ON BINARY QUARTICS AND THE CASSELS-TATE PAIRING

TOM FISHER

Abstract. We use the invariant theory of binary quartics to give a new formula for the Cassels-Tate pairing on the 2-Selmer group of an elliptic curve. Unlike earlier methods, our formula does not require us to solve any conics. An important role in our construction is played by a certain K3 surface defined by a (2, 2, 2)-form.

1. Introduction

Let E be an elliptic curve over a number field K. The Mordell-Weil theorem tells us that the abelian group E(K) is finitely generated, but there is no known algorithm guaranteed to compute its rank. Instead, for each integer n ≥ 2 there is an exact sequence of abelian groups

0 → E(K)/nE(K) → S^{(n)}(E/K) → � (E/K)[n] → 0.

The n-Selmer group S^{(n)}(E/K) is finite and effectively computable. Computing S^{(n)}(E/K) gives an upper bound for the rank of E(K), but this will be sharp only if the n-torsion of the Tate-Shafarevich group � (E/K) is trivial.

Cassels [4] showed that there is an alternating pairing

⟨ . ⟩_{CT} : S^{(n)}(E/K) × S^{(n)}(E/K) → ℚ/ℤ

whose kernel is the image of S^{(n^{2})}(E/K). By computing this pairing, our upper bound for the rank of E(K) improves from that obtained by n-descent to that obtained by n^{2}-descent. In view of the generalisation to abelian varieties, due to Tate, the pairing is known as the Cassels-Tate pairing.

Cassels [6] also described a method for computing the pairing in the case n = 2. His method involves solving conics over the field of definition of each 2-torsion point on E. More recently, Donnelly [10] found a method that only involves solving conics over K, and implemented this in Magma [3]. In this article we use the invariant theory of binary quartics to give a self-contained account of a version of his method that is relatively simple to implement.

Since this article was first written, Jiali Yan has written her PhD thesis [18], extending some of these ideas to Jacobians of genus 2 curves, and Bill Allombert has implemented our method for computing the pairing as part of the function ellrank in pari/gp [15]. I thank them both, and also Steve Donnelly and John Cremona, for useful discussions.

1

Date: 31st August 2022.

2. Binary quartics

A binary quartic over a field K is a homogeneous polynomial g ∈ K[x, z] of degree 4. Binary quartics g_{1} and g_{2} are K-equivalent if

g_{2}(x, z) = λ^{2}g_{1}(αx + γz, βx + δz)

for some λ, α, β, γ, δ ∈ K with λ(αδ − βγ) ≠ 0. They are properly K-equivalent if in addition λ(αδ − βγ) = ±1. The invariants of the binary quartic

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

are

I = 12ae − 3bd + c^{2},
J = 72ace − 27ad^{2} − 27b^{2}e + 9bcd − 2c^{3}.

The binary quartics g_{1} and g_{2} have invariants related by I(g_{2}) = λ^{4}(αδ−βγ)^{4}I(g_{1}) and J(g_{2}) = λ^{6}(αδ − βγ)^{6}J(g_{1}). In particular, properly equivalent binary quartics have the same invariants. The discriminant is Δ = 16(4I^{3} − J^{2})/27. We say that g is K-soluble if there exist x, z ∈ K, not both zero, such that g(x, z) is a square in K. The reason for this terminology is that if Δ(g) ≠ 0 then there is a smooth projective curve C of genus one with affine equation y^{2} = g(x, 1), and we are asking that C(K) ≠ ∅. As shown by Weil [17], the Jacobian of C is the elliptic curve

(2)  E_{I,J} : y^{2} = x^{3} − 27Ix − 27J.

Now let K be a number field, and M_{K} its set of places. A binary quartic over K is everywhere locally soluble if it is K_{v}-soluble for all places v ∈ M_{K}. We note that every elliptic curve over K can be written in the form (2) for some I, J ∈ K with 4I^{3} − J^{2} ≠ 0.

Lemma 2.1. Let I, J ∈ K with 4I^{3} − J^{2} ≠ 0. Then

S^{(2)}(E_{I,J}/K) = \{ \begin{matrix}everywhere locally soluble \\ binary quartics over K \\ with invariants I and J \end{matrix} \} /(proper K-equivalence).

Proof. The case K = ℚ is proved in [2], the only simplification in this case being that (since the only roots of unity in ℚ are ±1) equivalent quartics with the same invariants are always properly equivalent (even in the cases where I = 0 or J = 0). The general case is similar. □

Although Lemma 2.1 specifies S^{(2)}(E_{I,J}/K) as a set, the group law is not obvious. The following description is taken from [8], [9]. Let L be the étale algebra K[φ] where φ is a root of X^{3} − 3IX + J = 0. Then the binary quartic (1) has cubic invariant

z(g) = \frac{4aφ + 3b^{2} − 8ac}{3}.

By a change of coordinates (that is, replacing g by a properly equivalent quartic) we may assume that z(g) is a unit in L. The group law on S^{(2)}(E_{I,J}/K) is then given by multiplying the cubic invariants in L^{×}/(L^{×})^{2}. The method for converting an element of L^{×}/(L^{×})^{2} back to a binary quartic does, however, involve solving a conic over K.

3. Statement of results

In this section we state our new formula for the Cassels-Tate pairing on the 2-Selmer group of an elliptic curve. First we need some more invariant theory. The binary quartic (1) has Hessian

h(x, z) = (3b^{2} − 8ac)x^{4} + 4(bc − 6ad)x^{3}z + 2(2c^{2} − 24ae − 3bd)x^{2}z^{2}
+ 4(cd − 6be)xz^{3} + (3d^{2} − 8ce)z^{4}.

There are exactly three linear combinations of g(x, z) and h(x, z) that are singular (i.e. have repeated roots). Following [9] this prompts us to put

(3)  G(x, z) = \frac{1}{3} (4φg(x, z) + h(x, z)) ,
(4)  H(x, z) = \frac{1}{12} \frac{∂^{2}G}{∂x^{2}} + \frac{2}{9}(I − φ^{2})z^{2},

so that G(1, 0)G(x, z) = H(x, z)^{2}. We note that z(g) = G(1, 0) = H(1, 0).

Theorem 3.1. Let I, J ∈ K with 4I^{3} −J^{2} ≠ 0. Let g_{1}, g_{2}, g_{3} be everywhere locally soluble binary quartics over K with invariants I and J. Let H_{1}(x, z) be the binary quadratic form (4), with coefficients in L = K[φ], associated to g_{1}. Suppose that z(g_{1})z(g_{2})z(g_{3}) = m^{2} for some m ∈ L^{×}, and write

(5)  \frac{z(g_{2})z(g_{3})}{m}H_{1}(x, z) = α_{1}(x, z) + β_{1}(x, z)φ + γ_{1}(x, z)φ^{2}

where α_{1}, β_{1}, γ_{1} ∈ K[x, z]. For each v ∈ M_{K} we choose x_{v}, z_{v} ∈ K_{v} with g_{1}(x_{v}, z_{v}) a square in K_{v} and γ_{1}(x_{v}, z_{v}) ≠ 0. If g_{2}(1, 0) ≠ 0 then the Cassels-Tate pairing on S^{(2)}(E_{I,J}/K) is given by

(6)  ⟨[g_{1}], [g_{2}]⟩_{CT} = ∏_{v∈M_{K}}(g_{2}(1, 0), γ_{1}(x_{v}, z_{v}))_{v}

where ( , )_{v} : K_{v}^{×}/(K_{v}^{×})^{2} × K_{v}^{×}/(K_{v}^{×})^{2} → μ_{2} is the Hilbert norm residue symbol.

Remark 3.2. (i) If we wish to compute the pairing starting only with g_{1} and g_{2}, then we first change coordinates so that z(g_{1}) and z(g_{2}) are units in L, multiply these together, and then compute g_{3} by solving a conic over K. This conic is the same as the one that has to be solved in Donnelly’s method [10]. (ii) We show in Remark 8.3 that the binary quadratic form γ_{1} is not identically zero. Therefore, by our assumption that g_{1} is everywhere locally soluble, it is always possible to choose x_{v}, z_{v} ∈ K_{v} with the stated properties. (iii) The assumption that g_{2}(1, 0) ≠ 0 is no limitation, since if g_{2}(1, 0) = 0 then [g_{2}] = 0 in the 2-Selmer group, which certainly implies the pairing is trivial. (iv) By definition the Cassels-Tate pairing takes values in ℚ/ℤ. In our formula it takes values in μ_{2}. It should be understood that we have identified μ_{2} = \frac{1}{2}ℤ/ℤ.

(v) Since ⟨ , ⟩_{CT} is alternating and bilinear and [g_{1}] + [g_{2}] + [g_{3}] = 0 we have ⟨[g_{1}], [g_{2}]⟩_{CT} = ⟨[g_{1}], [g_{3}]⟩_{CT}. So we may equally write g_{2}(1, 0) or g_{3}(1, 0) in (6). Notice however that the binary quartics g_{2} and g_{3} do both contribute to the pairing via (5). Moreover we must use the exact formulae for z(g_{1}), z(g_{2}) and z(g_{3}), these being linear in φ. It is not enough just to know these quantities up to squares, since this would change the left hand side of (5). (vi) If E(K)[2] = 0 then m is uniquely determined up to sign. By the product formula for the Hilbert norm residue symbol this makes no difference to (6). If E(K)[2] ≠ 0 then there are more choices for m, but it turns out (see the proof of Theorem 8.2) that we may use any one of these to compute the pairing.

Remark 3.3. The product over all places in Theorem 3.1 is a finite product. Indeed, outside an easily determined finite set of places, we have

(i) v is a finite prime, with residue field of size at least 11.

(ii) g_{1} and γ_{1} have v-adically integral coefficients, with v ∤ Δ(g_{1}) content(γ_{1}). (iii) g_{2}(1, 0) is a v-adic unit and v ∤ 2.

Under conditions (i) and (ii) we can pick our local point (by Hensel lifting a smooth point on the reduction that is not a root of γ_{1}) such that γ_{1}(x_{v}, z_{v}) is a unit. It follows by (iii) that the local contribution at v is trivial.

Example 3.4. Let E/ℚ be the elliptic curve

y^{2} + y = x^{3} − x^{2} − 929x − 10595

labelled 571a1 in [7]. A 2-descent shows that S^{(2)}(E/ℚ) ≅ (ℤ/2ℤ)^{2}, and its nonzero elements are represented by

g_{1}(x, z) = −11x^{4} + 68x^{3}z − 52x^{2}z^{2} − 164xz^{3} − 64z^{4},
g_{2}(x, z) = −4x^{4} − 60x^{3}z − 232x^{2}z^{2} − 52xz^{3} − 3z^{4},
g_{3}(x, z) = −31x^{4} − 78x^{3}z + 32x^{2}z^{2} + 102xz^{3} − 53z^{4}.

Each of these binary quartics has invariants I = 44608 and J = 18842960, and discriminant Δ = −2^{12} · 571. By (5), with m = \frac{1}{9}(20φ^{2} − 8656φ + 936032), we get

γ_{1}(x, z) = \frac{4}{9}(5x^{2} − 16xz − 12z^{2}).

For each odd prime p there is a smooth 𝔽_{p}-point on the reduction of y^{2} = g_{1}(x, 1) mod p, whose x-coordinate is not a root of 5x^{2} − 16x − 12 = 0. Indeed we checked this claim directly for p = 3, 5, 7, 11 and 571, and for all other primes it follows by Hasse’s bound. Therefore the odd primes make no contribution to (6).

To compute the contribution at p = 2 we write g_{1}(x, 1) = x^{4} + 4q(x) where

q(x) = −3x^{4} + 17x^{3} − 13x^{2} − 41x − 16.

By Hensel’s lemma the equation q(x) = 0 has a root in ℤ_{2} with x = 2^{4} + O(2^{5}). But then γ_{1}(x, 1) ≡ 5 mod (ℚ_{2}^{×})^{2}, and since (5, −1)_{2} = 1 the contribution is again trivial. Finally, since g_{1}(15, 4) > 0 and γ_{1}(15, 4) < 0, there is a contribution from the real place. This shows that the Cassels-Tate pairing on S^{(2)}(E/ℚ) is non-trivial, and hence rank E(ℚ) = 0.

4. The Cassels-Tate pairing

There are two standard definitions of the Cassels-Tate pairing (in the case of elliptic curves) called in [11], [16] the homogeneous space definition and the Weil pairing definition. Both definitions appear in Cassels’ original paper [4], although the method in [6] (see also [12]) is a variant of the Weil pairing definition. In this section we review the homogeneous space definition, and highlight its connection with the Brauer-Manin obstruction.

Let K be a field with separable closure \overline{K}. We write H^{i}(K, −) for the Galois cohomology group H^{i}(Gal(\overline{K}/K), −). Let C/K be a smooth projective curve. We define

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

It is shown in the Appendix to [14] that this is equivalent to the usual definition Br(C) = H_{ét}^{2}(C, 𝔾_{m}). Identifying Br(K) = H^{2}(K, \overline{K}^{×}), there is a natural map

(8)  Br(K) → Br(C).

We will need the following two facts, whose proofs we give below.

(i) For P ∈ C(K) there is an evaluation map

Br(C) → Br(K) ; A ↦ A(P).

This is a group homomorphism, and a section to the map (8). Moreover the evaluation maps behave functorially with respect to all field extensions.

(ii) Suppose C is a smooth curve of genus one, with Jacobian elliptic curve E. If H^{3}(K, \overline{K}^{×}) = 0 then there is an isomorphism

(9)  Ψ_{C} : \frac{H^{1}(K, E)}{⟨[C]⟩} ⟶^{∼} \frac{Br(C)}{Br(K)}.

Now let E be an elliptic curve over a number field K. Let C and D be principal homogeneous spaces under E. Since H^{3}(K, \overline{K}^{×}) = 0 for K a number field, we have Ψ_{C}([D]) = A mod Br(K) for some A ∈ Br(C). Now suppose that C and D are everywhere locally soluble. For each place v ∈ M_{K} we pick a local point P_{v} ∈ C(K_{v}). The Cassels-Tate pairing � (E/K) × � (E/K) → ℚ/ℤ is defined by

(10)  ⟨[C], [D]⟩_{CT} = ∑_{v∈M_{K}} inv_{v}(A(P_{v}))

where inv_{v} : Br(K_{v}) → ℚ/ℤ is the local invariant map. As this form of the definition makes clear, if ⟨[C], [D]⟩_{CT} ≠ 0 then the genus one curve C is a counterexample to the Hasse Principle explained by the Brauer-Manin obstruction.

We check that the pairing is well defined, i.e. it does not depend on the choices of A and of the P_{v}. By class field theory there is an exact sequence

0 ⟶ Br(K) ⟶ ⨁_{v∈M_{K}} Br(K_{v}) \underrightarrow{∑ inv_{v}} ℚ/ℤ ⟶ 0.

It follows that if we change A by adding an element of Br(K) then the pairing (10) is unchanged. Next, since the class of D is trivial in H^{1}(K_{v}, E), the analogue of (9) over K_{v} shows that the restriction of A to the Brauer group of C/K_{v} is constant, i.e. it belongs to the image of Br(K_{v}). Therefore the pairing (10) does not depend on the choice of local points P_{v}.

We now prove the facts we quoted in (i) and (ii) above.

(i) For P ∈ C(K) there is a short exact sequence of Galois modules

0 ⟶ O_{P}^{×} ⟶ \overline{K}(C)^{×} ⟶^{ord_{P}} ℤ ⟶ 0

where O_{P} is the local ring at P. Taking Galois cohomology gives an exact sequence

0 ⟶ H^{2}(K, O_{P}^{×}) ⟶ H^{2}(K, \overline{K}(C)^{×}) ⟶^{ord_{P}} H^{2}(K, ℤ).

It follows by (7) that each element of Br(C) can be represented by a cocycle taking values in O_{P}^{×}, and so can be evaluated at P.

(ii) There is an exact sequence of Galois modules

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

where Div C and Pic C are the divisor group and Picard group for C over \overline{K}. Splitting into short exact sequences, and taking Galois cohomology, gives the following exact sequences

\xymatrix{ &  & H^{2}(K, \overline{K}^{×}) \ar[d] \\  &  & H^{2}(K, \overline{K}(C)^{×}) \ar[d] \\ H^{1}(K, Div C) \ar[r] & H^{1}(K, Pic C) \ar[r] & H^{2}(K, \overline{K}(C)^{×}/\overline{K}^{×}) \ar[r] \ar[d] & H^{2}(K, Div C) \\  &  & H^{3}(K, \overline{K}^{×})}

By Shapiro’s lemma and the fact that H^{1}(K, ℤ) = 0 we have H^{1}(K, Div C) = 0. It follows by (7) and a diagram chase that there is an exact sequence

(11)  Br(K) → Br(C) → H^{1}(K, Pic C) → H^{3}(K, \overline{K}^{×}).

In fact, had we started from the definition Br(C) = H_{ét}^{2}(C, 𝔾_{m}), then (11) would follow from the Hochschild–Serre spectral sequence.

If C is a smooth curve of genus one with Jacobian E, then taking Galois cohomology of the exact sequence

0 ⟶ Pic^{0} C ⟶ Pic C ⟶^{deg} ℤ ⟶ 0

gives

(12)  ℤ ⟶^{δ} H^{1}(K, E) ⟶ H^{1}(K, Pic C) ⟶ 0

with δ(1) = [C]. If H^{3}(K, \overline{K}^{×}) = 0 then from (11) and (12) we obtain the isomorphism Ψ_{C}.

5. Cyclic extensions

The definition of the map Ψ_{C} in the last section simplifies when we evaluate it on classes split by a cyclic extension L/K. Let G = Gal(L/K) be generated by σ of order n. We recall that for A a G-module, the Tate cohomology groups are

Ĥ^{0}(G, A) = \frac{ker(Δ|A)}{im(N|A)} and Ĥ^{1}(G, A) = \frac{ker(N|A)}{im(Δ|A)}

where Δ = 1 − σ and N = 1 + σ + … + σ^{n−1} satisfy ΔN = NΔ = 0 in ℤ[G].

For b ∈ K^{×} there is a cyclic K-algebra with basis 1, v, …, v^{n−1} as an L-vector space, and multiplication determined by v^{n} = b and vx = σ(x)v for all x ∈ L. We write (L/K, b) for the class of this algebra in Br(K) = H^{2}(K, \overline{K}^{×}). Likewise if f ∈ K(C)^{×} then (L/K, f) is an element of H^{2}(K, \overline{K}(C)^{×}).

Lemma 5.1. Suppose Ξ ∈ Div_{L}^{0} C with N_{L/K}(Ξ) = div(f) for some f ∈ K(C)^{×}. If ξ is the image of Ξ under

Ĥ^{1}(G, Pic_{L}^{0} C) ≅ H^{1}(G, Pic_{L}^{0} C) ⟶^{inf} H^{1}(K, E)

then Ψ_{C}(ξ) = (L/K, f).

Proof. We follow the construction of Ψ_{C} in Section 4. We start with the exact sequence of G-modules

0 → L^{×} → L(C)^{×} → Div_{L} C → Pic_{L} C → 0.

Splitting into short exact sequences, and taking Galois cohomology, gives a diagram as before. The connecting map

Ĥ^{1}(G, Pic_{L} C) → Ĥ^{0}(G, L(C)^{×}/L^{×})

is now given by Ξ ↦ f. Therefore Ψ_{C}(ξ) is the image of f under the map

\frac{K(C)^{×}}{N_{L/K}(L(C)^{×})} = Ĥ^{0}(G, L(C)^{×}) ≅ H^{2}(G, L(C)^{×}) ⟶^{inf} H^{2}(K, \overline{K}(C)^{×}).

This is the cyclic algebra (L/K, f) as required. □

6. Pairs of binary quartics and (2, 2)-forms

Let C be a smooth curve of genus one. First suppose, as in Section 2, that C is defined by a binary quartic g. Then C → ℙ^{1} is a double cover ramified over the 4 roots of g. We write H for the hyperplane section (i.e., fibre of the map C → ℙ^{1}), and ι for the involution on C with Q + ι(Q) ∼ H for all Q ∈ C.

Next we suppose that C ⊂ ℙ^{1} × ℙ^{1} is defined by a (2, 2)-form, i.e., a polynomial f(x_{1}, z_{1}; x_{2}, z_{2}) that is homogeneous of degree 2 in each of the sets of variables x_{1},z_{1} and x_{2},z_{2}. Projecting C to either factor gives a double cover of ℙ^{1}. The corresponding binary quartics are obtained by writing f as a binary quadratic form in one of the sets of variables, and taking its discriminant. We write pr_{1}, pr_{2} : C → ℙ^{1} for the projection maps. Let H_{1}, H_{2} and ι_{1}, ι_{2} be the corresponding hyperplane sections and involutions.

Lemma 6.1. Let C = {f(x_{1}, z_{1}; x_{2}, z_{2}) = 0} ⊂ ℙ^{1} × ℙ^{1} as above.

(i) The composite ι_{1}ι_{2} is translation by some P ∈ E = Jac(C). Moreover the isomorphism Pic^{0}(C) ≅ E sends [H_{1} − H_{2}] ↦ P.

(ii) If H_{i} = pr_{i}^{∗}(1 : 0) then

div(f(x_{1}, z_{1}; 1, 0)/z_{1}^{2}) = H_{2} + ι_{1}^{∗}H_{2} − 2H_{1}.

Proof. (i) If Q ∈ C then ι_{2}Q + ι_{1}ι_{2}Q ∼ H_{1} and Q + ι_{2}Q ∼ H_{2}. Subtracting one from the other gives [ι_{1}ι_{2}Q − Q] = [H_{1} − H_{2}] as required. (ii) The specified rational function on C factors via pr_{1} and is therefore invariant under pull back by ι_{1}. It has a zero at each point in the support of H_{2}, and a double pole at each point in the support of H_{1}. Since there are no other poles, and the divisor has degree 0, it must therefore be as stated. □

Remark 6.2. Lemma 6.1(i) is closely related to Poncelet’s Porism, as described in [13]. Our use of (2, 2)-forms was inspired by the treatment in [1].

We write disc_{k}(f) for the discriminant of f when it is viewed as a binary quadratic form in the kth set of variables.

Lemma 6.3. Let C = {f(x_{1}, z_{1}; x_{2}, z_{2}) = 0} ⊂ ℙ^{1} × ℙ^{1} as above. Let a ∈ K^{×} and let C_{1}, C_{2} be the following quadratic twists of C.

C_{1} :  ay^{2} = disc_{2}(f)
C_{2} :  ay^{2} = disc_{1}(f)

Then Ψ_{C_{1}}([C_{2}]) = (K(\sqrt{a})/K, f(x_{1}, z_{1}; 1, 0)/z_{1}^{2}).

Proof. If a ∈ (K^{×})^{2} then C_{1} and C_{2} are isomorphic over K and so by (9) we have Ψ_{C_{1}}([C_{2}]) = 0. We may therefore suppose that a ∉ (K^{×})^{2}. Let L = K(\sqrt{a}) and G = Gal(L/K) = {1, σ}. We claim there is a divisor Ξ ∈ Div_{L}^{0}(C_{1}) such that

(i) C_{2} is the twist of C_{1} by the class of Ξ in Ĥ^{1}(G, Pic_{L}^{0}(C_{1})), and (ii) N_{L/K}(Ξ) = div(f(x_{1}, z_{1}; 1, 0)/z_{1}^{2}).

Then by Lemma 5.1 we have Ψ_{C_{1}}([C_{2}] − [C_{1}]) = (L/K, f(x_{1}, z_{1}; 1, 0)/z_{1}^{2}). Since Ψ_{C_{1}} is a group homomorphism and Ψ_{C_{1}}([C_{1}]) = 0 this proves the lemma.

We construct Ξ as follows. We factor the projection map pr_{i} : C → ℙ^{1} as

C ⟶^{ϕ_{i}} C_{i} ⟶^{ξ_{i}} ℙ^{1}

where ϕ_{i} is the quadratic twist map (an isomorphism defined over L), and ξ_{i} = (x_{i} : z_{i}) is the natural double cover. Let D_{i} = ξ_{i}^{∗}(1 : 0) and H_{i} = ϕ_{i}^{∗}D_{i} = pr_{i}^{∗}(1 : 0). We put ϕ = ϕ_{1}ϕ_{2}^{−1} and Ξ = ϕ_{∗}D_{2} − D_{1}. We now prove (i) and (ii).

(i) Let ι_{1} and ι_{2} be the involutions on C defined before Lemma 6.1. Since σ(ϕ_{1}) = ϕ_{1}ι_{1} and σ(ϕ_{2}) = ϕ_{2}ι_{2} it follows that σ(ϕ)ϕ^{−1} = ϕ_{1}ι_{1}ι_{2}ϕ_{1}^{−1}. Identifying C and C_{1} via ϕ_{1}, and hence Ξ with H_{2} −H_{1}, it follows by Lemma 6.1 that σ(ϕ)ϕ^{−1} is translation by some P ∈ E = Jac(C_{1}), and the isomorphism Pic^{0}(C_{1}) ≅ E sends [Ξ] ↦ −P. The minus sign does not matter since |G| = 2.

(ii) By Lemma 6.1(ii) with H_{1} = ϕ_{1}^{∗}D_{1} and H_{2} = ϕ_{2}^{∗}D_{2} we have

div(f(x_{1}, z_{1}; 1, 0)/z_{1}^{2}) = ϕ_{1∗}(ϕ_{2}^{∗}D_{2} + ι_{1}^{∗}ϕ_{2}^{∗}D_{2} − 2ϕ_{1}^{∗}D_{1})
= ϕ_{∗}D_{2} + σ(ϕ_{∗}D_{2}) − 2D_{1}
= N_{L/K}(Ξ).  □

7. Triples of binary quartics and (2, 2, 2)-forms

Let E/K be an elliptic curve. An n-covering of E is a pair (C, ν) where C is a smooth curve of genus one, and ν : C → E is a morphism, such that, for some choice of isomorphism ψ : C → E defined over \overline{K}, there is a commutative diagram

\xymatrix{C \ar[d]_{ψ} \ar[dr]^{ν} \\ E \ar[r]_{×n} & E}

The n-coverings of E are parametrised by H^{1}(K, E[n]).

Suppose that C_{1}, C_{2}, C_{3} are 2-coverings of E that sum to zero in H^{1}(K, E[2]). We pick isomorphisms ψ_{i} : C_{i} → E as above, and let ε_{i} = (σ ↦ σ(ψ_{i})ψ_{i}^{−1}) be the corresponding cocycle in Z^{1}(Gal(\overline{K}/K), E[2]). Our hypothesis is that ε_{1} + ε_{2} + ε_{3} is a coboundary. However, by adjusting the choice of ψ_{3}, we may suppose that ε_{1} + ε_{2} + ε_{3} = 0. It may then be checked that the morphism

μ : C_{1} × C_{2} × C_{3} → E
(P_{1}, P_{2}, P_{3}) ↦ ψ_{1}(P_{1}) + ψ_{2}(P_{2}) + ψ_{3}(P_{3})

is defined over K.

Remark 7.1. We are still free to replace ψ_{3} by P ↦ ψ_{3}(P) + T for T ∈ E(K)[2], and for this reason there are #E(K)[2] choices for the map μ.

Suppose further that C_{1}, C_{2}, C_{3} are defined by binary quartics g_{1}, g_{2}, g_{3} with the same invariants I and J. Let π : C_{1} ×C_{2} ×C_{3} → ℙ^{1} ×ℙ^{1} ×ℙ^{1} be the map that projects to the x-coordinates. Then S = π(μ^{−1}(0_{E})) is a surface in ℙ^{1} × ℙ^{1} × ℙ^{1}. Geometrically it is the Kummer surface (E × E)/{±1}.

We write disc_{k}(F) for the discriminant of a (2, 2, 2) form F when it is viewed as a binary quadratic form in the kth set of variables.

Proposition 7.2. The surface S ⊂ ℙ^{1} × ℙ^{1} × ℙ^{1} is defined by a (2, 2, 2)-form F. Moreover we may scale F so that it has coefficients in K, and for all permutations i, j, k of 1, 2, 3 we have disc_{k}(F) = g_{i}g_{j}.

Proof. We first consider the special case where C_{1} = C_{2} = C_{3} = E. Suppose that P_{1}, P_{2}, P_{3} ∈ E satisfy P_{1} + P_{2} + P_{3} = 0_{E}. If we specify the x-coordinates of P_{1} and P_{2}, then in general this leaves two possibilities for the x-coordinate of P_{3}. The exceptional cases are when either P_{1} or P_{2} is a 2-torsion point.

Let Δ_{i} = ψ_{i}^{−1}(E[2]) be the set of ramification points for C_{i} → ℙ^{1}. We identify Δ_{i} with its image in ℙ^{1}, i.e., the set of roots of g_{i}. The observations in the last paragraph show that when we project onto the ith and jth factors, S → ℙ^{1} × ℙ^{1} is a double cover ramified over Δ_{i} × ℙ^{1} and ℙ^{1} × Δ_{j}. This shows that S is defined by a (2, 2, 2)-form F. Moreover disc_{k}(F) = λ_{k}g_{i}g_{j} for some λ_{1}, λ_{2}, λ_{3} ∈ K^{×}. We claim that (i) λ_{3} ∈ (K^{×})^{2} and (ii) λ_{1} = λ_{2} = λ_{3}. It is then clear we may rescale F so that λ_{1} = λ_{2} = λ_{3} = 1.

(i) Let C_{i} have equation y_{i}^{2} = g_{i}(x_{i}, z_{i}). We note that K(S) ⊂ K(C_{1} × C_{2}) is a quadratic extension of K(ℙ^{1} × ℙ^{1}) with Kummer generator

\frac{disc_{3}(F)}{z_{1}^{4}z_{2}^{4}} = \frac{λ_{3}g_{1}(x_{1}, z_{1})g_{2}(x_{2}, z_{2})}{z_{1}^{4}z_{2}^{4}} = λ_{3} (\frac{y_{1}y_{2}}{z_{1}^{2}z_{2}^{2}} )^{2} .

Since this is a square in K(C_{1} × C_{2}) it follows that λ_{3} ∈ (K^{×})^{2}.

(ii) Since g_{1}, g_{2}, g_{3} have the same invariants I and J, we may reduce by the action of SL_{2}(\overline{K}) × SL_{2}(\overline{K}) × SL_{2}(\overline{K}) to the case

g_{1}(x, z) = g_{2}(x, z) = g_{3}(x, z) = x^{3}z − \frac{1}{3}Ixz^{3} − \frac{1}{27}Jz^{4}.

The result then follows by symmetry. □

Corollary 7.3. Let C_{1}, C_{2}, C_{3} and F be as above. If a = g_{3}(1, 0) ≠ 0 then

Ψ_{C_{1}}([C_{2}]) = (K(\sqrt{a})/K, F (x_{1}, z_{1}; 1, 0; 1, 0)/z_{1}^{2}).

Proof. We put f(x_{1}, z_{1}; x_{2}, z_{2}) = F (x_{1}, z_{1}; x_{2}, z_{2}; 1, 0). By Proposition 7.2 we have disc_{1}(f) = ag_{2}(x_{2}, z_{2}) and disc_{2}(f) = ag_{1}(x_{1}, z_{1}). The curves C_{1} and C_{2} are therefore isomorphic to those considered in Lemma 6.3. Applying Lemma 6.3 gives the result. □

8. Computing the (2, 2, 2)-forms

To complete the proof of Theorem 3.1 we must explain how to compute the (2, 2, 2)-form F. As before it is helpful to first consider the special case where C_{1} = C_{2} = C_{3} = E.

Let E be the elliptic curve y^{2} = x^{3} + ax + b. We consider the maps

\xymatrix{E × E × E \ar[r]^{μ} \ar[d]^{π} & E \\ ℙ^{1} × ℙ^{1} × ℙ^{1}}

where μ(P_{1}, P_{2}, P_{3}) = P_{1} + P_{2} + P_{3} and π is the map taking the x-coordinate of each point. An equation for S = π(μ^{−1}(0_{E})) is computed as follows.

Let P_{i} = (x_{i}, y_{i}) for i = 1, 2, 3 be points on E with P_{1} + P_{2} + P_{3} = 0_{E}. These points lie on a line, say y = λx + ν. Then as polynomials in x we have

x^{3} + ax + b − (λx + ν)^{2} = (x − x_{1})(x − x_{2})(x − x_{3}).

Comparing the coefficients of the powers of x we obtain

λ^{2} = s_{1},
2λν = a − s_{2},
ν^{2} = b + s_{3},

where s_{1}, s_{2}, s_{3} are the elementary symmetric polynomials in x_{1}, x_{2}, x_{3}. Eliminating λ and ν gives the equation

(a − s_{2})^{2} − 4s_{1}(b + s_{3}) = 0.

The required (2, 2, 2)-form F is obtained by homogenising this equation, i.e. we replace x_{i} by x_{i}/z_{i} and multiply through by z_{1}^{2}z_{2}^{2}z_{3}^{2}.

Remark 8.1. We have F (x_{1}, 1; x_{2}, 1; x_{3}, 1) = W_{0}x_{3}^{2} − W_{1}x_{3} + W_{2} where

W_{0} = (x_{1} − x_{2})^{2},
W_{1} = 2(x_{1}x_{2} + a)(x_{1} + x_{2}) + 4b,
W_{2} = x_{1}^{2}x_{2}^{2} − 2ax_{1}x_{2} − 4b(x_{1} + x_{2}) + a^{2}.

These are the formulae used in [5, Chapter 17] to show that the height on an elliptic curve is a quadratic form.

We now turn to the general case. So let S ⊂ ℙ^{1} × ℙ^{1} × ℙ^{1} be as in Section 7. Let z(g_{i}) be the cubic invariant, and write H_{1}, H_{2}, H_{3} for the binary quadratic forms (4) over L = K[φ] associated to g_{1}, g_{2}, g_{3}.

Theorem 8.2. If z(g_{1})z(g_{2})z(g_{3}) = m^{2} for some m ∈ L^{×}, and

(13)  \frac{H_{1}H_{2}H_{3}}{m} = F_{0} + F_{1}φ + F_{2}φ^{2}

where F_{0}, F_{1}, F_{2} are (2, 2, 2)-forms defined over K, then S has equation F_{2} = 0.

Proof. Let P_{i} = (x_{i} : y_{i} : z_{i}) ∈ C_{i} for i = 1, 2, 3, with μ(P_{1}, P_{2}, P_{3}) = 0_{E}. Let Q_{i} be the image of P_{i} under the covering map C_{i} → E. By the formulae for the covering map coming from classical invariant theory (see for example [8, Proposition 4.2]), the x-coordinate of Q_{i} is

(14)  ξ_{i} = \frac{3h_{i}(x_{i}, z_{i})}{4g_{i}(x_{i}, z_{i})}.

We recall from Section 3 that

(15)  z(g_{i})\frac{4φg_{i} + h_{i}}{3} = H_{i}^{2}.

By (14), (15) and the equation y_{i}^{2} = g_{i}(x_{i}, z_{i}) for C_{i} we have

ξ_{i} + 3φ = \frac{9H_{i}^{2}}{4z(g_{i})y_{i}^{2}}

and hence

(16)  ∏_{i=1}^{3}(ξ_{i} + 3φ) = (\frac{27H_{1}H_{2}H_{3}}{8my_{1}y_{2}y_{3}} )^{2} .

Since Q_{1} + Q_{2} + Q_{3} = 0_{E} these points lie on a line, say y = λx + ν for some λ, ν ∈ K. Then as a polynomial in x we have

x^{3} − 27Ix − 27J − (λx + ν)^{2} = (x − ξ_{1})(x − ξ_{2})(x − ξ_{3}).

Putting x = −3φ gives

(17)  ∏_{i=1}^{3}(ξ_{i} + 3φ) = (ν − 3λφ)^{2}.

We first suppose E(K)[2] = 0. In this case L is a field, so comparing (16) and (17) we have

\frac{27H_{1}H_{2}H_{3}}{8my_{1}y_{2}y_{3}} = ±(ν − 3λφ),

in L(S). Taking the coefficient of φ^{2} we see that F_{2} vanishes on S. In general there are #E(K)[2] choices for the square root, up to sign, and these correspond to the #E(K)[2] choices in Remark 7.1.

It remains to check that F_{2} is not identically zero. For this we may work over an algebraically closed field. Then by a change of coordinates we may suppose that g_{i} and h_{i} are linear combinations of x_{i}^{4} + z_{i}^{4} and x_{i}^{2}z_{i}^{2}. The singular quartics in this pencil are (x_{i}^{2} − z_{i}^{2})^{2}, (x_{i}^{2} + z_{i}^{2})^{2} and (x_{i}z_{i})^{2}. Since L ≅ K × K × K we may identify H_{i} as a triple of binary quadratic forms. These are non-zero multiples of x_{i}^{2} − z_{i}^{2}, x_{i}^{2} + z_{i}^{2} and x_{i}z_{i}, in this order if we made a suitable change of coordinates. (This last claim may be checked without any calculation if we use stereographic projection to identify the roots of the binary quadratic forms with the vertices of an octahedron, and then rotate the octahedron.) Therefore the space of (2, 2, 2)- forms spanned by F_{0}, F_{1}, F_{2} contains the forms

(x_{1}^{2} − z_{1}^{2})(x_{2}^{2} − z_{2}^{2})(x_{3}^{2} − z_{3}^{2}), (x_{1}^{2} + z_{1}^{2})(x_{2}^{2} + z_{2}^{2})(x_{3}^{2} + z_{3}^{2}), x_{1}z_{1}x_{2}z_{2}x_{3}z_{3}.

Since these are linearly independent, it follows that F_{2} is non-zero. □

Proof of Theorem 3.1. Let F = F_{2} be the equation for S in Theorem 8.2. We specialise the last two sets of variables in (13) to (1, 0). Then comparing with (5) we have F (x, z; 1, 0; 1, 0) = γ_{1}(x, z). By Corollary 7.3 we have

Ψ_{C_{1}}([C_{2}]) = (K(\sqrt{a})/K, γ_{1}(x, z)/z^{2}),

where a = g_{3}(1, 0). Then by (10) we have

⟨[C_{1}], [C_{2}]⟩_{CT} = ∑_{v∈M_{K}} inv_{v}(K_{v}(\sqrt{a})/K_{v}, γ_{1}(x_{v}, z_{v})/z_{v}^{2}).

Subject to identifying μ_{2} = \frac{1}{2}ℤ/ℤ, the Hilbert norm residue symbol is given by

(a, b)_{v} = inv_{v}(K_{v}(\sqrt{a})/K_{v}, b).

This gives the formula in Theorem 3.1, except that we have g_{3}(1, 0) in place of g_{2}(1, 0). As noted in Remark 3.2(v), this change does not matter. □

Remark 8.3. To show that γ_{1}(x, z) is not identically zero we show more generally that F cannot be made to vanish by specialising two of the sets of variables. Indeed, by considering F as given in Remark 8.1, it suffices to show that the polynomials W_{0}, W_{1}, W_{2} never simultaneously vanish. This may be checked by setting x_{1} = x_{2} = x and computing that the resultant of W_{1} and W_{2} is 2^{8}(4a^{3} + 27b^{2})^{2}. This last expression is non-zero, by definition of an elliptic curve.

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 Email address: T.A.Fisher@dpmms.cam.ac.uk