DIAGONAL CUBIC EQUATIONS IN FOUR VARIABLES WITH PRIME COEFFICIENTS

Carmen Laura Basile^{∗}

Department of Mathematics, Imperial College of Science, Technology and Medicine, London

                         (laura.basile@ic.ac.uk)

Thomas Anthony Fisher

Department of Pure Mathematics and Mathematical Statistics, Sidney Sussex College, Cambridge

                       (T.A.Fisher@dpmms.cam.ac.uk)

November 2000

Abstract

The aim of this paper is to give an alternative proof of a theorem of R. Heath-Brown [3] regarding the existence of non-zero integral solutions of the equation p_{1}X_{1}^{3} + p_{2}X_{2}^{3} + p_{3}X_{3}^{3} + p_{4}X_{4}^{3} = 0 , where the p_{j} are prime integers ≡ 2 (mod 3).

We start by presenting the main result of this paper. This result has been proved by Roger Heath-Brown [3] under the conjecture that the difference s(A) − r(A) between the Selmer rank and the arithmetic rank of the elliptic curve X^{3} + Y^{3} = AZ^{3} is even. In this note we will show that we do not need this assumption and give a detailed proof of this result.

Theorem 1 Let p_{1}, p_{2}, p_{3}, p_{4} be prime integers such that p_{i} ≡ 2 (mod 3) (1 ≤ i ≤ 4). Then the equation

p_{1}X_{1}^{3} + p_{2}X_{2}^{3} + p_{3}X_{3}^{3} + p_{4}X_{4}^{3} = 0

has non-zero integral solutions, assuming the conjecture that the Tate-Shafarevich group of the elliptic curve X^{3} + Y^{3} = AZ^{3} over ℚ is finite.

A much stronger result has been recently proved by Sir Peter Swinnerton-Dyer [7].

Note that in order to prove the above theorem, it is sufficient to prove that the equations

\begin{matrix}p_{1}X_{1}^{3} + p_{2}X_{2}^{3} = p \\ p_{3}X_{3}^{3} + p_{4}X_{4}^{3} = −p \end{matrix}

∗Supported by INdAM (Istituto Nazionale di Alta Matematica “F.Severi”), Italy.

have non-zero rational solutions for some prime integer p. So it suffices to prove that the equation p_{1}X^{3} + p_{2}Y^{3} = p has non-zero solutions in ℚ or, equivalently, in a quadratic extension of ℚ.

Notation Let ω be a primitive cube root of unity and let k = ℚ(ω). Let A ∈ ℤ \ {−1, 0, 1} be a cube-free integer. We denote by E_{A} the elliptic curve

E_{A}: X^{3} + Y^{3} = AZ^{3}.

For any α ∈ k^{∗}, let C_{A,α} be the smooth projective curve given by the equation

C_{A,α}: αX^{3} + α^{−1}Y^{3} = AZ^{3}.

For α ∈ k^{∗}, the curves C_{A,α} are principal homogeneous spaces over E_{A}. Moreover, it is clear that if α and β belong to the same class modulo (k^{∗})^{3} then the curves C_{A,α} and C_{A,β} are isomorphic.

Let us consider the multiplication-by-\sqrt{−3} endomorphism on E_{A} (see [1]), given by

\begin{aligned}\sqrt{−3} : & E_{A} & ⟶ & E_{A} \\ & (X, Y, Z) & ⟼ & (ω^{2}X^{3} + ωY^{3} − AZ^{3}, ωX^{3} + ω^{2}Y^{3} − AZ^{3}, −3XY Z), \end{aligned}

and the following diagram with exact row

0 ⟶ E_{A}(k)/\sqrt{−3}E_{A}(k) ⟶ H^{1}(k, E_{A} [\sqrt{−3}]) ⟶ WC(E_{A}/k) [\sqrt{−3}] ⟶ 0
↘  ↓
∏_{v} WC(E_{A}/k_{v}) [\sqrt{−3}]

where WC(E_{A}/k) is the Weil-Châtelet group of E_{A}/k and v runs over all the places of k. It is readily verified that the group E_{A}(k̄) [\sqrt{−3}] is isomorphic to the group μ_{3}(k̄) of cube roots of unity, as a Gal(k̄/k)-module. It follows from Kummer theory that H^{1}(k, E_{A} [\sqrt{−3}]) is isomorphic to k^{∗}/(k^{∗})^{3} and so we get the exact sequence

0 ⟶ E_{A}(k)/\sqrt{−3}E_{A}(k) ⟶ k^{∗}/(k^{∗})^{3} ⟶^{f} WC(E_{A}/k) [\sqrt{−3}] ⟶ 0
g↘  ↓
∏_{v} WC(E_{A}/k_{v}) [\sqrt{−3}]

where the map f sends an element α(k^{∗})^{3} to the curve C_{A,α}. We denote by S(A) the Selmer group S^{(\sqrt{−3})}(E_{A}/k) (which is defined to be the kernel of the map g in the diagram above) and by C(A) the kernel of the map f. Obviously C(A) is a subgroup of S(A); we can write them explicitly

S(A) = {α(k^{∗})^{3} : C_{A,α} has k_{π}−points for any prime π ∈ k, α ∈ k^{∗}},
C(A) = {α(k^{∗})^{3} : C_{A,α} has k−points, α ∈ k^{∗}}.

Observe also that C(A) is isomorphic to E_{A}(k)/\sqrt{−3}E_{A}(k) (this follows immediately from the diagram above).

In the following two lemmas we will determine the structure of S(A) when A satisfies certain conditions.

Lemma 2 Let ρ ∈ k be a prime above 3, and let a, b, c ∈ k such that a, b, c are congruent to 1 modulo ρ^{2}. If the projective curve

E : aX^{3} + bωY^{3} + cω^{2}Z^{3} = 0

has a point over k_{ρ}, then abc ≡ 1 (mod ρ^{3}).

Proof. Let (x, y, z) be a k_{ρ}-point of the curve E and suppose that

min{val_{ρ}(x) , val_{ρ}(y) , val_{ρ}(z)} = 0.

Say

\begin{aligned}x & = & x_{0} + x_{1}ρ + x_{2}ρ^{2} + …, \\ y & = & y_{0} + y_{1}ρ + y_{2}ρ^{2} + …, \\ z & = & z_{0} + z_{1}ρ + z_{2}ρ^{2} + … \end{aligned}

with (x_{0}, y_{0}, z_{0}) ≠ (0, 0, 0).

From the condition ax^{3} + bωy^{3} + cω^{2}z^{3} = 0 it follows that x_{0}^{3} ≡ y_{0}^{3} ≡ z_{0}^{3} (mod ρ^{2}). Therefore we may assume (x_{0}, y_{0}, z_{0}) = (1, 1, 1) and hence we get

(x^{3}, y^{3}, z^{3}) ≡ (1, 1, 1) (mod ρ^{3}).

Let a ≡ 1 + a_{2}ρ^{2} (mod ρ^{3}), b ≡ 1 + b_{2}ρ^{2} (mod ρ^{3}) and c ≡ 1 + c_{2}ρ^{2} (mod ρ^{3}). Then from a + bω + cω^{2} ≡ 0 (mod ρ^{3}) it follows a_{2} + b_{2} + c_{2} ≡ 0 (mod ρ) and therefore abc ≡ 1 + (a_{2} + b_{2} + c_{2})ρ^{2} ≡ 1 (mod ρ^{3}). □

Lemma 3 Let A = p_{1}p_{2}N(π) where p_{1}, p_{2} ≡ 2 (mod 3) are integer primes, π ≡ 1

(mod 3) is a prime in ℤ[ω] such that A ≢ 1 (mod 9) and (\frac{π}{p_{1}} )_{3} = (\frac{π}{p_{2}} )_{3} ≠ 1. Then

S(A) is isomorphic to ℤ/3 × ℤ/3 as an abelian group.

Proof. We will prove that S(A) is generated by the elements A(k^{∗})^{3} and p_{1}p_{2}^{2}(k^{∗})^{3}.

The curve C_{A,A} has a k-point, namely (1, 0, 1); hence A(k^{∗})^{3} ∈ S(A). Suppose now α ≠ A. Note that, in order to determine for which elements α the coset α(k^{∗})^{3} belongs to S(A), it is sufficient to test the elements α where α is a cube-free integer in ℤ[ω] such that

(i) α is composed of primes dividing A;

(ii) we can fix a prime in {p_{1}, p_{2}, π, π̄}, say π̄, such that π̄ does not divide α.

Indeed:

(i) suppose α contains a prime factor ρ which does not divide A; it is then easy to verify that C_{A,α} contains no k_{ρ}-points. To get a contradiction, we suppose C_{A,α} contains a k_{ρ}-point (X, Y, Z); so we have

α^{2}X^{3} + Y^{3} = αAZ^{3}.

Considering the ρ-adic valuation of the right hand side and of the left hand side, and noting val_{ρ}(α^{2}X^{3}) ≠ val_{ρ}(Y^{3}), we get

val_{ρ}(αAZ^{3}) = val_{ρ}(α^{2}X^{3} + Y^{3}) = min{val_{ρ}(α^{2}X^{3}) , val_{ρ}(Y^{3})}.

But this is impossible since, on the one hand, val_{ρ}(αAZ^{3}) = i + 3j, where i ∈ {1, 2} is defined to be the ρ-adic valuation of α and j is an integer, and on the other hand

min{val_{ρ}(α^{2}X^{3}) , val_{ρ}(Y^{3})} = min{2i + 3h, 3l}

with h, l integers;

(ii) suppose π̄^{j} divides α (with j = 1, 2); then, since S(A) is a group and A(k^{∗})^{3} belongs to S(A), instead of α we may consider the cube-free integer representative of the coset (α/A^{j}) (k^{∗})^{3}.

As a result, we may assume α = ω^{m}p_{1}^{n_{1}}p_{2}^{n_{2}}π^{n} where m, n_{1}, n_{2}, n ∈ {0, 1, 2}. In fact it is not difficult to prove that if m ≠ 0 then α does not belong to S(A). Indeed, to get a contradiction suppose that m ≠ 0 and that C_{A,α} contains a k_{ρ}-point where ρ ∈ k is a prime above 3. Then from Lemma 2 it follows that A ≡ 1 (mod ρ^{3}) and hence A ≡ 1 (mod 9) which contradicts our hypotheses. Therefore we may assume m = 0.

Let ρ ∈ ℤ[ω] be a prime. If ρ does not divide 3A then C_{A,α} contains points over k_{ρ} for any α. Indeed, a smooth curve of genus 1 always contains points over any finite field (see [2]); moreover if ρ does not divide 3A then C_{A,α} is non-singular over ℤ[ω]/ρ and therefore it contains a non-singular point over ℤ[ω]/ρ which, by Hensel’s lemma, can be lifted to a point over k_{ρ}.

Suppose now that ρ divides 3A; hence we have to consider the two cases ρ | A and ρ | 3.

(1) Let ρ divide A, so ρ belongs to {p_{1}, p_{2}, π, π̄}. We can consider three further subcases.

(1a) Let ρ divide α exactly; then the projective curve

αX^{3} + α^{−1}Y^{3} = AZ^{3}

can also be written as

\frac{α^{2}}{ρ^{2}} (ρX)^{3} + ρY^{3} = \frac{αA}{ρ^{2}}(ρZ)^{3}.

Hence, considering the transformation ρX → X, ρZ → Z and reducing modulo ρ, we get the curve

X^{3} − \frac{A}{α}Z^{3} = 0

which contains a smooth point over 𝔽_{ρ} if and only if \frac{A}{α} ∈ (𝔽_{ρ}^{∗})^{3}, i.e. if and only if (\frac{Aα^{−1}}{ρ} )_{3} = 1. By Hensel’s lemma, we can lift this smooth point over 𝔽_{ρ} to a point over k_{ρ}.

(1b) Suppose now ρ^{2} divides exactly α. Similarly, considering a suitable transfor-

mation and reducing modulo ρ, we get the curve

Y^{3} − \frac{Aα}{ρ^{3}}Z^{3} = 0

which contains a smooth point over 𝔽_{ρ} if and only if \frac{Aα}{ρ^{3}} ∈ (𝔽_{ρ}^{∗})^{3}, i.e. if and

only if (\frac{Aαρ^{−3}}{ρ} )_{3} = 1.

(1c) Similarly, if ρ does not divide α, we get the curve

α^{2}X^{3} + Y^{3} = 0

which contains a smooth point over 𝔽_{ρ} if and only if α^{2} ∈ (𝔽_{ρ}^{∗})^{3}, i.e. (\frac{α}{ρ})_{3} = 1. Suppose ρ = p_{j} with j = 1, 2. If p_{j} divides exactly α, then by case (1a) we must have (\frac{Aα^{−1}}{p_{j}} )_{3} = 1 and hence

(\frac{αp_{j}^{−1}}{p_{j}} )_{3} = 1; indeed ^{†}

(\frac{Aα^{−1}}{p_{j}} )_{3} = (\frac{p_{h}}{p_{j}} )_{3} (\frac{p}{p_{j}} )_{3} (\frac{α^{−1}p_{j}}{p_{j}} )_{3} = (\frac{αp_{j}^{−1}}{p_{j}} )_{3}^{−1}

where h ∈ {1, 2}, h ≠ j and p = N(π). If p_{j}^{2} divides exactly α, then by case (1b) we must have (\frac{Aαp_{j}^{−3}}{p_{j}} )_{3} = 1 and therefore

(\frac{αp_{j}^{−2}}{p_{j}} )_{3} = 1, since

(\frac{Aαp_{j}^{−3}}{p_{j}} )_{3} = (\frac{p_{h}}{p_{j}} )_{3} (\frac{p}{p_{j}} )_{3} (\frac{αp_{j}^{−2}}{p_{j}} )_{3} = (\frac{αp_{j}^{−2}}{p_{j}} )_{3} ,

where again h ∈ {1, 2} and h ≠ j. If p_{j} does not divide α, then by case (1c) we must have (\frac{α}{p_{j}} )_{3} = 1.

†If q ≠ 3 is a prime integer with #𝔽_{q}^{∗} ≢ 0 (mod 3) then (\frac{n}{q})_{3} = 1 for any n ∈ ℤ such that (n, q) = 1.

Finally, we obtain the two following conditions which must be satisfied in order to have α(k^{∗})^{3} ∈ S(A):

(\frac{p_{2}^{n_{2}}π^{n}}{p_{1}} )_{3} = 1 and (\frac{p_{1}^{n_{1}}π^{n}}{p_{2}} )_{3} = 1.

Since

(\frac{p_{2}^{n_{2}}π^{n}}{p_{1}} )_{3} = (\frac{p_{2}}{p_{1}} )_{3}^{n_{2}} (\frac{π}{p_{1}} )_{3}^{n} = (\frac{π}{p_{1}} )_{3}^{n}

and (\frac{π}{p_{1}} )_{3} ≠ 1 by hypothesis, the first condition is satisfied if and only if n = 0.

It is easy to verify that for n = 0 the second condition is satisfied as well. Hence α = p_{1}^{n_{1}}p_{2}^{n_{2}}. So, by case (1c), it follows that we must have (\frac{α}{π})_{3} = 1; therefore, since

by the Cubic Reciprocity Law (\frac{p_{1}}{π} )_{3} = (\frac{p_{2}}{π} )_{3} ≠ 1, we get the condition n_{1} + n_{2} ≡ 0

(mod 3). We may suppose n_{1} = 1; then α = p_{1}p_{2}^{2} or, equivalently, α = p_{1}p_{2}^{−1}.

(2) Suppose now that ρ divides 3; in other words, since 3 = −ω^{2}(1 − ω)^{2}, suppose ρ = 1 − ω . Recall that now we have A = p_{1}p_{2}p and α = p_{1}p_{2}^{−1}; therefore the curve

αX^{3} + α^{−1}Y^{3} = AZ^{3}

is isomorphic to the curve

E: p_{1}X^{3} + p_{2}Y^{3} = pZ^{3}.

We have to prove that E contains a point over k_{ρ}. Since ℚ_{3} is contained in k_{ρ}, it suffices to find a point over ℚ_{3}.

By hypothesis p ≡ 1 (mod 3) and p_{1}, p_{2} ≡ 2 (mod 3), so let

p ≡ 1 + 3b (mod 9) and p_{j} ≡ −1 + 3a_{j} (mod 9)

for j = 1, 2; since A ≢ 1 (mod 9) by hypothesis, we have a_{1} + a_{2} − b ≢ 0 (mod 3). Thus the integers a_{1}, a_{2}, −b cannot be all different modulo 3; wlog we may suppose a_{1} = a_{2}. Therefore we have p_{1}p_{2}^{−1} ≡ 1 (mod 9); in other words p_{1}p_{2}^{−1} belongs to 1 + 9ℤ_{3} which is contained in (ℤ_{3}^{∗})^{3}. It follows that E contains the point (1, −y, 0) where y ∈ ℤ_{3} is a cube root of p_{1}p_{2}^{−1}. More precisely, if p_{1}p_{2}^{−1} ≡ 1 + 9l (mod 27), then we use Hensel’s lemma to construct y ∈ ℤ_{3} such that y ≡ 1 + 3l (mod 9) and y^{3} = p_{1}p_{2}^{−1}.

Hence E contains a k_{ρ}-point.

In conclusion, we have proved that S(A) is generated by the two elements of order 3 A(k^{∗})^{3} and p_{1}p_{2}^{2}(k^{∗})^{3} and thus it is isomorphic to ℤ/3 × ℤ/3. □

The following two lemmas allow us to conclude that, under the hypotheses of Lemma 3, the two groups C(A) and S(A) coincide. This provides a local-to-global principle for the curves C_{A,α} when A is as in Lemma 3. As in the statement of Theorem 1, our work here is conditional on the finiteness of the Tate-Shafarevich group � (E_{A}/ℚ).

To get a contradiction, let us suppose that C(A) is strictly included in S(A). Note that C(A) cannot be the trivial group as A(k^{∗})^{3} belongs to C(A); then C(A) has order 3. From the exactness of the sequence

0 ⟶ E_{A}(k)/\sqrt{−3}E_{A}(k) ⟶ k^{∗}/(k^{∗})^{3} ⟶^{f} WC(E_{A}/k) [\sqrt{−3}] ⟶ 0,

it follows that E_{A}(k)/\sqrt{−3}E_{A}(k) and ℤ/3 are isomorphic as abelian groups (recall that C(A) is the kernel of the map f). Hence from Lemma 3 and the exact sequence

0 ⟶ E_{A}(k)/\sqrt{−3}E_{A}(k) ⟶ S^{(\sqrt{−3})}(E_{A}/k) ⟶ � (E_{A}/k) [\sqrt{−3}] ⟶ 0

we deduce that � (E_{A}/k) [\sqrt{−3}] is isomorphic to ℤ/3 and this is impossible, as we will show in Lemma 5. But first we need one more result.

Lemma 4 Let E/L be an elliptic curve over a number field L. Let K be a Galois extension of L of degree n. Let m be a positive integer such that (m, n) = 1. Then:

� (E/L)[m] = � (E/K)[m]^{Gal(K/L)}.

In particular, assuming the finiteness of � (E/L), the order of � (E/K)[m]^{Gal(K/L)} must be a square.

Proof. Let us consider the following commutative diagram with exact rows and columns where the rows are obtained by the multiplication-by-m endomorphism and the columns are restriction-inflation sequences:

[figure: 0; E(K)/mE(K); H^{1}(K, E[m]); H^{1}(K, E)[m]; 0.; 0; E(L)/mE(L); H^{1}(L, E[m]); H^{1}(L, E)[m]; 0; H^{1}(Gal(K/L), E(K)[m]); H^{1}(Gal(K/L), E(K))[m]; 0; 0]

Since Gal(K/L) has order n, every element of H^{1}(Gal(K/L), E(K)) has order dividing n (this follows from properties of the restriction and corestriction maps; see [5]). Hence, as m and n are coprime, H^{1}(Gal(K/L), E(K))[m] = 0. Thus from the diagram above it follows that H^{1}(L, E)[m] injects into H^{1}(K, E)[m].

From the exactness of the second row of the diagram we get the exact sequence

0 ⟶ E(K)/mE(K)^{Gal(K/L)} ⟶ H^{1}(K, E[m])^{Gal(K/L)} ⟶
⟶ H^{1}(K, E)[m]^{Gal(K/L)} ⟶ H^{1}(Gal(K/L), E(K)/mE(K)) = 0

where H^{1}(Gal(K/L), E(K)/mE(K)) is the zero group because it is killed by m and by n which are coprime.

On the other hand, from the exact sequence of low degree terms of the Hochschild-Serre spectral sequence, we get the exact sequence

H^{1}(Gal(K/L), E(K)[m]) ⟶ H^{1}(L, E[m]) ⟶^{φ} H^{1}(K, E[m])^{Gal(K/L)} ⟶

⟶ H^{2}(Gal(K/L), E(K)[m])

where the first and the last term are trivial because, again, they are killed by coprime integers. Hence the map φ is an isomorphism. The following diagram of exact rows and columns summarizes the information we have obtained so far

0  0
↓  ↓
H^{1}(L, E[m]) ⟶ H^{1}(L, E)[m] ⟶ 0
↓ φ  ↓ φ′
H^{1}(K, E[m])^{Gal(K/L)} ⟶^{φ′′} H^{1}(K, E)[m]^{Gal(K/L)} ⟶ 0
↓
0.

It is immediate to verify that the injective map φ′ is also surjective because of the surjectivity of the maps φ′′ and φ. Therefore we obtain

H^{1}(L, E)[m] = H^{1}(K, E)[m]^{Gal(K/L)}.

Let us consider now a place v of L; since K is a Galois extension of L, for any place w of K over v the degrees of the local extensions K_{w}/L_{v} divide n and therefore they are coprime to m. Hence the reasoning above can be applied also to the extensions K_{w}/L_{v} and thus we obtain

H^{1}(L_{v}, E)[m] = H^{1}(K_{w}, E)[m]^{Gal(K_{w}/L_{v})}.

Considering the corresponding Tate-Shafarevich groups, we get

� (E/L)[m] = � (E/K)[m]^{Gal(K/L)}.

Furthermore, assuming the finiteness of � (E/L), it follows from the existence of the Cassels alternating bilinear pairing on � (E/L) that the order of � (E/L)[m] is a perfect square and hence the order of � (E/K)[m]^{Gal(K/L)} is a square too. □

Lemma 5 If � (E_{A}/ℚ) is finite, then � (E_{A}/k) [\sqrt{−3}] cannot have order 3.

Proof. To get a contradiction, assume that � (E_{A}/k) [\sqrt{−3}] and ℤ/3 are isomorphic as abelian groups.

Let Ẽ_{A} be the quadratic twist of E_{A} corresponding to the class of −3 in H^{1}(ℚ, ℤ/2) = ℚ^{∗}/ℚ^{∗^{2}}; Ẽ_{A} has equation

Ẽ_{A}: −3Y^{2}Z = X^{3} − 432A^{2}Z^{3}

and is isomorphic to E_{A} over k through the map

\begin{aligned}ψ : & E_{A} & ⟶ & Ẽ_{A} \\ & (X, Y, Z) & ⟼ & (12AZ, \frac{36A}{\sqrt{−3}}(X − Y), X + Y). \end{aligned}

Let us consider the dual isogenies ϕ_{1} : E_{A} ⟶ Ẽ_{A} and ϕ_{2} : Ẽ_{A} ⟶ E_{A} given by the compositions ϕ_{1} = ψ ∘ \sqrt{−3} and ϕ_{2} = −\sqrt{−3} ∘ ψ^{−1}; they are defined over ℚ and their composition ϕ_{2} ∘ ϕ_{1} gives the multiplication-by-3 map on E_{A}.

To obtain a contradiction we have assumed that � (E_{A}/k) [\sqrt{−3}] is isomorphic to ℤ/3 as an abelian group; this is equivalent to the assumption that � (E_{A}/k)[ϕ_{1}] is isomorphic to ℤ/3.

Let Gal(k/ℚ) = ⟨σ⟩. We have two possibilities: either σ acts trivially on � (E_{A}/k)[ϕ_{1}] and therefore � (E_{A}/k)[ϕ] and ℤ/3 are isomorphic as Gal(k/ℚ)-modules; or σ exchanges the two non-trivial elements of � (E_{A}/k)[ϕ_{1}] and so � (E_{A}/k)[ϕ_{1}] is isomorphic to μ_{3}.

If � (E_{A}/k)[ϕ_{1}] is isomorphic to ℤ/3 as a Gal(k/ℚ)-module then � (Ẽ_{A}/k)[ϕ_{2}] is isomorphic to μ_{3} as a Gal(k/ℚ)-module and vice versa. Indeed, if � (E_{A}/k)[ϕ_{1}] is composed of the cohomology classes ξ_{0}, ξ_{1}, ξ_{2}, then � (Ẽ_{A}/k)[ϕ_{2}] is composed of ψξ_{0}, ψξ_{1}, ψξ_{2}; moreover ^{σ}ψ = −ψ. So, if σ acts trivially on � (E_{A}/k)[ϕ_{1}] then it does not on � (Ẽ_{A}/k)[ϕ_{2}] and vice versa.

Suppose that � (E_{A}/k)[ϕ_{1}] ≅ ℤ/3 as a Gal(k/ℚ)-module and consider the exact sequence of Gal(k/ℚ)-modules

0 ⟶ � (E_{A}/k)[ϕ_{1}] ⟶ � (E_{A}/k)[3] ⟶ � (Ẽ_{A}/k)[ϕ_{2}]

where the first map is the natural inclusion and the second one is induced by ϕ_{1}. From this sequence we get the exact sequence

0 ⟶ � (E_{A}/k)[ϕ_{1}]^{Gal(k/ℚ)} ⟶ � (E_{A}/k)[3]^{Gal(k/ℚ)} ⟶ � (Ẽ_{A}/k)[ϕ_{2}]^{Gal(k/ℚ)}

where � (E_{A}/k)[ϕ_{1}]^{Gal(k/ℚ)} ≅ ℤ/3 and � (Ẽ_{A}/k)[ϕ_{2}]^{Gal(k/ℚ)} ≅ 0. If � (E_{A}/k)[ϕ_{1}] ≅ μ_{3}, it is sufficient to consider the sequence

0 ⟶ � (Ẽ_{A}/k)[ϕ_{2}] ⟶ � (E_{A}/k)[3] ⟶ � (E_{A}/k)[ϕ_{1}]

instead of that above.

In both cases, we can conclude that � (E_{A}/k)[3]^{Gal(k/ℚ)} is isomorphic to ℤ/3. This contradicts Lemma 4 which claims that the order of � (E_{A}/k)[3]^{Gal(k/ℚ)} must be a square. As a result, � (E_{A}/k) [\sqrt{−3}] cannot have order 3. □

In conclusion, we have proved that C(A) = S(A) for A = p_{1}p_{2}p. Moreover, in the proof of Lemma 3 we have shown p_{1}p_{2}^{2}(k^{∗})^{3} belongs to S(A); it follows that the curve p_{1}X^{3} + p_{2}Y^{3} = p has a k_{ρ}−point for any prime ρ of k and hence, by the local-to-global principle, it has a point over the quadratic extension k of ℚ. Therefore it has a point over ℚ.

In order to prove Theorem 1, it only remains to be shown that given the prime integers p_{1}, p_{2}, p_{3}, p_{4} ≡ 2 (mod 3) there exists a prime π such that the hypotheses of Lemma 3 are satisfied for each of the triples p_{1}, p_{2}, π and p_{3}, p_{4}, π.

Lemma 6 Let p_{1}, p_{2}, p_{3}, p_{4} be prime integers congruent to 2 modulo 3. Then there exists a prime π ∈ ℤ[ω] such that

(i) π ≡ 1 (mod 3);

(ii) p_{1}p_{2}N(π) and p_{3}p_{4}N(π) are not congruent to 1 modulo 9;

(iii) (\frac{π}{p_{1}} )_{3} = (\frac{π}{p_{2}} )_{3} = (\frac{π}{p_{3}} )_{3} = (\frac{π}{p_{4}} )_{3} ≠ 1.

Proof. Let B ∈ {1, 4, 7} such that

\{ \begin{matrix}p_{1}p_{2}B ≢ 1 (mod 9) \\ p_{3}p_{4}B ≢ 1 (mod 9). \end{matrix}

Take a prime π ∈ ℤ[ω] such that N(π) ≡ B (mod 9). This condition can be satisfied by taking π ≡ β (mod 9), where β is an element of ℤ[ω] congruent to 1, −2 or 1+3ω modulo 9 if B = 1, 4 or 7, respectively. Hence we have that p_{i}p_{j}B ≢ 1 (mod 9) if and only if p_{i}p_{j}N(π) ≢ 1 (mod 9). Therefore conditions (i) and (ii) are satisfied.

As far as condition (iii) is concerned, in order for π to satisfy

(\frac{π}{p_{1}} )_{3} = (\frac{π}{p_{2}} )_{3} = (\frac{π}{p_{3}} )_{3} = (\frac{π}{p_{4}} )_{3} ≠ 1

it is sufficient to take π belonging to a suitable congruence class modulo p_{1}p_{2}p_{3}p_{4}. The Chinese Remainder Theorem allows us to determine a suitable residue class γ modulo 9p_{1}p_{2}p_{3}p_{3} such that, if π ≡ γ (mod 9p_{1}p_{2}p_{3}p_{3}), then π satisfies the required conditions. The existence of such a prime π is assured by Dirichlet’s Theorem. □

Hence Theorem 1 is proved.

We thank Alexei Skorobogatov for his help in the preparation of this note.

References

[1] J.W.S. Cassels. Arithmetic on curves of genus 1. I. On a conjecture of Selmer. J. reine angew. Math. 202 (1959) 52–99.

[2] J.W.S. Cassels. Lectures on Elliptic Curves. Cambridge University Press, 1991.

[3] R. Heath-Brown. The solubility of diagonal cubic equations. Proc. London Math. Soc. (3) 79 (1999), 241–259.

[4] J.-P. Serre. A course in Arithmetic. New York: Springer–Verlag, 1973.

[5] J.-P. Serre. Galois cohomology. Springer–Verlag Berlin Heidelberg, 1997.

[6] J.H. Silverman. The Arithmetic of Elliptic Curves. New York: Springer–Verlag, 1986.

[7] H.P.F. Swinnerton–Dyer. The solubility of diagonal cubic surfaces. Preprint (August