A counterexample to a conjecture of Selmer

Tom Fisher

6 November 2002

Abstract We present a counterexample to a conjecture cited by Cassels [CaI] and attributed to Selmer. The issues raised have been given new significance by the recent work of Heath-Brown [HB] and Swinnerton-Dyer [SwD] on the arithmetic of diagonal cubic surfaces.

1 Introduction

Let E be an elliptic curve over a number field k, with complex multiplication by Z[ω] where ω is a primitive cube root of unity. Let K = k(ω), so that [K : k] = 1 or 2 according as ω ∈ k or ω ∉ k. In his work on cubic surfaces, Heath-Brown [HB] makes implicit use of

Theorem 1.1 If [K : k] = 2 and the Tate-Shafarevich group � (E/k) is finite, then the order of � (E/K)[\sqrt{−3}] is a perfect square.

We explain how this result follows from the work of Cassels [CaIV], and give an example to show that the condition [K : k] = 2 is necessary.

For the application to cubic surfaces, we only need a special case of the theorem, namely that � (E/K)[\sqrt{−3}] cannot have order 3. This result, still conditional on the finiteness of the Tate-Shafarevich group, has already appeared in [BF] and [SwD]. In fact Swinnerton-Dyer [SwD] vastly generalises Heath-Brown’s results. In the case [K : k] = 2 he proves the Hasse principle for diagonal cubic 3-folds over k, conditional only on the finiteness of the Tate-Shafarevich group for elliptic curves over k. The condition [K : k] = 2 is unnatural, and conjecturally should not appear. However, the counterexample presented in this article suggests that, if we are to follow the methods of Heath-Brown and Swinnerton-Dyer, then this condition on k is unavoidable.

In §2 we recall how it is possible to pass between the fields k and K. Then in §3 we give a modern treatment of the descent by 3-isogeny studied by Selmer [S1] and Cassels [CaI]. In §§4,5 we recall how the conjectures of Selmer may be deduced from properties of the Cassels-Tate pairing. This culminates in a proof of Theorem 1.1. Finally in §6 we present our new example.

2 Decomposition into Galois eigenspaces

Let E be an elliptic curve over k with complex multiplication by Z[ω]. The isogeny [\sqrt{−3}] : E → E is defined over K = k(ω). But the kernel E[\sqrt{−3}] is defined over k. It follows that there is a 3-isogeny ϕ : E → Ẽ defined over k with E[\sqrt{−3}] = E[ϕ]. Here Ẽ is a second elliptic curve defined over k, which we immediately recognise as the −3-twist of E. The dual isogeny ϕ̂: Ẽ → E satisfies ϕ ∘ ϕ̂ = [3] and ϕ̂ ∘ ϕ = [3]. Our notation for the Selmer groups and Tate-Shafarevich groups follows [Sil, Chapter X].

Lemma 2.1 If [K : k] = 2 then the exact sequence

0 ⟶ E(K)/\sqrt{−3}E(K) ⟶ S^{(\sqrt{−3})}(E/K) ⟶ � (E/K)[\sqrt{−3}] ⟶ 0 (1)

is the direct sum of the exact sequences

0 ⟶ Ẽ(k)/ϕE(k) ⟶ S^{(ϕ)}(E/k) ⟶ � (E/k)[ϕ] ⟶ 0  (2)

and

0 ⟶ E(k)/ϕ̂Ẽ(k) ⟶ S^{(ϕ̂)}(Ẽ/k) ⟶ � (Ẽ/k)[ϕ̂] ⟶ 0.  (3)

Proof. Since arguments of this type have already appeared in [BF], [N],

[SwD] and presumably countless other places in the literature, we will not dwell on the proof. Suffice to say that we decompose (1) into eigenspaces for the action of Gal(K/k), and then use the inflation-restriction exact sequence to identify these eigenspaces as (2) and (3). The observation that [K : k] = 2 is prime to deg ϕ = 3 is crucial throughout the proof. □

Remark 2.2 Each term of the exact sequence (1) is a Z/3Z-vector space with an action of Gal(K/k). So each term is a direct sum of the Galois modules Z/3Z and μ_{3}. If we replace E by Ẽ in (1) then we obtain the same exact sequence of abelian groups, but as Galois modules the summands Z/3Z and μ_{3} are interchanged.

3 Computation of Selmer groups

Let k be a number field. Let T [a_{0}, a_{1}, a_{2}] be the diagonal plane cubic

a_{0}x_{0}^{3} + a_{1}x_{1}^{3} + a_{2}x_{2}^{3} = 0  (4)

where a_{0}, a_{1}, a_{2} ∈ k^{∗}/k^{∗3}. Let E_{A} be the elliptic curve T [A, 1, 1] with identity element 0 = (0 : 1 : −1). It is well known [St] that E_{A} has Weierstrass equation y^{2} = x^{3} − 432A^{2}. An alternative proof of the following lemma may be found in [CaL, §18].

Lemma 3.1 The diagonal plane cubic T [a_{0}, a_{1}, a_{2}] is a smooth curve of genus 1 with Jacobian E_{A} where A = a_{0}a_{1}a_{2}.

Proof. There is an isomorphism T [a_{0}, a_{1}, a_{2}] ≃ E_{A} defined over k(\sqrt[3]{α}) where α = a_{1}a_{2}^{2}. It is given by

ψ : (x_{0} : x_{1} : x_{2}) ↦ (a_{2}x_{0} : α^{2/3}x_{1} : α^{1/3}a_{2}x_{2}).

The cocycle σ(ψ)ψ^{−1} takes values in the subgroup μ_{3} ⊂ Aut(E_{A}) generated by x_{i} ↦ ω^{i}x_{i}. But since μ_{3} acts on E_{A} without fixed points, this action belongs to the translation subgroup of Aut(E_{A}). It follows that T [a_{0}, a_{1}, a_{2}] is a torsor under E_{A} and that E_{A} is the Jacobian of T [a_{0}, a_{1}, a_{2}]. □

Temporarily working over K = k(ω) we note that E_{A} has complex multiplication by Z[ω] where ω : (x_{0} : x_{1} : x_{2}) ↦ (ωx_{0} : x_{1} : x_{2}) and that E_{A}[1 − ω] = E_{A}[\sqrt{−3}] is generated by (0 : ω : −ω^{2}). So as in §2 there is a map ϕ which gives an exact sequence of Galois modules

0 ⟶ μ_{3} ⟶ E_{A} ⟶^{ϕ} Ẽ_{A} ⟶ 0

where Ẽ_{A} is the −3-twist of E_{A}. Taking Galois cohomology we obtain an exact sequence

0 ⟶ Ẽ_{A}(k)/ϕE_{A}(k) ⟶^{δ} k^{∗}/k^{∗3} ⟶ H^{1}(k, E_{A})[ϕ] ⟶ 0.  (5)

The group H^{1}(k, E_{A}) parametrises the torsors under E_{A}. We write C_{A,α} for the torsor under E_{A} described by α ∈ k^{∗}/k^{∗3}. The proof of Lemma 3.1 shows that

T [a_{0}, a_{1}, a_{2}] ≃ C_{A,α}  for A = ∏ a_{ν} and α = ∏ a_{ν}^{ν}  (6)

where the products are over ν ∈ Z/3Z. Since T [a_{0}, a_{1}, a_{2}] ≃ T [a_{1}, a_{2}, a_{0}] it is clear that A ∈ im δ. If Ẽ_{A} has Weierstrass equation Y^{2}Z = −4AX^{3} + Z^{3} then the 3-covering map T [a_{0}, a_{1}, a_{2}] → Ẽ_{A} is given by

(x_{0} : x_{1} : x_{2}) ↦ (x_{0}x_{1}x_{2} : a_{1}x_{1}^{3} − a_{2}x_{2}^{3} : a_{0}x_{0}^{3}).

The Selmer group attached to ϕ is

S^{(ϕ)}(E_{A}/k) = { α ∈ k^{∗}/k^{∗3} | C_{A,α}(k_{𝔭}) ≠ ∅ for all primes 𝔭 }.

Since deg ϕ = 3 is odd we have ignored the infinite places. We write δ_{𝔭} for the local connecting map obtained when we apply (5) to the local field k_{𝔭}. Then the condition C_{A,α}(k_{𝔭}) ≠ ∅ may also be written α ∈ im δ_{𝔭}. Using (6) to give equations for C_{A,α} it is easy to prove

Lemma 3.2 Let k be a number field, and let 𝔭 be a prime not dividing 3. Let 𝔬_{𝔭} denote the ring of integers of k_{𝔭}. Then

im δ_{𝔭} = \{ \begin{matrix}𝔬_{𝔭}^{∗}/𝔬_{𝔭}^{∗3} & if ord_{𝔭}(A) ≡ 0 (mod 3) \\ ⟨A⟩ & if ord_{𝔭}(A) ≢ 0 (mod 3). \end{matrix}

If 𝔭 divides 3 the situation is more complicated, although we still have

im δ_{𝔭} ⊂ 𝔬_{𝔭}^{∗}/𝔬_{𝔭}^{∗3} if ord_{𝔭}(A) ≡ 0 (mod 3).  (7)

If ω ∈ k_{𝔭} then Tate local duality tells us that im δ_{𝔭} is a maximal isotropic subspace with respect to the Hilbert norm residue symbol

k_{𝔭}^{∗}/k_{𝔭}^{∗3} × k_{𝔭}^{∗}/k_{𝔭}^{∗3} → μ_{3}.  (8)

The next lemma treats the case k = Q(ω). This field has ring of integers Z[ω] and class number 1. The unique prime above 3 is π = ω − ω^{2}.

Lemma 3.3 Let A ∈ Z[ω] be non-zero and cube-free. Then

im δ_{π} = \{ \begin{matrix}⟨ A, (1 − A)/(1 + A) ⟩ & if ord_{π}(A) ≠ 0 \\ ⟨ A, 1 − π^{3} ⟩ & if ord_{π}(A) = 0 and A^{2} ≢ ±1 (π^{3}) \\ ⟨ ω(1 + 3a), 1 − π^{3} ⟩ & if A = ±(1 + aπ^{3}) for some a ∈ Z[ω]. \end{matrix}

Proof. We recall [CF, Exercise 2.13] that k_{π}^{∗}/k_{π}^{∗3} has basis π, ω, 1 − π^{2}, 1 − π^{3} and that these elements define a filtration compatible with the pairing (8). By Tate local duality it follows that im δ_{π} has order 9. So to prove the lemma it suffices to prove the inclusions ⊃. As always A ∈ im δ_{π}, whereas (7) and Tate local duality tell us that 1 − π^{3} ∈ im δ_{π}. There is at most one more element to find.

(i) Suppose ord_{π}(A) ≠ 0. If α satisfies α − α^{−1} = A then T [A, α, α^{−1}] is soluble. Splitting into the cases ord_{π}(A) = 1 or 2 we find

4A/(1 − A^{2}) ≡ A (mod π^{4}).

So α = (1 − A)/(1 + A) provides a solution mod π^{4}. (ii) Suppose A = 1 + aπ^{3} for some a ∈ Z[ω]. If α satisfies A + α + α^{−1} = 0 then T [A, α, α^{−1}] is soluble. In view of the identity

(1 + π^{3}a) + ω(1 + 3a) + ω^{2}(1 − 3a) = 0

we see that α = ω(1 + 3a) provides a solution mod π^{4}. □

4 Selmer’s conjectures

In this section we take k = Q, so that K = Q(ω). We consider the elliptic curves E_{A} and Ẽ_{A} over Q where A ≥ 2 is a cube-free integer.

Lemma 4.1 If A ≥ 3 then the torsion subgroups are

E_{A}(Q)_{tors} = 0 and Ẽ_{A}(Q)_{tors} ≃ Z/3Z.

Proof. See [St, §6] or [K, Chapter 1, Problem 7]. □

Lemma 2.1 gives a decomposition into Gal(K/Q)-eigenspaces

S^{(\sqrt{−3})}(E_{A}/K) ≃ S^{(ϕ)}(E_{A}/Q) ⊕ S^{(ϕ̂)}(Ẽ_{A}/Q).  (9)

The following examples were found by Selmer [S1], [S2].

Example 4.2 Let A = 60. Lemmas 3.2 and 3.3 tell us that

S^{(\sqrt{−3})}(E_{60}/K) ≃ ⟨2, 3, 5⟩ ⊂ K^{∗}/K^{∗3}.

Then (9) gives S^{(ϕ)}(E_{60}/Q) ≃ (Z/3Z)^{3} and S^{(ϕ̂)}(Ẽ_{60}/Q) = 0. But a 2-descent [CaL, §15], [Cr] shows that E_{60}(Q) has rank 0. We deduce

� (E_{60}/Q)[3] ≃ (Z/3Z)^{2}.

Example 4.3 Let A = 473. Lemmas 3.2 and 3.3 tell us that

S^{(\sqrt{−3})}(E_{473}/K) ≃ ⟨11, 1 − 6ω, 1 − 6ω^{2}⟩ ⊂ K^{∗}/K^{∗3}.

Then (9) gives S^{(ϕ)}(E_{473}/Q) ≃ (Z/3Z)^{2} and S^{(ϕ̂)}(Ẽ_{473}/Q) ≃ Z/3Z. But a 2-descent [S2], [Cr] shows that E_{473}(Q) has rank 0. We deduce

� (E_{473}/Q)[ϕ] ≃ Z/3Z and � (Ẽ_{473}/Q)[ϕ̂] ≃ Z/3Z.

Remark 4.4 According to the formulae and tables of Stephens [St], the above examples have L(E_{A}, 1) ≠ 0. So the claims rank E_{A}(Q) = 0 could equally be deduced from the work of Coates-Wiles [CW].

Example 4.2 tells us that each of the curves

\begin{aligned}T [3, 4, 5] : & 3x_{0}^{3} + 4x_{1}^{3} + 5x_{2}^{3} & = & 0 \\ T [1, 3, 20] : & x_{0}^{3} + 3x_{1}^{3} + 20x_{2}^{3} & = & 0 \\ T [1, 4, 15] : & x_{0}^{3} + 4x_{1}^{3} + 15x_{2}^{3} & = & 0 \\ T [1, 5, 12] : & x_{0}^{3} + 5x_{1}^{3} + 12x_{2}^{3} & = & 0 \end{aligned}  (10)

is a counterexample to the Hasse Principle for smooth curves of genus 1 defined over Q. Selmer proves this without the need for a 2-descent. Instead he shows that the equations (10) are insoluble over Q by writing them as norm equations. As Cassels explains [CaI, §11] this is equivalent to performing a second descent, i.e. computing the middle group in

Ẽ_{A}(Q)/ϕE_{A}(Q) ⊂ ϕ̂S^{(3)}(Ẽ_{A}/Q) ⊂ S^{(ϕ)}(E_{A}/Q).  (11)

In fact Selmer’s calculations suffice to show that � (E_{60}/Q)(3) ≃ (Z/3Z)^{2}. In other words � (E_{60}/Q) does not contain an element of order 9. More recent work of Rubin [M] improves this to � (E_{60}/Q) ≃ (Z/3Z)^{2}.

Selmer also gave practical methods for computing the two right hand groups in

E_{A}(Q)/ϕ̂Ẽ_{A}(Q) ⊂ ϕS^{(3)}(E_{A}/Q) ⊂ S^{(ϕ̂)}(Ẽ_{A}/Q).  (12)

Following Stephens [St] we write g_{1} + 1, λ′_{1} + 1, λ_{1} + 1 for the dimensions of the Z/3Z-vector spaces (11) and g_{2}, λ′_{2}, λ_{2} for the dimensions of the Z/3Zvector spaces (12). Trivially we have 0 ≤ g_{1} ≤ λ′_{1} ≤ λ_{1}, 0 ≤ g_{2} ≤ λ′_{2} ≤ λ_{2} and rank E_{A}(Q) = g_{1} + g_{2}. Based on a large amount of numerical evidence, Selmer [S3] made the following

Conjecture 4.5 Let A ≥ 2 be a cube-free integer. Let E_{A} be the elliptic curve x^{3} + y^{3} = Az^{3} defined over Q. Then Weak form. The second descent excludes an even number of generators, i.e. λ_{1} ≡ λ′_{1} (mod 2) and λ_{2} ≡ λ′_{2} (mod 2). Strong form. The number of generators of infinite order for E_{A}(Q) is an even number less than what is indicated by the first descent, i.e. λ_{1} + λ_{2} ≡ g_{1} + g_{2} (mod 2).

For A = 473, Selmer found λ_{1} = λ′_{1} = λ_{2} = λ′_{2} = 1 yet g_{1} = g_{2} = 0. He was thus aware of the need to combine the contributions from ϕ and ϕ̂ in the strong form of his conjecture.

Remark 4.6 In Heath-Brown’s notation [HB] we have

r(A) = rank E_{A}(Q) = g_{1} + g_{2} and s(A) = λ_{1} + λ_{2}.

By (9) the order of S^{(\sqrt{−3})}(E_{A}/K) is 3^{s(A)+1} and in fact it is this relation that Heath-Brown uses to define s(A). Naturally he writes the strong form of Selmer’s conjecture as r(A) ≡ s(A) (mod 2).

Now let k be any number field. Conjecture 4.5 is equivalent to the case k = Q of the following

Conjecture 4.7 Let A ∈ k^{∗} not a perfect cube. Let E_{A} be the elliptic curve x^{3} + y^{3} = Az^{3} defined over k. Then Weak form. The subgroup ϕ̂(� (Ẽ_{A}/k)[3]) ⊂ � (E_{A}/k)[ϕ] has index a perfect square. The same is true for ϕ(� (E_{A}/k)[3]) ⊂ � (Ẽ_{A}/k)[ϕ̂]. Strong form. The order of � (E_{A}/k)[ϕ] multiplied by the order of � (Ẽ_{A}/k)[ϕ̂] is a perfect square.

In the next section we recall how Conjecture 4.7 follows from the work of Cassels, the strong form being conditional on the finiteness of � (E_{A}/k).

5 The Cassels-Tate pairing

Let E be an elliptic curve over a number field k. For ϕ : E → E′ an isogeny of elliptic curves over k we shall write ϕ̂ : E′ → E for the dual isogeny. Cassels [CaIV] defines an alternating bilinear pairing

⟨ , ⟩ : � (E/k) × � (E/k) → Q/Z  (13)

with the following non-degeneracy property.

Theorem 5.1 Let ϕ : E → E′ be an isogeny of elliptic curves over k. Then x ∈ � (E/k) belongs to the image of ϕ̂: � (E′/k) → � (E/k) if and only if ⟨x, y⟩ = 0 for all y ∈ � (E/k)[ϕ].

Proof. This was proved by Cassels [CaIV] in the case ϕ = [m] for m a rational integer. The general case follows by his methods and is explained in [F]. □

The pairing was later generalised to abelian varieties by Tate, and so is known as the Cassels-Tate pairing. The most striking applications in the case of elliptic curves come from the following easy lemma.

Lemma 5.2 If a finite abelian group admits a non-degenerate alternating bilinear pairing, then its order must be a perfect square.

The weak form of Conjecture 4.7 is a special case of

Corollary 5.3 Let ϕ : E → E′ be an m-isogeny of elliptic curves over k. Then the subgroup ϕ̂(� (E′/k)[m]) ⊂ � (E/k)[ϕ] has index a perfect square.

Proof. According to Theorem 5.1 the pairing (13) restricted to � (E/k)[ϕ] has kernel ϕ̂(� (E′/k)[m]). We are done by Lemma 5.2. □

Let us assume that � (E/k) is finite. So by Theorem 5.1 and Lemma 5.2 the order of � (E/k) is a perfect square. If ϕ : E → E′ is an isogeny of elliptic curves over k then the same conclusions will hold for E′. We define

⟨ , ⟩_{ϕ} : � (E/k) × � (E′/k) → Q/Z ; (x, y) ↦ ⟨ϕx, y⟩ = ⟨x, ϕ̂y⟩ (14)

where the equality on the right is [CaVIII, Theorem 1.2]. The strong form of Conjecture 4.7 is a special case of Corollary 5.4 Let ϕ : E → E′ be an isogeny of elliptic curves over k. If � (E/k) is finite then the order of � (E/k)[ϕ] multiplied by the order of � (E′/k)[ϕ̂] is a perfect square.

Proof. According to Theorem 5.1 the left and right kernels of ⟨ , ⟩_{ϕ} are � (E/k)[ϕ] and � (E′/k)[ϕ̂]. We obtain a non-degenerate pairing

� (E/k)/� (E/k)[ϕ] × � (E′/k)/� (E′/k)[ϕ̂] → Q/Z.

We deduce that these quotients have the same order and are done since � (E/k) and � (E′/k) each have order a perfect square. □

Another well known consequence is

Corollary 5.5 Let E be an elliptic curve over k whose Tate-Shafarevich group is finite, and let m be a rational integer. Then the order of � (E/k)[m] is a perfect square.

Proof. According to Theorem 5.1 the kernel of ⟨ , ⟩_{m} is � (E/k)[m]. We obtain a non-degenerate alternating pairing

� (E/k)/� (E/k)[m] × � (E/k)/� (E/k)[m] → Q/Z.

We apply Lemma 5.2 to this pairing and are done since � (E/k) has order a perfect square. □

Remark 5.6 We could equally deduce Corollary 5.4 from Corollaries 5.3 and 5.5.

Proof of Theorem 1.1. Let E be an elliptic curve over k with complex multiplication by Z[ω] and suppose that [K : k] = 2. Lemma 2.1 tells us that

� (E/K)[\sqrt{−3}] ≃ � (E/k)[ϕ] ⊕ � (Ẽ/k)[ϕ̂].

Assuming � (E/k) is finite, Corollary 5.4 shows that the group on the right has order a perfect square. So the group on the left has order a perfect square, and this is precisely the statement of Theorem 1.1. □

In the first of his celebrated series of papers, Cassels [CaI] defines a pairing S^{(\sqrt{−3})}(E_{A}/K) × S^{(\sqrt{−3})}(E_{A}/K) → μ_{3}. It is of course a special case of the pairing (13). He uses it to prove the weak form of Conjecture 4.7 in the case [K : k] = 1. However in the introduction to the same paper he misquotes the strong form of Selmer’s conjecture. The statement he gives is equivalent to

• If [K : k] = 1 then the order of � (E_{A}/K)[\sqrt{−3}] is a perfect square.

It is this statement to which we have found a counterexample. It is possible that Cassels was mislead by earlier work of Selmer at a time when he did not appreciate the need to combine the contributions from ϕ and ϕ̂ in the strong form of his conjecture.

Remark 5.7 It is tempting to try and prove Theorem 1.1 in the case [K : k] = 1 by imitating the proof of Corollary 5.5. However the isogeny [\sqrt{−3}] has dual [−\sqrt{−3}] and this extra sign means that the pairing ⟨ , ⟩\sqrt{−3} is symmetric rather than alternating. Lemma 5.2 does not apply.

6 A new example

In this section we take K = Q(ω). Let E_{A} be the elliptic curve x^{3}+y^{3} = Az^{3}. We aim to find A ∈ K such that the order of � (E_{A}/K)[\sqrt{−3}] is not a perfect square. As in Example 4.3 our method is to compare a 3-descent with a 2descent. The form of the curves E_{A} makes the 3-descent easy. We use the results of §3 to compute the Selmer group S^{(\sqrt{−3})}(E_{A}/K). For the 2-descent we would like to use John Cremona’s program mwrank [Cr]. But mwrank is written specifically for elliptic curves over Q, whereas Theorem 1.1 tells us that there are no examples of the required form with A^{2} ∈ Q. Fortunately we were able to use a program of Denis Simon [Si1], [Si2], written using pari [BBBCO], that extends Cremona’s work on 2-descents to general number fields (in practice of degrees 1 to 5).

We consider all cube-free A ∈ Z[ω] with A^{2} ∉ Q and Norm(A) ≤ 150. We ignore repeats of the form ±σ(A) for σ ∈ Gal(K/Q). In all 123 cases a calculation based on Lemmas 3.2 and 3.3 shows that S^{(\sqrt{−3})}(E_{A}/K) is isomorphic to either Z/3Z or (Z/3Z)^{2}. In the 98 cases where S^{(\sqrt{−3})}(E_{A}/K) ≃ Z/3Z it follows immediately that rank E_{A}(K) = 0. In the remaining 25 cases we run Simon’s program. For 20 of these curves the program exhibits a point of infinite order. Since E_{A}(K) has the structure of Z[ω]-module, we are able to deduce that rank E_{A}(K) = 2. The remaining 5 cases are

A = ±(3 + 7ω), ±(9 + ω), ±(12 + 5ω), ±(6 + 13ω), ±(13 + 7ω)

and their Galois conjugates. In each case Simon’s program reports that rank E_{A}(K) = 0. Reducing modulo some small primes we find E_{A}(K) ≃ Z/3Z.

Thus

� (E_{A}/K)[\sqrt{−3}] ≃ Z/3Z.

For the remainder of this article we restrict attention to the first of these examples, namely A = 3 + 7ω, and give further details of the descent calculations involved. In particular we establish the counterexample of the title in a way that is independent of Simon’s program.

We begin by checking the above computation of S^{(\sqrt{−3})}(E_{A}/K) for A = 3 + 7ω. Since (A) is prime, Lemma 3.2 tells us that

S^{(\sqrt{−3})}(E_{A}/K) ⊂ ⟨ω, 3 + 7ω⟩.  (15)

We check the local conditions at the primes (π) and (A) above 3 and 37 respectively.

• Since 37 ≡ 1 (mod 9) we know that ω is a cube locally at (A).

• Lemma 3.3 gives im δ_{π} = ⟨A, 1 − π^{3}⟩ ⊂ K_{π}^{∗}/K_{π}^{∗3}. Since A = ω − π^{3} it is clear that ω belongs to this subgroup.

It follows that equality holds in (15) as required.

Given the provisional nature of Simon’s program we have taken the liberty of writing out the 2-descent for A = 3+7ω in the style of Cassels [CaL, p.72-73]. The curve E_{A} has Weierstrass form

Y^{2} = X^{3} − 2^{4}3^{3}(3 + 7ω)^{2}.  (16)

The 2-descent takes place over the field L = K(δ) where δ^{3} = 4(3 + 7ω). According to pari [BBBCO]^{1}, L has class number h = 3, and fundamental units

\begin{aligned}η_{1} & = & (−7 − 3ω) + (−3 − 2ω)δ + (−2 + ω)δ^{2}/2 \\ η_{2} & = & (−7 − 3ω) + (2 − ω)δ + (3 + 2ω)δ^{2}/2. \end{aligned}

Furthermore pari is able to certify these results, independent of any conjecture. We have chosen η_{1} and η_{2} to be K-conjugates. They have minimal polynomial

x^{3} + (21 + 9ω)x^{2} + (102 − 165ω)x − 1.

If (X, Y) = (r/t^{2}, s/t^{3}) is a solution of (16), with fractions in lowest terms, then a common prime divisor of any two of

r − 3δ^{2}t^{2}, r − 3ωδ^{2}t^{2}, r − 3ω^{2}δ^{2}t^{2}

1These calculations were performed using Version 2.0.20 (beta)

must divide 2(1 − ω)(3 + 7ω). Since 2, (1 − ω), (3 + 7ω) ramify completely, r − 3δ^{2}t^{2} must be a perfect ideal square. Since h is odd it follows that S^{(2)}(E/K) is a subgroup of ⟨−1, η_{1}, η_{2}⟩ ⊂ L^{∗}/L^{∗2}. We claim that S^{(2)}(E/K) is trivial. By considering norms from L to K, it suffices to show that the equation

r − 3δ^{2}t^{2} = ηα^{2} with η = η_{1}, η_{2} or 1/(η_{1}η_{2})

is insoluble for r, t ∈ K and α ∈ L. The action of Gal(L/K) shows that we need only consider the case η = η_{1}. Put α = u + vδ + wδ^{2}. Equating coefficients of powers of δ we obtain

0 = (−3 − 2ω)u^{2} + (−14 − 6ω)uv + (−26 − 36ω)v^{2}
+ (−52 − 72ω)uw + (40 − 104ω)vw + (−148ω)w^{2}
−3t^{2} = ((−2 + ω)/2)u^{2} + (−6 − 4ω)uv + (−7 − 3ω)v^{2}
+ (−14 − 6ω)uw + (−52 − 72ω)vw + (20 − 52ω)w^{2}.

On putting

\begin{aligned}u & = & (−8 + 6ω)e + (−6 − 34ω)f + (−20 + 15ω)g \\ v & = & (−4 − 4ω)e + (12 + 4ω)f + (−10 − 11ω)g \\ w & = & (1 − ω)e + (1 + 4ω)f + (2 − 2ω)g \end{aligned}

in the first equation, it becomes

0 = (3 + 7ω)g^{2} − 16ef.

Hence there are m, n such that

e : f : g = m^{2} : (3 + 7ω)n^{2} : 4mn.

On substituting into the second equation, we get

\begin{aligned}−3t^{2} & = & 2(−1 − 4ω)m^{4} + 8(−4 + 3ω)m^{3}n + 4(21 + 12ω)m^{2}n^{2} \\ & & + 8(4 − 3ω)mn^{3} + 2(−33 − 40ω)n^{4}. \end{aligned}

But this is impossible in K_{2}. Hence S^{(2)}(E_{A}/K) is trivial and rank E_{A}(K) = 0 as claimed.

Acknowledgements

The author would like to thank Laura Basile, Alexei Skorobogatov and Sir Peter Swinnerton-Dyer for useful conversations.

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