Journal de Théorie des Nombres de Bordeaux 00 (XXXX), 000–000

More cubic surfaces violating the Hasse principle

par Jörg Jahnel

Résumé. Nous généralisons la construction due à L. J. Mordell des surfaces cubiques pour lesquelles la principe de Hasse est fausse.

Abstract. We generalize L. J. Mordell’s construction of cubic surfaces for which the Hasse principle fails.

1. Introduction and main result

Sir Peter Swinnerton-Dyer [4] was the first to construct a cubic surface over ℚ for which the Hasse principle provably fails. Swinnerton-Dyer’s construction had soon been generalized by L. J. Mordell [3] who found two series of such examples. The starting points of Mordell’s construction are the cubic number fields contained in ℚ(ζ_{p}) for p = 7 and p = 13, respectively.

In this note, we will show that Mordell’s construction may be generalized to an arbitrary prime p ≡ 1 (mod 3).

Notation. i) We denote by K/ℚ the unique cubic field extension contained in the cyclotomic extension ℚ(ζ_{p})/ℚ. ii) We fix the explicit generator θ ∈ K given by θ := tr_{ℚ(ζ_{p})/K} (ζ_{p} − 1). More concretely, θ = −n + ∑_{i∈(𝔽_{p}^{∗})^{3}}ζ_{p}^{i} for n := \frac{p−1}{3}.

Theorem 1.1. Consider the cubic surface X ⊂ P_{ℚ}^{3} , given by

T_{3}(a_{1}T_{0} + d_{1}T_{3})(a_{2}T_{0} + d_{2}T_{3}) = ∏_{i=1}^{3} (T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2}) .

Here, a_{1}, a_{2}, d_{1}, d_{2} are integers and θ^{(i)} are the images of θ under Gal(K/ℚ).

i) Then, the reduction X_{p} of X at p is given by

T_{3}(a_{1}T_{0} + d_{1}T_{3})(a_{2}T_{0} + d_{2}T_{3}) = T_{0}^{3} .

Over the algebraic closure, X_{p} is the union of three planes. These are given by

T_{3}/T_{0} = s_{1}, T_{3}/T_{0} = s_{2}, T_{3}/T_{0} = s_{3}

for s_{i} the zeroes of T (a_{1} + d_{1}T)(a_{2} + d_{2}T) − 1, considered as a polynomial over 𝔽_{p}. ii) Suppose p ∤ d_{1}d_{2} and gcd(d_{1}, d_{2}) = 1. Then, for every

(t_{0} : t_{1} : t_{2} : t_{3}) ∈ X(ℚ)

the term s := (t_{3}/t_{0} mod p) admits the property that

\frac{a_{1} + d_{1}s}{s}

is a cube in 𝔽_{p}^{∗}. In particular, if (a_{1} + d_{1}s_{i})/s_{i} ∈ 𝔽_{p}^{∗} is a non-cube for every i such that s_{i} ∈ 𝔽_{p} then X(ℚ) = ∅. iii) Assume that p ∤ d_{1}d_{2} and that gcd(a_{1}, d_{1}) and gcd(a_{2}, d_{2}) contain only prime factors that completely split in K. Suppose further that T (a_{1} + d_{1}T)(a_{2} + d_{2}T) − 1 ∈ 𝔽_{p}[T] has at least one simple zero in 𝔽_{p}. Then, X(𝔸_{ℚ} ) ≠ ∅.

Remarks. i) K/ℚ is an abelian cubic field extension. It is totally ramified at p and unramified at all other primes. A prime q ≠ p is split in K if and only if q is a cube modulo p. ii) We will write 𝔭 for the prime ideal in K lying above (p). Note that 𝔭 = (θ) by virtue of our definition of θ. iii) We have ∏_{i=1}^{3}(T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2}) = N_{K/ℚ} (T_{0} + θT_{1} + θ^{2}T_{2}) .

Remark. For p = 7 and 13, we recover exactly the result of L. J. Mordell. The original example of Sir Peter Swinnerton-Dyer reappears for p = 7, a_{1} = d_{1} = a_{2} = 1, and d_{2} = 2.

2. The proofs

Observations 2.1. i) For s any solution of T (a_{1}+d_{1}T)(a_{2}+d_{2}T)−1 = 0, the expression (a_{1} + d_{1}s)/s is well defined and non-zero. ii) No ℚ_{p}-valued point on X reduces to the triple line “ T_{0} = T_{3} = 0”. iii) For every (t_{0} : t_{1} : t_{2} : t_{3}) ∈ X(ℚ_{p}), the fraction (a_{1}t_{0} + d_{1}t_{3})/t_{3} is a p-adic unit.

Proof. i) By assumption, we have s ≠ 0 and a_{1} + d_{1}s ≠ 0. ii) Suppose, (t_{0} : t_{1} : t_{2} : t_{3}) ∈ X(ℚ_{p}) is a point reducing to the triple line. We may assume t_{0}, t_{1}, t_{2}, t_{3} ∈ ℤ_{p} are coprime. Then ν_{p}(t_{0}) ≥ 1 and ν_{p}(t_{3}) ≥ 1 together imply that ν_{p}(t_{3}(a_{1}t_{0} + d_{1}t_{3})(a_{2}t_{0} + d_{2}t_{3})) ≥ 3. On the other hand,

ν_{p}( ∏_{i=1}^{3}(t_{0} + θ^{(i)}t_{1} + (θ^{(i)})^{2}t_{2}))

equals 1 or 2 since t_{1} or t_{2} is a unit and ν_{p}(θ^{(i)}) = \frac{1}{3}. iii) Again, assume t_{0}, t_{1}, t_{2}, t_{3} ∈ ℤ_{p} to be coprime. Assertion ii) implies that t_{3} is a p-adic unit. Hence, (a_{1}t_{0} + d_{1}t_{3})/t_{3} ∈ ℤ_{p}. Further,

(\frac{a_{1}t_{0} + d_{1}t_{3}}{t_{3}} mod p) = \frac{a_{1} + d_{1}s}{s}

for s := (t_{3}/t_{0} mod p) a solution of T (a_{1} + d_{1}T)(a_{2} + d_{2}T) − 1 = 0. □

Lemma 2.1. Let ν be any valuation of ℚ different from ν_{p} and w an extension of ν to K. Further, let X be as in Theorem 1.1.ii) and (t_{0} : t_{1} : t_{2} : t_{3}) ∈ X(ℚ_{ν}). Then, (a_{1}t_{0}+d_{1}t_{3})/t_{3} ∈ ℚ_{ν}^{∗} is in the image of the norm map N: K_{w} → ℚ_{ν}.

Proof. First step: Elementary cases. If q is a prime split in K then every element of ℚ_{q}^{∗} is a norm. The same applies to the infinite prime.

Second step: Preparations. It remains to consider the case that q remains prime in K. Then, an element x ∈ ℚ_{q}^{∗} is a norm if and only if 3|ν(x) for ν := ν_{q}.

It might happen that θ is not a unit in K_{w}. However, as K_{w}/ℚ_{q} is unramified, there exists some t ∈ ℚ_{q}^{∗} such that \underline{θ} := tθ ∈ K_{w} is a unit. The surface X̃ given by

T_{3}(a_{1}T_{0} + d_{1}T_{3})(a_{2}T_{0} + d_{2}T_{3}) = ∏_{i=1}^{3} (T_{0} + \underline{θ}^{(i)}T_{1} + (\underline{θ}^{(i)})^{2}T_{2})

is isomorphic to X ×_{Spec ℚ} Spec ℚ_{q}. Even more, the map

ι: X ×_{Spec ℚ} Spec ℚ_{q} → X̃, (t_{0} : t_{1} : t_{2} : t_{3}) ↦ (t_{0} : \frac{t_{1}}{t} : \frac{t_{2}}{t^{2}} : t_{3})

is an isomorphism that leaves the rational function (a_{1}T_{0} + d_{1}T_{3})/T_{3} unchanged. Hence, we may assume without restriction that θ ∈ K_{w} is a unit.

Third step: The case that θ is a unit. Assume that t_{0}, t_{1}, t_{2}, t_{3} ∈ ℤ_{q} are coprime. If

ν(t_{3}(a_{1}t_{0} + d_{1}t_{3})(a_{2}t_{0} + d_{2}t_{3})) = 0

then (a_{1}t_{0} + d_{1}t_{3})/t_{3} is a q-adic unit, hence clearly a norm. Otherwise, we have

ν( ∏_{i=1}^{3}(t_{0} + θ^{(i)}t_{1} + (θ^{(i)})^{2}t_{2})) > 0 .

This means that one of the factors t_{0} + θ^{(i)}t_{1} + (θ^{(i)})^{2}t_{2} vanishes after reduction to the residue field 𝔽_{q^{3}}. As θ^{(i)} is reduced to a generator of the extension 𝔽_{q^{3}}/𝔽_{q}, this implies that ν(t_{0}), ν(t_{1}), ν(t_{2}) > 0. Consequently, t_{3} must be a unit.

From the equation of X, we deduce ν(d_{1}d_{2}) > 0. If ν(d_{2}) > 0 then, according to the assumption, d_{1} is a unit. This shows ν(a_{1}t_{0} + d_{1}t_{3}) = 0, from which the assertion follows.

Thus, assume ν(d_{1}) > 0. Then, d_{2} is a unit and, therefore, ν(a_{2}t_{0} + d_{2}t_{3}) = 0. Further, we note that

3 | ν(∏_{i=1}^{3}(t_{0} + θ^{(i)}t_{1} + (θ^{(i)})^{2}t_{2}))

since the product is a norm. By consequence,

3 | ν(t_{3}(a_{1}t_{0} + d_{1}t_{3})(a_{2}t_{0} + d_{2}t_{3})).

Altogether, we see that 3 | ν(a_{1}t_{0} +d_{1}t_{3}) and 3 | ν((a_{1}t_{0} +d_{1}t_{3})/t_{3}). The assertion follows. □

Proof of Theorem 1.1.ii). According to Lemma 2.1, (a_{1}t_{0}+d_{1}t_{3})/t_{3} ∈ ℚ^{∗} is a local norm at every prime except p. Global class field theory [5, Theorem 5.1 together with 6.3] shows that it must necessarily be a norm at that prime, too.

By Observation 2.1.iii), (a_{1}t_{0} + d_{1}t_{3})/t_{3} is automatically a p-adic unit.

(p) = 𝔭^{3} is a totally ramified prime. A p-adic unit u is a norm if and only if \overline{u} := (u mod p) is a cube in 𝔽_{p}^{∗}. As (a_{1}+d_{1}s)/s = ((a_{1}t_{0}+d_{1}t_{3})/t_{3} mod p), this is exactly the assertion. □

Proof of Theorem 1.1.iii). We have to show that X(ℚ_{ν}) ≠ ∅ for every valuation of ℚ. X(ℝ) ≠ ∅ is obvious. For a prime number q, in order to prove X(ℚ_{q}) ≠ ∅, we use Hensel’s lemma. It is sufficient to verify that the reduction X_{q} has a smooth 𝔽_{q}-valued point. Thereby, we may replace X by a ℚ_{q}-scheme X̃ isomorphic to X ×_{Spec ℚ} Spec ℚ_{q}.

Case 1: q = p. Then, the reduction X_{p} is the union of three planes meeting in the line given by T_{0} = T_{3} = 0. By assumption, one of the planes appears with multiplicity one and is defined over 𝔽_{p}. It contains p^{2} smooth points.

Case 2: q ≠ p. Assume without restriction that θ is a w-adic unit. There are two subcases.

a) q ∤ d_{1}d_{2}. It suffices to show that there is a smooth 𝔽_{q}-valued point on the intersection X′_{q} of X_{q} with the hyperplane “T_{0} = 0”. This curve is given by

\overline{d}_{1}\overline{d}_{2}T_{3}^{3} = \overline{θ}^{(1)}\overline{θ}^{(2)}\overline{θ}^{(3)} ∏_{i=1}^{3}(T_{1} + \overline{θ}^{(i)}T_{2}) .

If q ≠ 3 then this equation defines a smooth genus one curve. It has an 𝔽_{q}-valued point by Hasse’s bound.

If q = 3 then the projection X′_{q} → P^{1} given by (T_{1} : T_{2} : T_{3}) ↦ (T_{1} : T_{2}) is one-to-one on 𝔽_{q}-valued points. At least one of them is smooth since ∏_{i=1}^{3}(T + \overline{θ}^{(i)}) is a separable polynomial.

b) q|d_{1}d_{2}. Then, X′_{q} := X_{q} ∩ “T_{0} = 0” is given by

0 = \overline{θ}^{(1)}\overline{θ}^{(2)}\overline{θ}^{(3)} ∏_{i=1}^{3}(T_{1} + \overline{θ}^{(i)}T_{2}) .

In particular, x = (0 : 0 : 0 : 1) ∈ X_{q}(𝔽_{q}). We may assume that x is singular.

Then, X_{q} is given as Q(T_{0}, T_{1}, T_{2})T_{3}+K(T_{0}, T_{1}, T_{2}) = 0 for Q a quadratic form and K a cubic form. If Q ≢ 0 then there is an 𝔽_{q}-rational line ℓ through x such that Q|_{ℓ} ≠ 0. Hence, ℓ meets X_{q} twice in x and once in another 𝔽_{q}-valued point that is smooth.

Otherwise, (F mod q) does not depend on T_{3}. I.e., the left hand side of the equation of X vanishes modulo q. This means that one of the factors must vanish. We have, say, a_{1} ≡ d_{1} ≡ 0 (mod q). Then, by assumption, q splits completely in K. At such a prime, X′_{q} is the union of three lines that are defined over 𝔽_{q}, different from each other, and meet in one point. There are plenty of smooth points on X′_{q}. These points are smooth on X_{q}, as well. □

3. Examples

Example. For p = 19, a counterexample to the Hasse principle is given by

T_{3}(19T_{0} + 5T_{3})(19T_{0} + 4T_{3}) = ∏_{i=1}^{3} (T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2})
= T_{0}^{3} − 19T_{0}^{2}T_{1} + 133T_{0}^{2}T_{2} + 114T_{0}T_{1}^{2}
− 1 539 T_{0}T_{1}T_{2} + 5 054 T_{0}T_{2}^{2} − 209T_{1}^{3}

+ 3 971 T_{1}^{2}T_{2} − 23 826 T_{1}T_{2}^{2} + 43 681 T_{2}^{3} .

Indeed, in 𝔽_{19}, the cubic equation T^{3} − 1 = 0 has the three solutions 1, 7, and 11. However, in any case (a_{1} + d_{1}s)/s = 5, which is a non-cube.

Example. Put p = 19. Consider the cubic surface X given by

T_{3}(T_{0} + T_{3})(12T_{0} + T_{3}) = ∏_{i=1}^{3} (T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2}) .

Then, X(𝔸_{ℚ} ) ≠ ∅ but X(ℚ) = ∅. X violates the Hasse principle.

Indeed, in 𝔽_{19}, the cubic equation T (1 + T)(12 + T) − 1 = 0 has the three solutions 12, 15, and 17. However, in 𝔽_{19}, 13/12 = 9, 16/15 = 15, and 18/17 = 10, which are three non-cubes.

Example. For p = 19, consider the cubic surface X given by

T_{3}(T_{0} + T_{3})(2T_{0} + T_{3}) = ∏_{i=1}^{3} (T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2}) .

Then, for X, the Hasse principle fails.

Indeed, in 𝔽_{19}, the cubic equation T (1 + T)(2 + T) − 1 = 0 has T = 5 as its only solution. The two other solutions are conjugate to each other in 𝔽_{19^{2}}. However, in 𝔽_{19}, 6/5 = 5 is a non-cube.

Example. Put p = 19 and consider the cubic surface X given by

T_{3}(T_{0} + T_{3})(6T_{0} + T_{3}) = ∏_{i=1}^{3} (T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2}) .

There are ℚ-rational points on X but weak approximation fails.

Indeed, in 𝔽_{19}, the cubic equation T (1 + T)(6 + T) − 1 = 0 has the three solutions 8, 9, and 14. However, in 𝔽_{19}, 10/9 = 18 is a cube while 9/8 = 13 and 15/14 = 16 are non-cubes. The smallest ℚ-rational point on X is (14 : 15 : 2 : (−7)). Observe that, in fact, T_{3}/T_{0} = −7/14 ≡ 9 (mod 19).

Remark. From each of the examples given, by adding multiples of p to the coefficients a_{1}, d_{1}, a_{2}, and d_{2}, a family of surfaces arises, which are of similar nature.

Remark (Lattice basis reduction). The norm form in the p = 19 examples produces coefficients that are rather large. An equivalent form with smaller coefficients may be obtained using lattice basis reduction. In its simplest form, this means the following.

For the rank 2 lattice in ℝ^{3}, generated by v_{1} := (θ^{(1)}, θ^{(2)}, θ^{(3)}) and v_{2} := ((θ^{(1)})^{2}, (θ^{(2)})^{2}, (θ^{(3)})^{2}), in fact {v_{1}, v_{2} + 7v_{1}} is a reduced basis. Therefore, the substitution T′_{1} := T_{1}−7T_{2} simplifies the norm form. We find

∏_{i=1}^{3}(T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2}) = T_{0}^{3} − 19T_{0}^{2}T′_{1} + 114T_{0}T′_{1}^{2} + 57T_{0}T′_{1}T_{2}

− 133T_{0}T_{2}^{2} − 209T′_{1}^{3} − 418T′_{1}^{2}T_{2} + 1045T′_{1}T_{2}^{2} − 209T_{2}^{3} .

Remark (Brauer-Manin obstruction). It was observed by Yu. I. Manin [2] that a class α ∈ Br(X) in the Grothendieck-Brauer group may be responsible for the failure of the Hasse principle or of weak approximation. In fact, all the examples given above may be explained more conceptually in this way.

In short, this may be seen is as follows. We consider the rational function f ∈ ℚ(X) given by (a_{1}T_{0}+d_{1}T_{3})/T_{3}. The principal divisor div(f) is the norm of a divisor E ∈ Div(X_{K}). Indeed, it is the sum over the three conjugate lines given by a_{1}T_{0} + d_{1}T_{3} = T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2} = 0 minus the sum of the three conjugate lines given by T_{3} = T_{0} + θ^{(i)}T_{1} + (θ^{(i)})^{2}T_{2} = 0.

By Manin’s formula [2, Proposition 31.3], such a principal divisor induces a class in H^{1}(Gal(K/ℚ), Pic(X_{K})) ⊆ H^{1}(Gal(\overline{ℚ}/ℚ), Pic(X\overline{ℚ} )). Furthermore, for cubic surfaces, the Hochschild-Serre spectral sequence shows that the latter Galois cohomology group is isomorphic to Br(X)/ Br(ℚ).

Hence, associated to f, we have a Brauer class α ∈ Br(X), which is unique up to addition of some element from Br(ℚ). To evaluate α ∈ Br(X) at an adelic point x ∈ X(𝔸_{ℚ} ) essentially means to evaluate f and to apply the norm-residue-homomorphisms ℚ_{ν}^{∗}/NK_{w}^{∗} → ℚ/ℤ [2, 45.2]. This is exactly what we did in the proof of Theorem 1.1.ii).

More details on this approach are given in the author’s Habilitation thesis [1, Sec. III.5].

References

[1] J. Jahnel Brauer groups, Tamagawa measures, and rational points on algebraic varieties. Habilitation thesis, Göttingen 2008

[2] Yu. I. Manin Cubic forms, algebra, geometry, arithmetic. North-Holland Publishing Co. and American Elsevier Publishing Co., Amsterdam-London and New York 1974

[3] L. J. Mordell On the conjecture for the rational points on a cubic surface. J. London Math. Soc. 40 (1965), 149–158

[4] Sir Peter Swinnerton-Dyer Two special cubic surfaces. Mathematika 9 (1962), 54–56 [5] J. Tate Global class field theory. In: Algebraic number theory, Edited by J. W. S. Cassels

and A. Fröhlich, Academic Press and Thompson Book Co., London and Washington 1967

Jörg Jahnel FB6 Mathematik Walter-Flex-Str. 3, D-57068 Siegen, Germany

E-mail : ~jjahnel@mathematik.uni-siegen.de
URL: http://wwwuser.gwdg.de/~jjahnel