Theorem 1.2

In this section we recall the definition of the isomorphism (1), and then prove that the construction of A_{f} from C_{f} is an explicit realisation of this map.

6.1. Galois cohomology. Let E/K be an elliptic curve. Writing \overline{K} for a separable closure of K, the short exact sequences of Galois modules

(9)  0 → \overline{K}^{×} → \overline{K}(E)^{×} → \overline{K}(E)^{×}/\overline{K}^{×} → 0,

and

0 → \overline{K}(E)^{×}/\overline{K}^{×} → Div E → Pic E → 0,

give rise to long exact sequences

(10)  H^{2}(K, \overline{K}^{×}) → H^{2}(K, \overline{K}(E)^{×}) → H^{2}(K, \overline{K}(E)^{×}/\overline{K}^{×}),

and

(11) H^{1}(K, Div E) → H^{1}(K, Pic E) → H^{2}(K, \overline{K}(E)^{×}/\overline{K}^{×}) → H^{2}(K, Div E).

Since H^{1}(K, ℤ) = 0 it follows by Shapiro’s lemma that H^{1}(K, Div E) = 0. We may identify H^{1}(K, Pic E) = H^{1}(K, Pic^{0} E) = H^{1}(K, E) and H^{2}(K, \overline{K}^{×}) = Br(K). As shown in [14, Appendix] we may identify

Br(E) = ker (H^{2}(K, \overline{K}(E)^{×}) → H^{2}(K, Div E)) .

We fix a local parameter t at 0 ∈ E(K). The left hand map in (9) is split by the map sending a Laurent power series in t to its leading coefficient. It follows that the right hand map in (10) is surjective, and hence H^{1}(K, E) ≅ Br(E)/ Br(K). Since the natural map Br(K) → Br(E) is split by evaluation at 0 ∈ E(K) this also gives the isomorphism (1).

6.2. Cyclic algebras. Let L/K be a Galois extension with Gal(L/K) cyclic of order n, generated by σ. For b ∈ K^{×} the cyclic algebra (L/K, b) is the K-algebra with basis 1, v, …, v^{n−1} as an L-vector space, and multiplication determined by v^{n} = b and vλ = σ(λ)v for all λ ∈ L. This is a central simple algebra over K of dimension n^{2}. It is split by L and so determines a class in Br(L/K).

We compute cohomology of C_{n} = ⟨σ|σ^{n} = 1⟩ relative to the resolution

… ⟶ℤ[C_{n}] ⟶^{Δ} ℤ[C_{n}] ⟶^{N} ℤ[C_{n}] ⟶^{Δ} ℤ[C_{n}] ⟶ 0,

where Δ = σ − 1 and N = 1 + σ + … + σ^{n−1}. Thus for A a Gal(L/K)-module,

H^{i}(Gal(L/K), A) = \{ \begin{matrix}ker(N|A)/im(Δ|A) & if i ≥ 1 odd, \\ ker(Δ|A)/im(N|A) & if i ≥ 2 even. \end{matrix}

In particular K^{×}/N_{L/K}(L^{×}) ≅ H^{2}(Gal(L/K), L^{×}) = Br(L/K). This isomorphism is realised by sending b ∈ K^{×} to the class of (L/K, b).

Let E/K be an elliptic curve, and fix a local parameter t at 0 ∈ E(K). If g ∈ K(E)^{×} then we write (L/K, g) for the cyclic algebra (L(E)/K(E), g). We may describe the isomorphism (1) in terms of cyclic algebras as follows.

Lemma 6.1. Let C/K be a smooth curve of genus one curve with Jacobian E, and suppose Q ∈ C(L). Let P be the image of [σQ − Q] under Pic^{0}(C) ≅ E. Then the isomorphism (1) sends the class of C to the class of (L/K, g) where g ∈ K(E)^{×} has divisor (P) + (σP) + … + (σ^{n−1}P) − n(0), and is scaled to have leading coefficient 1 when expanded as a Laurent power series in t.

Proof: We identify E ≅ Pic^{0}(E) via T ↦ (T) − (0). Then the class of C in H^{1}(K, Pic E) is represented by (P)−(0), and its image under the connecting map in (11) is represented by g ∈ K(E)^{×} where div g = N_{L/K}((P) − (0)). Finally to lift to an element of ker(ev_{0} : Br(E) → Br(K)) we scale g as indicated. □

6.3. Binary quartics. We prove Theorem 1.2 in the case n = 2. By a change of coordinates we may assume^{1} that a ≠ 0 and b = 0, i.e.

f(x, z) = ax^{4} + cx^{2}z^{2} + dxz^{3} + ez^{4}.

Let E be the Jacobian of C_{f}, with Weierstrass equation as specified in Section 3.1. We know by Theorem 1.1 that the centre Z of A_{f} is a Dedekind domain with field of fractions K(E). Therefore the natural map Br(Z) → Br(K(E)) is injective, and so it suffices for us to consider the class of the quaternion algebra A_{f} ⊗_{Z} K(E) in Br(K(E)). This algebra is generated by r and s subject to the rules r^{2} = a, rs + sr = 0 and s^{2} = ξ + c. It is therefore the cyclic algebra (L/K, g) where L = K(\sqrt{a}) and g ∈ K(E)^{×} is the rational function g(ξ, η) = ξ + c.

By inspection of the Weierstrass equation for E in Section 3.1, we see that div g = (P) + (σP) − 2(0) where P = (−c, d\sqrt{a}) ∈ E(L). Let C_{f} have equation y^{2} = f(x_{1}, x_{2}), and let Q ∈ C(L) be the point (x_{1} : x_{2} : y) = (1 : 0 : \sqrt{a}). Let π : C_{f} → E be the covering map, i.e. the map T ↦ [2(T) − H] where H is the fibre of the double cover C_{f} → ℙ^{1}. Using the formulae in [1] we find that π(Q) = −P. Therefore [σQ − Q] = [H − 2(Q)] = P. It follows by Lemma 6.1 that the isomorphism (1) sends the class of C to the class of the cyclic algebra (L/K, g). This completes the proof of Theorem 1.2 in the case n = 2.

6.4. Ternary cubics. We prove Theorem 1.2 in the case n = 3. Since 2 and 3 are coprime, we are free to replace our field K by a quadratic extension. We may therefore suppose that ζ_{3} ∈ K and that f(x, 0, z) = ax^{3} − z^{3} with a ≠ 0. Further substitutions of the form x ← x + λy and z ← z + λ′y reduce us to the case

f(x, y, z) = ax^{3} + by^{3} − z^{3} + b_{1}xy^{2} + b_{3}y^{2}z + mxyz.

The algebra A_{f} ⊗_{Z} K(E) is generated by x and v = yx − ζ_{3}xy − \frac{1}{3}(1 − ζ_{3})m subject to the rules x^{3} = a, xv = ζ_{3}vx and v^{3} = g(ξ, η) where

g(ξ, η) = η − ζ_{3}^{2}mξ − 3(1 − ζ_{3})ab + \frac{1}{9}(ζ_{3} − ζ_{3}^{2})m^{3}.

It is therefore the cyclic algebra (L/K, g) where L = K(\sqrt[3]{a}).

1It is incorrectly claimed in [10, Section 5] that we may further assume d = 0.

Let E be given by the Weierstrass equation specified in Section 3.2. We find that div g = (R) + (σR) + (σ^{2}R) − 3(0) for a certain point R ∈ E(L) with x-coordinate −(1/3)m^{2} + b_{1}\sqrt[3]{a} − b_{3}(\sqrt[3]{a})^{2}. Let Q = (1 : 0 : \sqrt[3]{a}) ∈ C_{f}(L). Let π : C_{f} → E be the covering map, i.e. the map T ↦ [3(T) − H] where H is the hyperplane section. Using the formulae in [1] we find that π(Q) = σR − σ^{2}R. We compute

3[σQ − Q] = π(σQ) − π(Q) = σ(σR − σ^{2}R) − (σR − σ^{2}R) = 3σ^{2}R.

Since generically E has no 3-torsion, it follows that [σQ − Q] = σ^{2}R. Taking P = σ^{2}R in Lemma 6.1 completes the proof.

6.5. Dihedral algebras. Let L/K be a Galois extension with Gal(L/K) ≅ D_{2n} where D_{2n} = ⟨σ, τ|σ^{n} = τ^{2} = (στ)^{2} = 1⟩ is the dihedral group of order 2n. Let K_{1}, F and F̃ be the fixed fields of σ, τ and στ. For (b, ε, ε̃) ∈ K_{1}^{×} ×F^{×} ×F̃^{×} satisfying N_{K_{1}/K}(b)N_{F/K}(ε) = N_{F̃/K}(ε̃) we define the dihedral algebra (L/K, b, ε, ε̃) to be the K-algebra with basis 1, v, …, v^{n−1}, w, vw, …, v^{n−1}w as an L-vector space, and multiplication determined by v^{n} = b, w^{2} = ε, (vw)^{2} = ε̃, vλ = σ(λ)v and wλ = τ(λ)w for all λ ∈ L. As we explain below, this is a special case of a crossed product algebra. In particular it is a central simple algebra over K of dimension (2n)^{2}. It is split by L and so determines a class in Br(L/K).

Let N = 1 + σ + … + σ^{n−1} ∈ ℤ[D_{2n}]. We compute cohomology of D_{2n} relative to the resolution

(12)  … ⟶ℤ[D_{2n}]^{4} ⟶^{Δ_{3}} ℤ[D_{2n}]^{3} ⟶^{Δ_{2}} ℤ[D_{2n}]^{2} ⟶^{Δ_{1}} ℤ[D_{2n}] ⟶ 0

where

Δ_{3} = (\begin{matrix}σ − 1 & 0 & 0 \\ 0 & τ − 1 & 0 \\ 0 & 0 & στ − 1 \\ τ + 1 & N & −N \end{matrix}) , Δ_{2} = (\begin{matrix}N & 0 \\ 0 & τ + 1 \\ στ + 1 & σ + τ \end{matrix}) , Δ_{1} = (\begin{matrix}σ − 1 \\ τ − 1 \end{matrix}) ,

and our convention is that Δ_{m} acts by right multiplication on row vectors. This resolution is a special case of that defined in [16], except that we have applied some row and column operations to simplify Δ_{2} and Δ_{3}. Using this resolution to compute Br(L/K) = H^{2}(Gal(L/K), L^{×}) we find

(13) \frac{\{(b, ε, ε̃) ∈ K_{1}^{×} × F^{×} × F̃^{×} | N_{K_{1}/K}(b)N_{F/K}(ε) = N_{F̃/K}(ε̃)\}}{\{(N_{L/K_{1}}(λ_{1}), N_{L/F} (λ_{2}), N_{L/F̃}(λ_{1}λ_{2})) | λ_{1}, λ_{2} ∈ L^{×}\}} ≅ Br(L/K).

This isomorphism is realised by sending (b, ε, ε̃) to the class of the dihedral algebra (L/K, b, ε, ε̃). Our claim that dihedral algebras are crossed product algebras is justified by comparing this description of Br(L/K) with that obtained from the standard resolution.

In more detail, there is a commutative diagram of free ℤ[D_{2n}]-modules

[figure: ⨁_{(g,h)∈D_{2n}^{2}} ℤ[D_{2n}]; ϕ_{2}; d_{2}; ⨁_{g∈D_{2n}} ℤ[D_{2n}]; d_{1}; ϕ_{1}; ℤ[D_{2n}]; 0; ℤ[D_{2n}]^{3}; Δ_{2}; ℤ[D_{2n}]^{2}; Δ_{1}; ℤ[D_{2n}]; 0]

where the first row is the standard resolution, i.e. d_{1}(e_{g}) = g − 1 and d_{2}(e_{g,h}) = g(e_{h})−e_{gh} +e_{g}, and the second row is the resolution (12). We choose ϕ_{1} such that

ϕ_{1}(e_{1}) = (0, 0),  ϕ_{1}(e_{σ^{i}}) = (1 + σ + … + σ^{i−1}, 0) for 0 < i < n,
ϕ_{1}(e_{τ}) = (0, 1),  ϕ_{1}(e_{σ^{i}τ}) = (1 + σ + … + σ^{i−1}, σ^{i}) for 0 < i < n.

We further choose ϕ_{2} such that for 0 ≤ i, j < n we have

ϕ_{2}(e_{σ^{i},σ^{j}}) = ϕ_{2}(e_{σ^{i},σ^{j}τ}) = \{ \begin{matrix}(0, 0, 0) & if i + j < n, \\ (1, 0, 0) & if i + j ≥ n, \end{matrix}

and ϕ_{2}(e_{τ,τ}) = (0, 1, 0), ϕ_{2}(e_{στ,στ}) = (0, 0, 1). The 2-cocycle ξ ∈ Z^{2}(D_{2n}, L^{×}) corresponding to (b, ε, ε̃) is now the unique 2-cocycle satisfying

ξ_{σ^{i},σ^{j}} = ξ_{σ^{i},σ^{j}τ} = \{ \begin{matrix}1 & if i + j < n, \\ b & if i + j ≥ n, \end{matrix}

and ξ_{τ,τ} = ε, ξ_{στ,στ} = ε̃. The cross product algebra associated to ξ is the K-algebra with basis {v_{g} : g ∈ D_{2n}} as an L-vector space, and multiplication determined by v_{g}v_{h} = ξ_{g,h}v_{gh} and v_{g}λ = g(λ)v_{g} for all λ ∈ L. Identifying v_{σ^{i}} = v^{i} and v_{σ^{i}τ} = v^{i}w, we recognise this as the dihedral algebra (L/K, b, ε, ε̃).

Let E/K be an elliptic curve, and fix a local parameter t at 0 ∈ E(K). We may describe the isomorphism (1) in terms of dihedral algebras as follows.

Lemma 6.2. Let C/K be a smooth curve of genus one with Jacobian E, and suppose Q ∈ C(F). Let P be the image of [σQ − Q] under Pic^{0}(C) ≅ E. Then the isomorphism (1) sends the class of C to the class of (L/K, g, 1, h) where g ∈ K_{1}(E)^{×} and h ∈ F̃(E)^{×} have divisors (P) + (σP) + … + (σ^{n−1}P) − n(0) and (P) + (−P) − 2(0), and are scaled to have leading coefficient 1 when expanded as Laurent power series in t.

Proof: We have [σQ − Q] = P and [τQ − Q] = 0. We identify E ≅ Pic^{0}(E) via T ↦ (T) − (0). Then the class of C in H^{1}(K, Pic E) is represented by the pair ((P) − (0), 0). Reading down the first column of Δ_{2}, the image of this class under the connecting map in (11) is represented by a triple (g, 1, h) where div g = N_{L/K_{1}}((P) − (0)) and div h = (στ + 1)((P) − (0)) = (P) + (−P) − 2(0). Finally to lift to an element of ker(ev_{0} : Br(E) → Br(K)) we scale g and h as indicated. □

6.6. Quadric intersections. We prove Theorem 1.2 in the case n = 4. We are free to make field extensions of odd degree. We may therefore suppose that C_{f} meets the plane {x_{4} = 0} in four points in general position, and that one of the three singular fibres in the pencil of quadrics vanishing at these points is defined over K. In other words, we may assume that f_{1}(x_{1}, x_{2}, x_{3}, 0) = q_{1}(x_{1}, x_{3}) where q_{1} is a binary quadratic form. Then f_{2} must have a term x_{2}^{2}, and so by completing the square f_{2}(x_{1}, x_{2}, x_{3}, 0) = x_{2}^{2} + q_{2}(x_{1}, x_{3}). Adding a suitable multiple of f_{1} to f_{2} we may suppose that q_{2} factors over K, and so without loss of generality q_{2}(x_{1}, x_{3}) = −x_{1}x_{3}. Making linear substitutions of the form x_{i} ← x_{i} + λx_{4} for i = 1, 2, 3 brings us to the case

f_{1}(x_{1}, x_{2}, x_{3}, x_{4}) = ax_{1}^{2} + bx_{1}x_{3} + cx_{3}^{2} + (d_{1}x_{1} + d_{2}x_{2} + d_{3}x_{3} + d_{4}x_{4})x_{4},
f_{2}(x_{1}, x_{2}, x_{3}, x_{4}) = x_{2}^{2} − x_{1}x_{3} − ex_{4}^{2}.

Let L/K be the splitting field of G(X) = aX^{4} + bX^{2} + c. Then Gal(L/K) is a subgroup of D_{8}. We suppose it is equal to D_{8}, the other cases being similar. We have L = K(θ, \sqrt{δ}) where θ is a root of G and δ = ac(b^{2} − 4ac). The generators σ and τ of D_{8} act as

σ : θ ↦ \frac{1}{\sqrt{δ}} (abθ^{3} + (b^{2} − 2ac)θ),  σ : \sqrt{δ} ↦ \sqrt{δ},
τ : θ ↦ θ,  τ : \sqrt{δ} ↦ −\sqrt{δ}.

The fixed fields of σ, τ and στ are K_{1} = K(\sqrt{δ}), F = K(θ) and F̃ = K(ϕ) where ϕ = a(θ + σ(θ)).

Let A = A_{f} ⊗_{Z} K(E). The second generator q of A_{f} satisfies aq^{4} + bq^{2} + c = 0. We may therefore embed F ⊂ A via θ ↦ q, and hence L ⊂ A_{1} = A ⊗_{K} K_{1}. We find that A_{1} is generated as a K_{1}(E)-algebra by q and

v = a(rσ(q) − qr) + \frac{a}{\sqrt{δ}} (aq^{3}σ(q) − c)(d_{1}q + d_{2} + d_{3}q^{−1})

subject to the rules aq^{4} + bq^{2} + c = 0, vq = σ(q)v and v^{4} = g(ξ, η), for some g ∈ K_{1}(E). It is therefore the cyclic algebra (L/K_{1}, g). Writing ξ for the element that was denoted ξ + c_{0} in Section 5.3, we have

g(ξ, η) = ξ^{2} − \frac{4acd_{2}}{\sqrt{δ}} η + \frac{2(bm + 2acd_{2}^{2})}{b^{2} − 4ac}ξ + 8aced_{2}^{2} + \frac{m^{2} + d_{2}^{2}n}{b^{2} − 4ac}.

where m = cd_{1}^{2} − bd_{1}d_{3} + ad_{3}^{2} + (b^{2} − 4ac)d_{4} and n = bcd_{1}^{2} + ac(d_{2}^{2} − 4d_{1}d_{3}) + abd_{3}^{2}.

Let Q = (θ^{2} : θ : 1 : 0) ∈ C_{f}(F), and let P = [σQ − Q] under the usual identification Pic^{0}(C_{f}) ≅ E. We compute the point P as follows. We put

(\begin{matrix}T z_{1} \\ z_{2} \\ z_{1} \\ T z_{2} \end{matrix}) = (\begin{matrix}a & ϕ & (ϕ^{2} + ab)/2a & 0 \\ a & −ϕ & (ϕ^{2} + ab)/2a & 0 \\ −ad_{1} & −ad_{2} & −ad_{3} & −ad_{4} − eϕ^{2} \\ 0 & 0 & 0 & 1 \end{matrix}) (\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}) .

Inverting this 4 by 4 matrix M gives

(det M)f_{2}(x_{1}, x_{2}, x_{3}, x_{4}) = α(z_{1}, z_{2})T^{2} + β(z_{1}, z_{2})T + γ(z_{1}, z_{2})

for some binary quadratic forms α, β, γ. Relacing f_{2} by f_{1} gives a scalar multiple of the same equation. Therefore C_{f} has equation y^{2} = β(z_{1}, z_{2})^{2} −4α(z_{1}, z_{2})γ(z_{1}, z_{2}). The points Q and σ(Q) are given by (z_{1} : z_{2} : y) = (1 : 0 : ±a(θ − σ(θ))). Exactly as in Section 6.3, we compute P using the formulae for the covering map. Relative to the Weierstrass equation for E specified in Section 5.3, this point has x-coordinate

(14)  x(P) = \frac{2aϕ^{2}(d_{2}d_{3}ϕ + m) − d_{2}(d_{1}ϕ + ad_{2})(bϕ^{2} + a(b^{2} − 4ac))}{2a^{2}(b^{2} − 4ac)} ∈ F̃.

We find that P and its Galois conjugates are zeros of g. Therefore div g = (P) + (σP) + (σ^{2}P) + (σ^{3}P) − 4(0). It follows by Lemma 6.1 that the class of A_{f}, and the image of the class of C_{f} under (1), agree after restricting to Br(E ⊗_{K} K_{1}). It remains to show that the same conclusion holds without the quadratic extension.

For a ∈ A_{1} = A⊗_{K} K_{1} let \overline{a} = (1⊗τ)a. We find that v\overline{v} = ξ −x(P) where x(P) is given by (14). Now let A_{2} = A_{1}⊕A_{1}w with multiplication determined by w^{2} = 1 and wa = \overline{a}w for all a ∈ A_{1}. This is the dihedral algebra (L/K, g, 1, ξ − x(P)). The subalgebra generated by K_{1} and w is a trivial cyclic algebra. Therefore A_{2} ≅ A ⊗_{K} Mat_{2}(K). In particular A and A_{2} have the same class in Br(K(E)). Lemma 6.2 now completes the proof.

7. Geometric interpretation

Let C be a smooth curve of genus one with Jacobian elliptic curve E. Let H and H′ be K-rational divisors on C of degree n ≥ 2. We assume that H and H′ are not linearly equivalent, and so their difference corresponds to a non-zero point P ∈ E(K). The complete linear systems |H| and |H′| define an embedding C → ℙ^{n−1} × ℙ^{n−1}. Assuming n ∈ {2, 3, 4}, the composite of this map with the first and second projections is described by genus one models f and f′.

In this section we investigate the following problem.

Given f and P, how can we compute f′?

The answers we give might be viewed as explicitly realising the connection between the Tate pairing and the obstruction map, as studied in [6, 15, 18]. Our answers also serve to motivate the definition of A_{f}, and indeed (however much it might seem an obvious guess in hindsight) this is how we actually found the correct definition of A_{f} in the case n = 4.

We give no proofs in this section. However all our claims may be verified by generic calculations.

7.1. Binary quartics. The image of C → ℙ^{1} × ℙ^{1} is defined by a (2, 2)-form, say

F (x, z; x′, z′) = f_{1}(x, z)x^{′2} + 2f_{2}(x, z)x′z′ + f_{3}(x, z)z^{′2}.

Then f = f_{2}^{2} − f_{1}f_{3}, and f′ is obtained in the same way, after switching the two sets of variables. Thus, given a binary quartic f, we seek to find binary quadratic forms f_{1}, f_{2}, f_{3} such that

(\begin{matrix}f_{2} & −f_{1} \\ f_{3} & −f_{2} \end{matrix})^{2} = (\begin{matrix}f & 0 \\ 0 & f \end{matrix}) .

Equivalently, we look for matrices M_{1}, M_{2}, M_{3} ∈ Mat_{2}(K) satisfying

(α^{2}M_{1} + αβM_{2} + β^{2}M_{3})^{2} = f(α, β)I_{2}.

This reduces the problem of finding f′ from f to that of finding a K-algebra homomorphism A_{f} → Mat_{2}(K). By Theorem 1.1 any such homomorphism must factor via A_{f,P} for some 0 ≠ P ∈ E(K). This point P turns out to be the same as the point P considered at the start of Section 7. In conclusion, if A_{f,P} ≅ Mat_{2}(K) and we can find this isomorphism explicitly, then we can write down a (2, 2)-form, and hence a binary quartic f′, such that C_{f} and C_{f′} are isomorphic as genus one curves, but their hyperplane sections differ by P.

7.2. Ternary cubics. The image of C → ℙ^{2} × ℙ^{2} is defined by three (1, 1)- forms. The coefficients may be arranged as a 3 × 3 × 3 cube. As explained in [3], slicing this cube in three different ways gives rise to three ternary cubics. Two of these are f and f′. Thus, given a ternary cubic f, we seek to find matrices M_{1}, M_{2}, M_{3} ∈ Mat_{3}(K) satisfying

f(α, β, γ) = det(αM_{1} + βM_{2} + γM_{3}).

If f(0, 0, 1) ≠ 0 then we may assume (after rescaling f and multiplying each M_{i} on the left by the same invertible matrix) that M_{3} = −I_{3}. Then αM_{1} + βM_{2} has characteristic polynomial γ ↦ f(α, β, γ), and so by the Cayley-Hamilton theorem

f(α, β, αM_{1} + βM_{2}) = 0.

This reduces the problem of finding f′ from f to that of finding a K-algebra homomorphism A_{f} → Mat_{3}(K). By Theorem 1.1 any such homomorphism must factor via A_{f,P} for some 0 ≠ P ∈ E(K). This point P turns out to be the same as the point P considered at the start of Section 7. In conclusion, if A_{f,P} ≅ Mat_{3}(K) and we can find this isomorphism explicitly, then we can write down a 3 × 3 × 3 cube, and hence a ternary cubic f′, such that C_{f} and C_{f′} are isomorphic as genus one curves, but their hyperplane sections differ by P.

7.3. Quadric intersections. The image of C → ℙ^{3} × ℙ^{3} is defined by an 8dimensional vector space V of (1, 1)-forms in variables x_{1}, …, x_{4} and y_{1}, …, y_{4}. Let W be the vector space of 4 by 4 alternating matrices B = (b_{ij}) of linear forms in y_{1}, …, y_{4} such that

∑_{i=1}^{4} x_{i}b_{ij}(y_{1}, …, y_{4}) ∈ V for all j = 1, …, 4.

We find that W is 4-dimensional. We choose a basis, and let M be a generic linear combination of the basis elements, say with coefficients z_{1}, …, z_{4}. Then M = (m_{ij}) is a 4 by 4 alternating matrix of (1, 1)-forms in y_{1}, …, y_{4} and z_{1}, …, z_{4}. The Pfaffian of this matrix is a (2, 2)-form, which turns out to be

f_{1}^{+}(y_{1}, …, y_{4})f_{2}^{−}(z_{1}, …, z_{4}) − f_{2}^{+}(y_{1}, …, y_{4})f_{1}^{−}(z_{1}, …, z_{4}),

where f^{±} = (f_{1}^{±}, f_{2}^{±}) describes the image of C → ℙ^{3} via |H^{±}|, and [H−H^{±}] = ±P. To tie in with our earlier notation, H^{+} = H′ and f^{+} = f′.

We write m_{ij} = (y_{1}, …, y_{4})M_{ij}(z_{1}, …, z_{4})^{T} where M_{ij} ∈ Mat_{4}(K). Assuming C_{f} does not meet the line {x_{3} = x_{4} = 0} we have det(M_{12}) ≠ 0, and so we may choose our basis for W such that M_{12} = I_{4}. The matrices M_{ij} then satisfy f_{i}(αM_{23} + βM_{24}, −(αM_{13} + βM_{14}), α, β) = 0 for i = 1, 2, where the first two arguments commute, and M_{34} = M_{13}M_{24} − M_{23}M_{14} = M_{24}M_{13} − M_{14}M_{23}.

This reduces the problem of finding f′ from f to that of finding a K-algebra homomorphism A_{f} → Mat_{4}(K). By Theorem 1.1 any such homomorphism must factor via A_{f,P} for some 0 ≠ P ∈ E(K). Again this point P turns out to correspond to the difference of hyperplane sections for C_{f} and C_{f′}.

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University of Cambridge, DPMMS, Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK

E-mail address: T.A.Fisher@dpmms.cam.ac.uk