VISIBLE 2-TORSION IN THE TATE-SHAFAREVICH GROUP OF AN ELLIPTIC CURVE

TOM FISHER

Abstract. We show that every pair of 2-torsion elements in the Tate-Shafarevich group of an elliptic curve are visible in the same abelian surface. This was previously only known for a single 2-torsion element. Our result explains some of the original observations on visibility in the paper of Cremona and Mazur.

1. Introduction

Let E be an elliptic curve defined over a number field k. A torsor (or principal homogeneous space) under E is a pair (C, μ) where C is a smooth curve of genus one, and μ : E × C → C is morphism (both defined over k) inducing a simply transitive action on \overline{k}-points. The Weil-Châtelet group H^{1}(k, E) parametrises the torsors under E up to k-isomorphism. The Tate-Shafarevich group

� (E/k) = ker (H^{1}(k, E) → ∏_{v} H^{1}(k_{v}, E))

is the subgroup of torsors that are everywhere locally soluble. This group is conjecturally finite, and when finite, Cassels [7] showed it has order a square. Its non-zero elements correspond to genus one curves with Jacobian E that are counterexamples to the Hasse principle. Mazur introduced the notion of visibility as a way to study elements of � (E/k).

Definition 1.1. An element ξ ∈ H^{1}(k, E) is visible in an abelian variety A if there is an inclusion of abelian varieties ι : E → A such that ξ is in the kernel of the induced map

ι_{∗} : H^{1}(k, E) → H^{1}(k, A).

Equivalently, ξ is visible in A if the corresponding torsor may be realised as a coset of E inside A.

Cremona and Mazur [13, 2] gave examples, for each n ∈ {2, 3, 4, 5}, of elliptic curves E/ℚ where all of � (E/ℚ) ≅ (ℤ/nℤ)^{2} is visible in the same abelian surface. Mazur [20] showed that every 3-torsion element in � (E/k) is visible in an abelian surface. Klenke [19] showed that every 2-torsion element in H^{1}(k, E) is visible in an abelian surface. An argument using restriction of scalars, first recorded by Agashe and Stein [1], shows that every n-torsion element in � (E/k) is visible in an abelian variety of dimension n. Fisher [15, 16] gave examples of 6 and 7torsion elements in � (E/ℚ) that are not visible in an abelian surface, and also of 7-torsion elements that are visible in an abelian threefold.

Date: 29th May 2026.

It follows from the restriction of scalars argument that � (E/k) is finite if and only if all its elements are visible in the same abelian variety.

We prove the following.

Theorem 1.2. Every pair of 2-torsion elements in � (E/k) are visible in the same abelian surface.

The proof works by showing that certain complete intersections of two quadrics in ℙ^{5} satisfy the smooth Hasse principle.

In Section 2 we rephrase Theorem 1.2 as a statement about 2-Selmer groups. In Section 3 we introduce the relevant quadric intersections and describe some of their geometry. In Sections 4 and 5 we give two different proofs that these satisfy the smooth Hasse principle. We complete the proof of Theorem 1.2 in Section 6. In the final two sections we give some examples and discuss a variant of the proof.

We thank Shiva Chidambaram for sharing some data in connection with [4] which prompted us to revisit this problem. A Magma [6] file checking some of the calculations and examples in this paper is available on the author’s webpage.

2. The 2-Selmer group

Let E be an elliptic curve defined over a number field k, and let n ⩾ 2 be an integer. We write E[n] for the n-torsion subgroup. The n-Selmer group

S^{(n)}(E/k) = ker (H^{1}(k, E[n]) → ∏_{v} H^{1}(k_{v}, E))

parametrises the n-coverings of E that are everywhere locally soluble. It is the middle term in the Kummer exact sequence

0 ⟶ E(k)/nE(k) ⟶ S^{(n)}(E/k) ⟶ � (E/k)[n] ⟶ 0.

Lemma 2.1. Let F/k be a second elliptic curve and π : E[n] → F [n] an isomorphism of Gal(\overline{k}/k)-modules. If ξ, η ∈ S^{(n)}(E/k) are in the kernel of the composition

H^{1}(k, E[n]) ⟶^{π_{∗}} H^{1}(k, F [n]) ⟶ H^{1}(k, F)

then their images in � (E/k) are visible in the same abelian surface.

Proof. We identify E[n] and F [n] via π and write Δ = E[n] = F [n]. We embed Δ diagonally in E × F and let A = (E × F)/Δ. There is a commutative diagram

\xymatrix{H^{1}(k, Δ) \ar[d] \ar[r] & H^{1}(k, F) \ar[d] \\ H^{1}(k, E) \ar[r] & H^{1}(k, A).}

We are given that ξ and η are in the kernel of the top horizontal map. By the commutativity of the diagram, their images in � (E/k) are in the kernel of the bottom horizontal map, and so visible in A. □

We now take n = 2. Elements of the 2-Selmer group S^{(2)}(E/k) may be represented by binary quartics. See for example [3, 5, 10, 12, 26].

Definition 2.2. A binary quartic is a homogeneous polynomial f ∈ k[x, z] of degree 4. The invariants of the binary quartic

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

are

(1)  \begin{aligned}I & = 12ae − 3bd + c^{2}, \\ J & = 72ace − 27ad^{2} − 27b^{2}e + 9bcd − 2c^{3}. \end{aligned}

A binary quartic is non-singular if it has no repeated roots (i.e. linear factors) over \overline{k}, equivalently if its discriminant 4I^{3} − J^{2} is non-zero.

A non-singular binary quartic f defines a smooth projective curve of genus one with affine equation y^{2} = f(x, 1). The projective curve is defined either by gluing affine pieces, or directly as the curve with equation y^{2} = f(x, z) in the weighted projective plane ℙ(1, 2, 1). This curve is a 2-covering of its Jacobian

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

A binary quartic is soluble if the corresponding genus one curve has a k-point, and everywhere locally soluble if it has a k_{v}-point for all places v.

Definition 2.3. Let K/k be any field extension. Binary quartics f and g are properly K-equivalent if

f(x, z) = \frac{1}{(ps − qr)^{2}} g(px + rz, qx + sz)

for some p, q, r, s ∈ K with ps − qr ≠ 0.

Binary quartics that are properly k-equivalent have the same invariants. The converse is true for non-singular binary quartics over an algebraically closed field. We may write any elliptic curve E defined over k in the form E_{I,J} for some I, J ∈ k with 4I^{3} − J^{2} ≠ 0. The 2-Selmer group S^{(2)}(E_{I,J}/k) may then be identified with the set of proper k-equivalence classes of everywhere locally soluble binary quartics with invariants I and J.

The genus one curve defined by a non-singular binary quartic f is a double cover of ℙ^{1} ramified over 4 points. These 4 points are the fibre of the 2-covering above the identity on E, and form a torsor under E[2]. In Section 6 we will use the fact that the class of f in H^{1}(k, E[2]) is the class of this torsor.

3. Quadric intersections

We work over a field k with char(k) ≠ 2, 3, with algebraic closure \overline{k}. Let F (x_{1}, …, x_{n}) and G(x_{1}, …, x_{n}) be a pair of quadratic forms in n variables. Let A and B be the n × n symmetric matrices of second partial derivatives of F and G. Then det(λA + μB) is a binary form of degree n. By abuse of notation we write this binary form as det(λF + μG). The next two lemmas are well known. The first is [23, Proposition 2.1].

Lemma 3.1. The quadric intersection X = {F = G = 0} ⊂ ℙ^{n−1} is smooth (and of codimension 2) if and only if det(λF + μG) has no repeated factors.

Proof. By the Jacobian criterion for smoothness, any singular point on X must be a singular point on some quadric in the pencil spanned by F and G. It therefore gives a root of det(λF + μG). This is a repeated root, as may be seen by reducing to the case where F is a quadratic form in x_{2}, …, x_{n} only, and G has no term x_{1}^{2}. Conversely, a root of the binary form is a repeated root if and only if the corresponding quadric has rank less than n − 1, or it has rank exactly n − 1 and its singular point lies on X. In both these cases X has a singular point. □

Lemma 3.2. Let X be as in Lemma 3.1 with n = 4. Then X is a smooth curve of genus one. If rank F = 3 then X is a 2-covering of its Jacobian elliptic curve y^{2} = det(xF + G).

Proof. This is a special case of a standard construction used in 4-descent. See for example [3] or [21]. □

Given a binary quartic form

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

we define the associated quadratic form

Q_{f}(x_{0}, x_{1}, x_{2}) = ax_{0}^{2} + bx_{0}x_{1} + \frac{c}{3}(x_{0}x_{2} + 2x_{1}^{2}) + dx_{1}x_{2} + ex_{2}^{2}.

This choice is made so that

Q_{f}(x^{2}, xz, z^{2}) = f(x, z)

and in addition the following two properties hold.

Lemma 3.3. Let Q(x_{0}, x_{1}, x_{2}) = x_{0}x_{2} − x_{1}^{2}. Let f be a binary quartic form, with invariants I and J as defined by (1).

(i) We have det(xQ − 9Q_{f}) = 2(x^{3} − 27Ix − 27J).

(ii) If f̃(x, z) = f(px + rz, qx + sz) then Qf̃(x_{0}, x_{1}, x_{2}) = Q_{f}(x̃_{0}, x̃_{1}, x̃_{2}) where

x̃_{0} = p^{2}x_{0} + 2prx_{1} + r^{2}x_{2},
x̃_{1} = pqx_{0} + (ps + qr)x_{1} + rsx_{2},
x̃_{2} = q^{2}x_{0} + 2qsx_{1} + s^{2}x_{2}.

Moreover Q(x̃_{0}, x̃_{1}, x̃_{2}) = (ps − qr)^{2}Q(x_{0}, x_{1}, x_{2}).

Proof. A direct calculation. □

Definition 3.4. Let f and g be a pair of non-singular binary quartics with the same invariants I and J. The associated quadric intersection is

X_{f,g} = \begin{cases}u_{0}u_{2} − u_{1}^{2} & = v_{0}v_{2} − v_{1}^{2} \\ Q_{f}(u_{0}, u_{1}, u_{2}) & = Q_{g}(v_{0}, v_{1}, v_{2}) \end{cases}\} ⊂ ℙ^{5}.

In Section 4 we show that quadric intersections of the form X_{f,g} satisfy the smooth Hasse principle. First however we consider their geometry.

By Lemma 3.3(i) the quadratic forms F and G defining X_{f,g} satisfy

(3)  det(xF − 9G) = −4(x^{3} − 27Ix − 27J)^{2}.

In particular the pencil of quadrics defining X_{f,g} has 3 singular fibres. Each is singular along a line of the form

{(λu_{0} : λu_{1} : λu_{2} : μv_{0} : μv_{1} : μv_{2}) | (λ : μ) ∈ ℙ^{1}}

for some (u_{0} : u_{1} : u_{2}), (v_{0} : v_{1} : v_{2}) ∈ ℙ^{2} with u_{0}u_{2} − u_{1}^{2} ≠ 0 and v_{0}v_{2} − v_{1}^{2} ≠ 0. Such a line meets X_{f,g} in 2 distinct points swapped by the involution

(4)  (u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) ↦ (u_{0} : u_{1} : u_{2} : −v_{0} : −v_{1} : −v_{2}).

Therefore X_{f,g} has exactly 6 singular points.

Remark 3.5. We may change coordinates over \overline{k} so that X_{f,g} has equations {x_{1}x_{2} = x_{3}x_{4} = x_{5}x_{6}} ⊂ ℙ^{5}. It follows that X_{f,g} contains exactly 8 planes.

We recall from Section 2 that binary quartics f and g are properly \overline{k}-equivalent if there exist p, q, r, s ∈ \overline{k} with ps − qr ≠ 0 and

f(x, z) = \frac{1}{(ps − qr)^{2}} g(px + rz, qx + sz).

This relation is unchanged if we multiply p, q, r, s through by the same non-zero scalar. It is natural to regard such equivalences as the same. With this convention, there are exactly 4 proper \overline{k}-equivalences relating f and g, and these may be computed by factoring a (4, 4)-form as a product of (1, 1)-forms, as described in [12, Theorem 12]. Corresponding to each proper equivalence there is a plane Λ ⊂ X_{f,g} with equations

(ps − qr)v_{0} = p^{2}u_{0} + 2pru_{1} + r^{2}u_{2},
(ps − qr)v_{1} = pqu_{0} + (ps + qr)u_{1} + rsu_{2},
(ps − qr)v_{2} = q^{2}u_{0} + 2qsu_{1} + s^{2}u_{2}.

Lemma 3.6. The planes corresponding to distinct proper equivalences intersect only at singular points of X_{f,g}.

Proof. By Lemma 3.3(ii) we may reduce to the case where f = g. The proper equivalences then give the action of the 2-torsion of the Jacobian. We may take one to be given by the identity matrix, and the other by a matrix ( \begin{matrix}p & q \\ r & s \end{matrix}). Since the latter has order 2 in PGL_{2}, it has trace 0. A calculation then shows that the planes have unique point of intersection

(5)  (u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) = (−r : p : q : −r : p : q)

and that this is a singular point on X_{f,f} = X_{f,g}. □

The next lemma shows that it is easy to find a k-rational conic on X_{f,g}. By a conic, we shall always mean a smooth conic.

Lemma 3.7. Let F = u_{0}u_{2} −u_{1}^{2} −v_{0}v_{2} +v_{1}^{2} be the first of the two quadrics defining X_{f,g}. Then there are two algebraic families of planes contained in {F = 0}, each parametrised by ℙ^{3}. The planes in each family that meet X_{f,g} in a conic are parametrised by the complement of 4 planes in ℙ^{3}.

Proof. We change coordinates over k so that F = z_{12}z_{34}−z_{13}z_{24}+z_{14}z_{23}, this being the formula for the Pfaffian of a 4 × 4 alternating matrix Z. The two algebraic families are the orbits of the planes {z_{12} = z_{13} = z_{14} = 0} and {z_{23} = z_{24} = z_{34} = 0} under the action of GL_{4} via A : Z ↦ AZA^{T}. We may describe one of these families as follows. Let Φ : ℙ^{3} × ℙ^{3} − → ℙ^{5} be the rational map given by

((x_{1} : x_{2} : x_{3} :x_{4}), (y_{1} : y_{2} : y_{3} : y_{4}))
↦ (z_{12} : …: z_{34}) = (x_{1}y_{2} − x_{2}y_{1} : …: x_{3}y_{4} − x_{4}y_{3}).

Then for each point x ∈ ℙ^{3} the image of y ↦ Φ(x, y) is a plane Π_{x} ⊂ {F = 0}.

As suggested by Remark 3.5 we now change coordinates over \overline{k} so that X_{f,g} is defined by F = z_{12}z_{34} − z_{13}z_{24} + z_{14}z_{23} and G = λz_{12}z_{34} + μz_{13}z_{24} where λ, μ ∈ \overline{k} with λ, μ, λ + μ ≠ 0. After substituting z_{ij} = x_{i}y_{j} − x_{j}y_{i} into G, its matrix of second partial derivatives (with respect to y_{1}, …, y_{4}) is

(\begin{matrix}0 & μx_{3}x_{4} & λx_{2}x_{4} & −(λ + μ)x_{2}x_{3} \\ μx_{3}x_{4} & 0 & −(λ + μ)x_{1}x_{4} & λx_{1}x_{3} \\ λx_{2}x_{4} & −(λ + μ)x_{1}x_{4} & 0 & μx_{1}x_{2} \\ −(λ + μ)x_{2}x_{3} & λx_{1}x_{3} & μx_{1}x_{2} & 0 \end{matrix}) .

The plane Π_{x} meets X_{f,g} in a conic if and only if this matrix has rank 3. This is the case for all x ∈ ℙ^{3} \ {x_{1}x_{2}x_{3}x_{4} = 0}. □

4. The Hasse principle

We take k a number field, and show that the quadric intersections X_{f,g} in Definition 3.4 satisfy the smooth Hasse principle. (The Hasse principle itself is not true. See Example 7.2 for a counterexample.)

Theorem 4.1. If X_{f,g} has a smooth k_{v}-point for all places v then it has a smooth k-point.

We know by Lemma 3.7 that X_{f,g} contains a k-rational conic. So if X_{f,g} were smooth then it would satisfy the Hasse principle by a theorem of Salberger [24]. We initially tried to adapt his proof, to allow for the fact that X_{f,g} is singular. However, we found that a simpler argument is possible in our situation. In particular, we appeal to standard facts about del Pezzo surfaces, instead of Salberger’s results on conic bundle surfaces.

Remark 4.2. If E_{I,J} has no k-rational 2-torsion points then (3) shows that X_{f,g} is a quadric intersection of type (F), as defined by Colliot Thélène, Sansuc and Swinnerton-Dyer. This case is specifically excluded from [8, Theorem 3.20]. According to [8, Remark 3.20.1] it is not known in general whether quadric intersections of this type have trivial Brauer-Manin obstruction.

We have found two different proofs of Theorem 4.1. We give details of both, since the first motivates the second. Readers uninterested in this motivation may skip to the start of the next section.

We start our first proof of Theorem 4.1 by showing that every smooth hyperplane section of X_{f,g} satisfies the Hasse principle.

Lemma 4.3. Let H ⊂ ℙ^{5} be a k-rational hyperplane. If X_{f,g} ∩ H is a smooth surface then it satisfies the Hasse principle.

Proof. We are given that Y = X_{f,g} ∩ H is a smooth intersection of two quadrics in ℙ^{4}. It is therefore a del Pezzo surface of degree 4. As explained in Section 3 the proper equivalences between f and g determine planes Λ_{1}, …, Λ_{4} ⊂ X_{f,g}. We cannot have Λ_{i} ⊂ H as this would contradict that Y is smooth. Intersecting with H we therefore get lines ℓ_{1}, …, ℓ_{4} ⊂ Y. These lines are jointly defined over k, by which we mean that they are permuted by the action of Gal(\overline{k}/k). By Lemma 3.6 they are disjoint. We may therefore blow them down to give a del Pezzo surface of degree 8. It is well known that del Pezzo surfaces of degree 8 satisfy the Hasse principle. This completes the proof. □

Remark 4.4. Applying the involution (4) to the planes Λ_{1}, …, Λ_{4} gives a further 4 planes on X_{f,g} and hence a further 4 lines ℓ′_{1}, …, ℓ′_{4} on the del Pezzo surface Y. It may be checked using Remark 3.5 that ℓ_{i} meets ℓ′_{j} if and only if i ≠ j. The two sets of four lines therefore constitute a “double-four” in the terminology of Swinnerton-Dyer [25, Section 7]. It follows that the del Pezzo surface of degree 8 (considered at the end of the last proof) has no −1-curves, and so is isomorphic over \overline{k} to ℙ^{1} × ℙ^{1} (instead of ℙ^{2} blown up at one point).

We could use Bertini’s theorem to choose the hyperplane H in Lemma 4.3. However the following more explicit lemma simplifies the next steps.

Lemma 4.5. Let Π ⊂ {u_{0}u_{2} − u_{1}^{2} = v_{0}v_{2} − v_{2}^{2}} ⊂ ℙ^{5} be a k-rational plane. Let H ⊂ ℙ^{5} be a k-rational hyperplane that contains Π, but does not contain any of the singular points on X_{f,g}. Then the surface X_{f,g} ∩ H is either smooth or has a k-rational singular point.

Proof. The pencil of quadrics defining X_{f,g} contains F = u_{0}u_{2} − u_{1}^{2} − v_{0}v_{2} + v_{2}^{2} of rank 6, and three singular quadrics G_{1}, G_{2}, G_{3} each of rank 4. We claim that on restricting to the hyperplane H, each of these quadrics has rank 4. Indeed the restriction of F has rank at most 4 since Π ⊂ {F = 0}, and rank at least 4 since the rank can drop by at most 2. If the rank of G_{i} dropped then H would have to contain the singular locus of G_{i}. However this singular locus is a line meeting X_{f,g} in two singular points. The claim follows since we assumed that H does not contain any of the singular points on X_{f,g}.

The singular quadrics in the pencil defining X_{f,g} ∩ H correspond to the roots of a binary quintic form. We may account for 4 of the roots as corresponding to the restrictions of F, G_{1}, G_{2}, G_{3}. If any of these are repeated roots then the corresponding rank 4 quadric is defined over k, and its singular point is a krational singular point on X_{f,g} ∩ H. The same argument applies when the binary quintic form is identically zero (a case also covered by [8, Lemma 1.14]), provided we select the restriction of F as our rank 4 quadric, the problem with G_{1}, G_{2}, G_{3} being that these might not be defined over k. In the remaining case the binary quintic form has distinct roots, in which case X_{f,g}∩H is smooth by Lemma 3.1. □

The following lemma will help us choose the hyperplane H in Lemma 4.5 so that X_{f,g} ∩ H is everywhere locally soluble.

Lemma 4.6. Let X = {F = G = 0} ⊂ ℙ^{5} be a quadric intersection. Suppose that rank F ⩾ 5 and Π ⊂ {F = 0} ⊂ ℙ^{5} is a plane. Then for every P ∈ X_{sm} \ Π there is a hyperplane H containing both Π and P such that P is a smooth point on the intersection X ∩ H.

Proof. We change coordinates so that Π = {(∗ : ∗ : ∗ : 0 : 0 : 0)} and P = (0 : 0 : 0 : 1 : 0 : 0). A hyperplane of the required form is given by H = {λx_{5} + μx_{6} = 0}, provided that the linear form λx_{5} + μx_{6} does not vanish on T_{P} X. So we are done unless T_{P} X = {x_{5} = x_{6} = 0}. However in that case the first 4 partial derivatives of F vanish at P. This implies that F vanishes on the linear span of Π and P, and so rank F ⩽ 4. This contradicts our assumption that rank F ⩾ 5. □

The remainder of the proof is based on arguments in [24, Section 2].

Proof of Theorem 4.1. Let X_{f,g} ⊂ ℙ^{5} be as given in Definition 3.4. By Lemma 3.7 we may pick a k-rational plane Π ⊂ {u_{0}u_{2}−u_{1}^{2} = v_{0}v_{2}−v_{2}^{2}} such that C = X_{f,g} ∩Π is a conic. Let S be the finite set of places v with C(k_{v}) = ∅.

For each place v ∈ S, we use Lemma 4.6, and our assumption that X_{f,g} has a smooth k_{v}-point, to pick a k_{v}-rational hyperplane H_{v} containing Π and meeting X_{f,g} transversely at a k_{v}-point. The implicit function theorem (specifically [8, Proposition 6.2]) shows that there is a v-adic neighbourhood U_{v} of H_{v} in (ℙ^{5})^{∨}(k_{v}) such that every hyperplane in U_{v} meets X_{f,g} transversely at a k_{v}-point. Since C is smooth, none of the singular points of X_{f,g} are contained in Π. The hyperplanes containing Π, but not containing any of the 6 singular points on X_{f,g}, form a Zariski open subset W ⊂ ℙ^{2} (the complement of at most 6 lines). By weak approximation we may pick a hyperplane H in W (k) that belongs to U_{v} for each v ∈ S. We claim that (X_{f,g} ∩ H)(k_{v}) ≠ ∅ for all places v. Indeed, this is true for places v ∉ S since C ⊂ X_{f,g} ∩ H and it is true for places v ∈ S since H ∈ U_{v}.

We know by Lemma 4.5 that X_{f,g} ∩ H is either smooth or has a k-point. It follows by Lemma 4.3 that X_{f,g} ∩ H has a k-point. By our choice of H, this is not a singular point on X_{f,g}. □

Remark 4.7. The conclusion of Theorem 4.1 may be strengthened to say that the k-points on X_{f,g} are Zariski dense. Indeed, in Lemma 4.3 the corresponding strengthening follows from standard facts about del Pezzo surfaces, and in the proof just given there are clearly infinitely many choices for the hyperplane H that satisfy the required hypotheses.

5. An alternative proof

In the previous section we proved that the quadratic intersections X_{f,g} satisfy the smooth Hasse principle. Our proof worked by considering hyperplane sections which were del Pezzo surfaces. However it turns out that a more direct approach is possible.

Proposition 5.1. Let X = X_{f,g} be as in Definition 3.4. Let Λ_{1}, …, Λ_{4} be the planes contained in X corresponding to the proper equivalences between f and g, and let H be the hyperplane section. Then the linear system |3H−Λ_{1}−Λ_{2}−Λ_{3}−Λ_{4}| defines a birational map ϕ : X −→ Y where Y ⊂ ℙ^{9} is a Brauer-Severi threefold. Moreover, the birational map ϕ is regular at all smooth points of X.

Proof. The planes Λ_{1}, …, Λ_{4} are jointly defined over k, by which we mean that they are permuted by the action of Gal(\overline{k}/k). Having made this observation, we are free for the remainder of the proof to work over \overline{k}. In particular, we make the change of coordinates suggested in Remark 3.5.

We label our coordinates on ℙ^{5} as x_{I} = x_{ij} where I = {i, j} runs over the subsets of {1, 2, 3, 4} of size 2. By Lemma 3.6 we may assume that

X = {x_{12}x_{34} = x_{13}x_{24} = x_{14}x_{23}} ⊂ ℙ^{5}

and each Λ_{i} is defined by the vanishing of the x_{I} with i ∈ I. The space of cubic forms vanishing on Λ_{1} ∪ … ∪ Λ_{4} is spanned by the monomials x_{I}x_{J}x_{K} with #(I ∪ J ∪ K) = 4. Inside this space, a complement to the space of cubic forms vanishing on X is spanned by the entries on and above the leading diagonal of the following symmetric matrix, where for simplicity we put α = x_{12}x_{34}.

(\begin{matrix}x_{12}x_{13}x_{14} & αx_{12} & αx_{13} & αx_{14} \\ αx_{12} & x_{12}x_{23}x_{24} & αx_{23} & αx_{24} \\ αx_{13} & αx_{23} & x_{13}x_{23}x_{34} & αx_{34} \\ αx_{14} & αx_{24} & αx_{34} & x_{14}x_{24}x_{34} \end{matrix})

This matrix defines a rational map ϕ : X −→ Y where Y ⊂ ℙ^{9} is the locus of all 4 × 4 symmetric matrices of rank 1. Let ψ : ℙ^{3} −→ X be given by

(6)  (x_{1} : x_{2} : x_{3} : x_{4}) ↦ (x_{12} : …: x_{34}) = (x_{1}x_{2} : …: x_{3}x_{4}).

The composite ϕ ∘ ψ : ℙ^{3} → Y is the 2-uple embedding, which is clearly an isomorphism. It follows that ϕ is birational and Y is a Brauer-Severi variety. The final statement is checked by direct calculation. □

Our alternative proof of Theorem 4.1 is immediate from Proposition 5.1 and the Hasse principle for Brauer-Severi varieties. Remark 4.7 is also immediate.

6. Application to visibility

The results of Section 2 reduce the proof of Theorem 1.2 to showing that if f and g are non-singular binary quartics with the same invariants I and J, and both are everywhere locally soluble, then their classes ξ and η in S^{(2)}(E_{I,J}/k) satisfy the hypotheses of Lemma 2.1.

Lemma 6.1. Let f and g be non-singular binary quartics over k with the same invariants I and J. If both f and g are everywhere locally soluble then X_{f,g} has a smooth k_{v}-point for all places v.

Proof. By assumption there exist x, y, z, X, Y, Z ∈ k_{v} with y^{2} = f(x, z) and Y^{2} = g(X, Z). Since any smooth curve with a k_{v}-point has infinitely many k_{v}-points we may further assume that yY ≠ 0. Then X_{f,g} has k_{v}-point

(u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) = (x^{2}Y : xzY : z^{2}Y : X^{2}y : XZy : Z^{2}y).

This is a smooth point since, as noted in Section 3, the singular points on X_{f,g} all have u_{0}u_{2} − u_{1}^{2} ≠ 0. □

Lemma 6.1 and Theorem 4.1 together show that X_{f,g}(k) ≠ ∅. The proof of Theorem 1.2 is completed by the following proposition. In the case that E has a rational 2-torsion point, we also appeal to Remark 4.7.

Proposition 6.2. Let f and g be non-singular binary quartics over k with the same invariants I and J. Let E = E_{I,J} be given by (2). Suppose that either

(i) X_{f,g}(k) ≠ ∅ and E(k)[2] = 0, or (ii) the k-points on X_{f,g} are Zariski dense.

Then there exists an elliptic curve F over k and π : E[2] → F [2] an isomorphism of Gal(\overline{k}/k)-modules such that the classes of f and g are in the kernel of the composition

(7)  H^{1}(k, E[2]) ⟶^{π_{∗}} H^{1}(k, F [2]) ⟶ H^{1}(k, F).

Proof. Let (u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) be a k-point on X_{f,g}. We put

(8)  λ = u_{0}u_{2} − u_{1}^{2} and μ = Q_{f}(u_{0}, u_{1}, u_{2}).

If λ = μ = 0 then either f or g has a k-rational root (i.e. linear factor) and so is trivial in H^{1}(k, E[2]). We comment on this case at the end of the proof. So for now we may suppose that (λ : μ) ∈ ℙ^{1}(k). We pick λ̃, μ̃ ∈ k with λμ̃ − μλ̃ = 1.

We consider the quadric intersection

C_{f} = \begin{cases}μ(x_{0}x_{2} − x_{1}^{2}) − λQ_{f}(x_{0}, x_{1}, x_{2}) & = 0 \\ μ̃(x_{0}x_{2} − x_{1}^{2}) − λ̃Q_{f}(x_{0}, x_{1}, x_{2}) & = y^{2} \end{cases}\} ⊂ ℙ^{3}.

If E(k)[2] = 0 then by Lemma 3.3(i) and (2) the first quadric has rank 3. If E(k)[2] ≠ 0 then we use our additional hypothesis (ii) to adjust our choice of k-point so that the same conclusion holds. In other words, our threefold X_{f,g} is fibred over the modular curve X_{E}(2) ≅ ℙ^{1} (which parametrises elliptic curves 2-congruent to E) and we want to make sure that our k-point is not above a cusp.

Lemmas 3.2 and 3.3(i) now show that C_{f} is a 2-covering of the elliptic curve

(9) F : y^{2} = −(9μx + 9μ̃)^{3} + 27I(9μx + 9μ̃)(λx + λ̃)^{2} + 27J(λx + λ̃)^{3}.

An explicit isomorphism of Gal(\overline{k}/k)-modules π : E[2] → F [2] is given, relative to the Weierstrass equations (2) and (9), by

(10)  (x, 0) ↦ (−(λ̃x − 9μ̃)/(λx − 9μ), 0).

The first quadratic form in the definition of C_{f} defines a conic with k-rational point (x_{0} : x_{1} : x_{2}) = (u_{0} : u_{1} : u_{2}). By parametrising this conic, we may rewrite C_{f} as a double cover of ℙ^{1}. The set of ramification points, as a set with Gal(\overline{k}/k)- action, is isomorphic to

(11)  {x_{0}x_{2} − x_{1}^{2} = Q_{f}(x_{0}, x_{1}, x_{2}) = 0} ⊂ ℙ^{2},

and hence also isomorphic to the set of ramification points for y^{2} = f(x, z). We upgrade this to an isomorphism of torsors, when E[2] and F [2] are identified via π. To do this we first note that E[2] \ {0} and F [2] \ {0} are naturally in bijection with the singular fibres of the pencil of quadrics defining (11). This gives a more conceptual explanation for (10). Next, since each singular fibre is a pair of lines, each element of E[2] \ {0} = F [2] \ {0} naturally determines a partition of the 4 points (11) into 2 subsets of size 2. This determines the torsor structure.

Our discussion shows that the induced map π_{∗} : H^{1}(k, E[2]) → H^{1}(k, F [2]) sends the class of y^{2} = f(x, z) (a 2-covering of E) to the class of C_{f} (a 2-covering of F). The latter has trivial image in H^{1}(k, F) since C_{f} has k-rational point (x_{0} : x_{1} : x_{2} : y) = (u_{0} : u_{1} : u_{2} : 1). The class of f is therefore in the kernel of the composition (7).

The proof is completed by repeating the same arguments for the binary quartic g. In view of the definition of X_{f,g} we may rewrite (8) as

λ = v_{0}v_{2} − v_{1}^{2} and μ = Q_{g}(v_{0}, v_{1}, v_{2}).

We then obtain a 2-covering

C_{g} = \begin{cases}μ(x_{0}x_{2} − x_{1}^{2}) − λQ_{g}(x_{0}, x_{1}, x_{2}) & = 0 \\ μ̃(x_{0}x_{2} − x_{1}^{2}) − λ̃Q_{g}(x_{0}, x_{1}, x_{2}) & = y^{2} \end{cases}\} ⊂ ℙ^{3}

of the same elliptic curve F. The isomorphism π : E[2] → F [2] is given by (10) and so is also unchanged. Since C_{g} has k-rational point (x_{0} : x_{1} : x_{2} : y) = (v_{0} : v_{1} : v_{2} : 1), its class in H^{1}(k, F [2]) has trivial image in H^{1}(k, F). The class of g is therefore in the kernel of the composition (7).

Finally we remark that if one of the binary quartics (say g) is trivial in H^{1}(k, E[2]), then the above argument is only needed for the other one (say f). For this we start (following what is essentially the proof of Klenke [19]) with any sufficiently general k-point (u_{0} : u_{1} : u_{2}) on ℙ^{2} in place of a k-point on X_{f,g}. This avoids the problem that the latter could have u_{0} = u_{1} = u_{2} = 0. □

7. Examples

We give an example to illustrate the proof of Theorem 1.2.

Example 7.1. Let E/ℚ be the elliptic curve

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

labelled 571a1 in [9]. A 2-descent shows that S^{(2)}(E/ℚ) ≅ (ℤ/2ℤ)^{2}, and its nonzero elements are represented by

f(x, z) = −11x^{4} + 68x^{3}z − 52x^{2}z^{2} − 164xz^{3} − 64z^{4},
g(x, z) = −4x^{4} − 60x^{3}z − 232x^{2}z^{2} − 52xz^{3} − 3z^{4},
h(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. In fact it may be shown (see [18]) that E(ℚ) = 0 and � (E/ℚ) ≅ (ℤ/2ℤ)^{2}.

On the quadric intersection X_{f,g} ⊂ ℙ^{5} we find the point

(u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) = (3 : 1 : 0 : −1 : 0 : 1).

Following the proof of Proposition 6.2, with (λ, μ, λ̃, μ̃) = (−1, 211/3, 0, −1), shows that all of � (E/ℚ) is visible in an abelian surface isogenous to E ×F where F/ℚ is the elliptic curve

y^{2} + y = x^{3} + x^{2} − 4x + 2

labelled 571b1 in [9]. This is the first pair of elliptic curves in [13, Table 2].

We now give examples to show that our proof of Theorem 1.2 for elements of � (E/ℚ)[2] does not extend to elements of H^{1}(ℚ, E)[2]. In other words, we really do need the hypothesis in Lemma 6.1 that both binary quartics are everywhere locally soluble.

Example 7.2. Let E/ℚ be the elliptic curve y^{2} = (x − 1)(x − 3)(x + 3) labelled 192a2 in [9]. The non-singular binary quartics

f(x, z) = −x^{4} − 4x^{2}z^{2} − z^{4},
g(x, z) = x^{4} + 2x^{3}z + 2x^{2}z^{2} + 4xz^{3} + 4z^{4},

have the same invariants I = 28 and J = −160, and represent 2-coverings of E. The associated quadric intersection is

X_{f,g} = \begin{cases}(u_{0} + u_{2})^{2} + 2u_{1}^{2} + (v_{0} + v_{1})^{2} + (v_{1} + 2v_{2})^{2} & = 0 \\ (u_{0} − u_{2})^{2} + 6u_{1}^{2} + (v_{0} + v_{1} + 2v_{2})^{2} − 3v_{1}^{2} & = 0 \end{cases}\} ⊂ ℙ^{5}.

The 6 singular points on X_{f,g} are (\sqrt{2} : 0 : −\sqrt{2} : 2 : −2 : 1), (\sqrt{−2} : 0 : \sqrt{−2} : −2 : 0 : 1), (0 : \sqrt{−1} : 0 : 2 : −1 : 1) and their Galois conjugates. For p ⩾ 5 a prime, the genus one curves y^{2} = f(x, z) and y^{2} = g(x, z) have good reduction, and so are soluble over ℚ_{p}. It follows by Lemma 6.1 that X_{f,g} has a smooth ℚ_{p}-point. There is also a smooth ℚ_{2}-point given by

(u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) = (α^{2} − 1 : 1 : 0 : α^{2} − 1 : α : 1)

where α ∈ ℤ_{2} is a root of x^{4} + x^{3} − x^{2} + x + 4 = 0. However, our first and second equations for X_{f,g} show that X_{f,g}(ℝ) and X_{f,g}(ℚ_{3}) consist only of the singular points defined over ℚ(\sqrt{2}) and ℚ(\sqrt{−2}) respectively.

In conclusion X_{f,g} is a counterexample to the Hasse principle, but not the smooth Hasse principle.

Finally we give some examples with X_{f,g}(ℚ_{p}) = ∅.

Proposition 7.3. Let p ⩾ 7 be a prime with p ≡ 2, 3 (mod 5). The binary quartics

f(x, z) = 20x^{2}z^{2} + p(x^{4} + 30x^{2}z^{2} + 5z^{4}),
g(x, z) = (x^{2} − 5z^{2})^{2} + p(x^{4} + 4x^{3}z − 30x^{2}z^{2} + 20xz^{3} + 25z^{4}),

are non-singular and have the same invariants. However, the associated quadric intersection X_{f,g} ⊂ ℙ^{5} has no ℚ_{p}-points.

Proof. Both f and g have invariants

I = 2^{4} · 5 · (12p^{2} + 15p + 5),
J = −2^{6} · 5^{2} · (3p + 2)(9p^{2} + 15p + 5),

and discriminant

4I^{3} − J^{2} = 2^{12} · 3^{3} · 5^{3} · p^{2}(11p^{2} + 15p + 5)^{2} ≠ 0.

One of the quadrics in the pencil defining X_{f,g} ⊂ ℙ^{5} reduces mod p to 20u_{1}^{2} −(v_{0} − 5v_{2})^{2}. So if (u_{0} : u_{1} : u_{2} : v_{0} : v_{1} : v_{2}) is a ℚ_{p}-point on X_{f,g}, with all coordinates in ℤ_{p}, but not all in pℤ_{p}, then (since 5 is a quadratic nonresidue mod p) we must have u_{1} ≡ v_{0} − 5v_{2} ≡ 0 (mod p). We may therefore write

(12)  (u_{0}, u_{1}, u_{2}, v_{0}, v_{1}, v_{2}) = (x_{1}, px_{5}, x_{2}, 5x_{4} + px_{6}, x_{3}, x_{4})

for some x_{1}, …, x_{6} ∈ ℤ_{p}. These then satisfy

x_{1}x_{2} + x_{3}^{2} − 5x_{4}^{2} ≡ 0 (mod p),
x_{1}^{2} + 10x_{1}x_{2} + 5x_{2}^{2} + 20x_{3}^{2} − 40x_{3}x_{4} ≡ 0 (mod p).

Since p ≡ 2, 3 (mod 5) we have [𝔽_{p}(ζ) : 𝔽_{p}] = 4, where ζ is a primitive 5th root of unity. We put \sqrt{5} = 1 + 2(ζ + ζ^{4}), and write x̃_{i} for x_{i} (mod p). Taking a linear combination of the last two equations gives

(x̃_{1} + \sqrt{5}x̃_{2})^{2} = (ζ − ζ^{4})^{2}(2x̃_{3} − (5 − \sqrt{5})x̃_{4})^{2}.

It follows that x_{1}, …, x_{4} all vanish mod p. Combined with (12) this gives that u_{0}, u_{1}, u_{2}, v_{0}, v_{1}, v_{2} all vanish mod p, contradicting our earlier assumption. □

8. Relation to the obstruction map

The obstruction map Ob_{n} : H^{1}(k, E[n]) → Br(k) is defined in [11, 22]. Taking n = 2, the significance of this map is that a 2-covering ξ ∈ H^{1}(k, E[2]) may be represented by a binary quartic if and only if Ob_{2}(ξ) = 0.

Let ξ, η ∈ H^{1}(k, E[2]) be represented by binary quartics f and g. We saw in Section 5 that X_{f,g} is birational to a Brauer-Severi variety Y. It is natural to ask: what is the class of Y in Br(k)? We guess it should be Ob_{2}(ξ + η). This suggested the following result, which gives an alternative (more computational) proof of Theorem 1.2. We work over a field k of characteristic 0.

Proposition 8.1. Let E/k be an elliptic curve. Let ξ, η ∈ H^{1}(k, E[2]) with Ob_{2}(ξ) = Ob_{2}(η) = 0. The following statements are equivalent.

(i) There exists an elliptic curve F/k, and π : E[2] → F [2] an isomorphism of Gal(\overline{k}/k)-modules, such that ξ and η are in the kernel of the composition

H^{1}(k, E[2]) ⟶^{π_{∗}} H^{1}(k, F [2]) ⟶ H^{1}(k, F).

(ii) Ob_{2}(ξ + η) = 0.

Proof. “(i) ⇒ (ii)” We write ∪ for the Tate pairing

H^{1}(k, E[n]) × H^{1}(k, E[n]) → Br(k)

given by the Weil pairing and cup product. It is known (see [22, 27]) that for any ξ, η ∈ H^{1}(k, E[n]) we have

ξ ∪ η = Ob_{n}(ξ + η) − Ob_{n}(ξ) − Ob_{n}(η).

Compatibility with another pairing, also due to Tate (see for example [14, Proposition 2.1] or [22, Section 5]), shows that ξ ∪ η = 0 whenever ξ and η are both in the image of the connecting map E(k)/nE(k) → H^{1}(k, E[n]).

As in the statement of the proposition, we now take ξ, η ∈ H^{1}(k, E[2]) with Ob_{2}(ξ) = Ob_{2}(η) = 0. Then Ob_{2}(ξ + η) = ξ ∪ η = π_{∗}(ξ) ∪ π_{∗}(η) = 0, where the first and last equality follow from the two facts we just quoted, and the middle equality holds since π : E[2] → F [2] preserves the Weil pairing. This must be the case since there is only one nondegenerate alternating pairing on (ℤ/2ℤ)^{2}.

“(ii) ⇒ (i)” We are given that Ob_{2}(ξ) = Ob_{2}(η) = Ob_{2}(ξ + η) = 0. We may therefore represent ξ, η, ξ + η by binary quartics f, g, h, all with the same invariants. Our aim is to show that the k-points on X_{f,g} are Zariski dense, since (i) then follows by Proposition 6.2. We are free to replace f, g, h by λf, λg, λh for any λ ∈ k^{×}, as this does not change X_{f,g}. We may therefore suppose that y^{2} = h(x, z) has a k-point, equivalently y^{2} = f(x, z) and y^{2} = g(x, z) are k-isomorphic as curves. Let C be this genus one curve, equipped with two different maps C → ℙ^{1}. For the same reasons as we explained at the end of the proof of Proposition 6.2, we may ignore the case where f and g are equivalent. The image of the map C → ℙ^{1} × ℙ^{1} is therefore defined by a (2, 2)-form:

F (x_{1}, x_{2}; y_{1}, y_{2}) = (x_{1}^{2} x_{1}x_{2} x_{2}^{2}) (\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}) (\begin{matrix}y_{1}^{2} \\ y_{1}y_{2} \\ y_{2}^{2} \end{matrix}) .

We may compute this (2, 2)-form by specialising the (2, 2, 2)-form in [17, Theorem 8.2] at a k-point on y^{2} = h(x, z). The binary quartics f(x_{1}, x_{2}) and g(y_{1}, y_{2}) are the discriminant of F, when this is viewed as a binary quadratic form in the other set of variables. A short calculation reveals that

X_{f,g} = \begin{cases}u_{0}u_{2} − u_{1}^{2} & = v_{0}v_{2} − v_{1}^{2} \\ l_{2}^{2} − 4l_{1}l_{3} & = m_{2}^{2} − 4m_{1}m_{3} \end{cases}\} ⊂ ℙ^{5}

where

(\begin{matrix}l_{1} \\ l_{2} \\ l_{3} \end{matrix}) = (\begin{matrix}a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{matrix}) (\begin{matrix}u_{0} \\ u_{1} \\ u_{2} \end{matrix}) and (\begin{matrix}m_{1} \\ m_{2} \\ m_{3} \end{matrix}) = (\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}) (\begin{matrix}v_{0} \\ v_{1} \\ v_{2} \end{matrix}) .

The rational parametrisation (a twisted form of (6))

u_{0} = −a_{21}x^{2} − a_{22}xy − a_{23}y^{2} − 2a_{31}xz − a_{32}(xt + yz) − 2a_{33}yt,
u_{1} = a_{11}x^{2} + a_{12}xy + a_{13}y^{2} − a_{31}z^{2} − a_{32}zt − a_{33}t^{2},
u_{2} = a_{21}z^{2} + a_{22}zt + a_{23}t^{2} + 2a_{11}xz + a_{12}(xt + yz) + 2a_{13}yt,
v_{0} = −a_{12}x^{2} − a_{22}xz − a_{32}z^{2} − 2a_{13}xy − a_{23}(xt + yz) − 2a_{33}zt,
v_{1} = a_{11}x^{2} + a_{21}xz + a_{31}z^{2} − a_{13}y^{2} − a_{23}yt − a_{33}t^{2},
v_{2} = a_{12}y^{2} + a_{22}yt + a_{32}t^{2} + 2a_{11}xy + a_{21}(xt + yz) + 2a_{31}zt,

then shows that the k-points on X_{f,g} are Zariski dense. □

We now take k a number field. If ξ, η ∈ S^{(2)}(E/k) then a standard argument, using the local-global principle for the Brauer group, shows that Ob_{2}(ξ) = Ob_{2}(η) = Ob_{2}(ξ + η) = 0. Combining Lemma 2.1 and Proposition 8.1 therefore gives an alternative (more computational) proof of Theorem 1.2.

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

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