Journal de Théorie des Nombres de Bordeaux

Jörg Jahnel

Jörg JAHNEL
More cubic surfaces violating the Hasse principle
Tome 23, no 2 (2011), p. 471-477.
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Journal de Théorie des Nombres
de Bordeaux 23 (2011), 471-477
More cubic surfaces violating the Hasse principle
par Jörg JAHNEL
Résumé. Nous généralisons la construction due à L. J. Mordell
de surfaces cubiques pour lesquelles le principe de Hasse est faux.
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 con-
struction had soon been generalized by L. J. Mordell [3] who found two se-
ries of such examples. The starting points of Mordell’s construction are the
cubic number fields contained in Q(ζp) 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 con-
tained in the cyclotomic extension Q(ζp)/Q.
ii) We fix the explicit generator θ ∈ K given by θ := trQ(ζp)/K(ζp − 1).
More concretely, θ = −n+
∑
i∈(F∗p )3
ζip for n :=
p−1
3 .
Theorem 1.1. Consider the cubic surface X ⊂ P3Q, given by
T3(a1T0 + d1T3)(a2T0 + d2T3) =
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
.
Here, a1, a2, d1, d2 are integers and θ(i) are the images of θ under
Gal(K/Q).
i) Then, the reduction Xp of X at p is given by
T3(a1T0 + d1T3)(a2T0 + d2T3) = T 30 .
Over the algebraic closure, Xp is the union of three planes. These are
given by
T3/T0 = s1, T3/T0 = s2, T3/T0 = s3
for si the zeroes of T (a1 + d1T )(a2 + d2T )− 1, considered as a polynomial
over Fp.
Manuscrit reçu le 10 août 2009.
472 Jörg Jahnel
ii) Suppose p - d1d2 and gcd(d1, d2) = 1. Then, for every
(t0 : t1 : t2 : t3) ∈ X(Q)
the term s := (t3/t0 mod p) admits the property that
a1 + d1s
s
is a cube in F∗p .
In particular, if (a1 + d1si)/si ∈ F∗p is a non-cube for every i such that
si ∈ Fp then X(Q) = ∅.
iii) Assume that p - d1d2 and that gcd(a1, d1) and gcd(a2, d2) contain
only prime factors that completely split in K. Suppose further that
T (a1 + d1T )(a2 + d2T )− 1 ∈ Fp[T ] has at least one simple zero in Fp.
Then, X(AQ) 6= ∅.
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 6= 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
∏3
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
= NK/Q(T0 + θT1 + θ2T2) .
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,
a1 = d1 = a2 = 1, and d2 = 2.
2. The proofs
Observations 2.1. i) For s any solution of T (a1+d1T )(a2+d2T )−1 = 0,
the expression (a1 + d1s)/s is well defined and non-zero.
ii) No Qp-valued point on X reduces to the triple line “T0 = T3 = 0”.
iii) For every (t0 : t1 : t2 : t3) ∈ X(Qp), the fraction (a1t0 + d1t3)/t3 is a
p-adic unit.
Proof. i) By assumption, we have s 6= 0 and a1 + d1s 6= 0.
ii) Suppose, (t0 : t1 : t2 : t3) ∈ X(Qp) is a point reducing to the triple line.
We may assume t0, t1, t2, t3 ∈ Zp are coprime. Then νp(t0) ≥ 1 and
νp(t3) ≥ 1 together imply that νp
(
t3(a1t0 + d1t3)(a2t0 + d2t3)
)
≥ 3. On the
other hand,
νp
( 3∏
i=1
(t0 + θ(i)t1 + (θ(i))2t2)
)
equals 1 or 2 since t1 or t2 is a unit and νp(θ(i)) = 13 .
More cubic surfaces violating the Hasse principle 473
iii) Again, assume t0, t1, t2, t3 ∈ Zp to be coprime. Assertion ii) implies that
t3 is a p-adic unit. Hence, (a1t0 + d1t3)/t3 ∈ Zp. Further,(a1t0 + d1t3
t3
mod p
)
= a1 + d1s
s
for s := (t3/t0 mod p) a solution of T (a1 + d1T )(a2 + d2T )− 1 = 0. 
Lemma 2.1. Let ν be any valuation of Q different from νp and w an
extension of ν to K. Further, let X be as in Theorem 1.1.ii) and
(t0 : t1 : t2 : t3) ∈ X(Qν).
Then, (a1t0 + d1t3)/t3 ∈ Q∗ν is in the image of the norm map
N : Kw → Qν .
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 := νq.
It might happen that θ is not a unit in Kw. However, as Kw/Qq is
unramified, there exists some t ∈ Q∗q such that θ := tθ ∈ Kw is a unit.
The surface X̃ given by
T3(a1T0 + d1T3)(a2T0 + d2T3) =
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
is isomorphic to X ×SpecQ SpecQq. Even more, the map
ι : X ×SpecQ SpecQq → X̃, (t0 : t1 : t2 : t3) 7→ (t0 : t1t :
t2
t2 : t3)
is an isomorphism that leaves the rational function (a1T0 + d1T3)/T3 un-
changed. Hence, we may assume without restriction that θ ∈ Kw is a unit.
Third step: The case that θ is a unit.
Assume that t0, t1, t2, t3 ∈ Zq are coprime. If
ν
(
t3(a1t0 + d1t3)(a2t0 + d2t3)
)
= 0
then (a1t0 + d1t3)/t3 is a q-adic unit, hence clearly a norm. Otherwise, we
have
ν
( 3∏
i=1
(t0 + θ(i)t1 + (θ(i))2t2)
)
> 0 .
This means that one of the factors t0 + θ(i)t1 + (θ(i))2t2 vanishes after
reduction to the residue field Fq3 . As θ(i) is reduced to a generator of the
extension Fq3/Fq, this implies that ν(t0), ν(t1), ν(t2) > 0. Consequently,
t3 must be a unit.
474 Jörg Jahnel
From the equation of X, we deduce ν(d1d2) > 0. If ν(d2) > 0 then,
according to the assumption, d1 is a unit. This shows ν(a1t0 + d1t3) = 0,
from which the assertion follows.
Thus, assume ν(d1) > 0. Then, d2 is a unit and, therefore,
ν(a2t0 + d2t3) = 0. Further, we note that
3 | ν
( 3∏
i=1
(t0 + θ(i)t1 + (θ(i))2t2)
)
since the product is a norm. By consequence,
3 | ν
(
t3(a1t0 + d1t3)(a2t0 + d2t3)
)
.
Altogether, we see that 3 | ν(a1t0 +d1t3) and 3 | ν
(
(a1t0 +d1t3)/t3
)
. The as-
sertion follows. 
Proof of Theorem 1.1.ii). According to Lemma 2.1, (a1t0 + d1t3)/t3 ∈ Q∗
is a local norm at every prime except p. Global class field theo-
ry [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), (a1t0 + d1t3)/t3 is automatically a p-adic unit.
(p) = p3 is a totally ramified prime. A p-adic unit u is a norm if and only if
u := (u mod p) is a cube in F∗p . As (a1 +d1s)/s =
(
(a1t0 +d1t3)/t3 mod p
)
,
this is exactly the assertion. 
Proof of Theorem 1.1.iii). We have to show that X(Qν) 6= ∅ for every val-
uation of Q. X(R) 6= ∅ is obvious. For a prime number q, in order to prove
X(Qq) 6= ∅, we use Hensel’s lemma. It is sufficient to verify that the re-
duction Xq has a smooth Fq-valued point. Thereby, we may replace X by
a Qq-scheme X̃ isomorphic to X ×SpecQ SpecQq.
Case 1: q = p.
Then, the reduction Xp is the union of three planes meeting in the line
given by T0 = T3 = 0. By assumption, one of the planes appears with
multiplicity one and is defined over Fp. It contains p2 smooth points.
Case 2: q 6= p.
Assume without restriction that θ is a w-adic unit. There are two sub-
cases.
a) q - d1d2. It suffices to show that there is a smooth Fq-valued point on the
intersection X ′q of Xq with the hyperplane “T0 = 0”. This curve is given by
d1d2T
3
3 = θ
(1)
θ
(2)
θ
(3)
3∏
i=1
(T1 + θ
(i)
T2) .
If q 6= 3 then this equation defines a smooth genus one curve. It has an
Fq-valued point by Hasse’s bound.
More cubic surfaces violating the Hasse principle 475
If q = 3 then the projection X ′q → P1 given by (T1 : T2 : T3) 7→ (T1 : T2)
is one-to-one on Fq-valued points. At least one of them is smooth since∏3
i=1(T + θ
(i)) is a separable polynomial.
b) q|d1d2. Then, X ′q := Xq ∩ “T0 = 0” is given by
0 = θ(1)θ(2)θ(3)
3∏
i=1
(T1 + θ
(i)
T2) .
In particular, x = (0 : 0 : 0 : 1) ∈ Xq(Fq). We may assume that x is singular.
Then,Xq is given as Q(T0, T1, T2)T3+K(T0, T1, T2) = 0 for Q a quadratic
form and K a cubic form. If Q 6≡ 0 then there is an Fq-rational line `
through x such that Q|` 6= 0. Hence, ` meets Xq twice in x and once in
another Fq-valued point that is smooth.
Otherwise, (F mod q) does not depend on T3. 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, a1 ≡ d1 ≡ 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 Fq, different from each other, and meet in one point.
There are plenty of smooth points on X ′q. These points are smooth on Xq,
as well. 
3. Examples
Example. For p = 19, a counterexample to the Hasse principle is given by
T3(19T0 + 5T3)(19T0 + 4T3) =
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
= T 30 − 19T 20 T1 + 133T 20 T2 + 114T0T 21
− 1 539T0T1T2 + 5 054T0T 22 − 209T 31
+ 3 971T 21 T2 − 23 826T1T 22 + 43 681T 32 .
Indeed, in F19, the cubic equation T 3 − 1 = 0 has the three solutions 1, 7,
and 11. However, in any case (a1 + d1s)/s = 5, which is a non-cube.
Example. Put p = 19. Consider the cubic surface X given by
T3(T0 + T3)(12T0 + T3) =
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
.
Then, X(AQ) 6= ∅ but X(Q) = ∅. X violates the Hasse principle.
Indeed, in F19, the cubic equation T (1 + T )(12 + T ) − 1 = 0 has the
three solutions 12, 15, and 17. However, in F19, 13/12 = 9, 16/15 = 15,
and 18/17 = 10, which are three non-cubes.
476 Jörg Jahnel
Example. For p = 19, consider the cubic surface X given by
T3(T0 + T3)(2T0 + T3) =
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
.
Then, for X, the Hasse principle fails.
Indeed, in F19, 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 F192 . However, in F19, 6/5 = 5 is a non-cube.
Example. Put p = 19 and consider the cubic surface X given by
T3(T0 + T3)(6T0 + T3) =
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
.
There are Q-rational points on X but weak approximation fails.
Indeed, in F19, the cubic equation T (1 +T )(6 +T )− 1 = 0 has the three
solutions 8, 9, and 14. However, in F19, 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, T3/T0 = −7/14 ≡ 9 (mod 19).
Remark. From each of the examples given, by adding multiples of p to
the coefficients a1, d1, a2, and d2, 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 R3, generated by v1 := (θ(1), θ(2), θ(3)) and
v2 :=
(
(θ(1))2, (θ(2))2, (θ(3))2
)
, in fact {v1, v2+7v1} is a reduced basis. There-
fore, the substitution T ′1 := T1 − 7T2 simplifies the norm form. We find
3∏
i=1
(
T0 + θ(i)T1 + (θ(i))2T2
)
= T 30 − 19T 20 T ′1 + 114T0T ′12 + 57T0T ′1T2
− 133T0T 22 − 209T ′13 − 418T ′12T2
+ 1045T ′1T 22 − 209T 32 .
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 re-
sponsible 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 func-
tion f ∈ Q(X) given by (a1T0+d1T3)/T3. The principal divisor div(f) is the
More cubic surfaces violating the Hasse principle 477
norm of a divisor E ∈ Div(XK). Indeed, it is the sum over the three con-
jugate lines given by a1T0 + d1T3 = T0 + θ(i)T1 + (θ(i))2T2 = 0 minus the
sum of the three conjugate lines given by T3 = T0 + θ(i)T1 + (θ(i))2T2 = 0.
By Manin’s formula [2, Proposition 31.3], such a principal divisor in-
duces a class in H1(Gal(K/Q),Pic(XK)) ⊆ H1(Gal(Q/Q),Pic(XQ)). Fur-
thermore, 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(AQ) essentially means to evaluate f and to
apply the norm-residue-homomorphismsQ∗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: jahnel@mathematik.uni-siegen.de
URL: http://www.uni-math.gwdg.de/jahnel


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