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

Jahnel, Jörg

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Jörg Jahnel

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Journal de Théorie des Nombres de Bordeaux

More cubic surfaces violating the Hasse principle

Numérisation de Documents Anciens Mathématiques

Jörg JAHNELMore cubic surfaces violating the Hasse principleTome 23, no 2 (2011), p. 471-477.<http://jtnb.cedram.org/item?id=JTNB_2011__23_2_471_0>© Société Arithmétique de Bordeaux, 2011, tous droits réservés.L’accès aux articles de la revue « Journal de Théorie des Nom-bres de Bordeaux » (http://jtnb.cedram.org/), implique l’accordavec les conditions générales d’utilisation (http://jtnb.cedram.org/legal/). Toute reproduction en tout ou partie cet article sousquelque forme que ce soit pour tout usage autre que l’utilisation àfin strictement personnelle du copiste est constitutive d’une infrac-tion pénale. Toute copie ou impression de ce fichier doit contenir laprésente mention de copyright.cedramArticle mis en ligne dans le cadre duCentre de diffusion des revues académiques de mathématiqueshttp://www.cedram.org/Journal de Théorie des Nombresde Bordeaux 23 (2011), 471-477More cubic surfaces violating the Hasse principlepar Jörg JAHNELRésumé. Nous généralisons la construction due à L. J. Mordellde surfaces cubiques pour lesquelles le principe de Hasse est faux.Abstract. We generalize L. J. Mordell’s construction of cubicsurfaces for which the Hasse principle fails.1. Introduction and main resultSir Peter Swinnerton-Dyer [4] was the first to construct a cubic surfaceover 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 thecubic 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 generalizedto 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 )3ζip for n :=p−13 .Theorem 1.1. Consider the cubic surface X ⊂ P3Q, given byT3(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 underGal(K/Q).i) Then, the reduction Xp of X at p is given byT3(a1T0 + d1T3)(a2T0 + d2T3) = T 30 .Over the algebraic closure, Xp is the union of three planes. These aregiven byT3/T0 = s1, T3/T0 = s2, T3/T0 = s3for si the zeroes of T (a1 + d1T )(a2 + d2T )− 1, considered as a polynomialover Fp.Manuscrit reçu le 10 août 2009.472 Jörg Jahnelii) 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 thata1 + d1ssis a cube in F∗p .In particular, if (a1 + d1si)/si ∈ F∗p is a non-cube for every i such thatsi ∈ Fp then X(Q) = ∅.iii) Assume that p - d1d2 and that gcd(a1, d1) and gcd(a2, d2) containonly prime factors that completely split in K. Suppose further thatT (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 ramifiedat p and unramified at all other primes. A prime q 6= p is split in K if andonly if q is a cube modulo p.ii) We will write p for the prime ideal in K lying above (p). Note thatp = (θ) by virtue of our definition of θ.iii) We have∏3i=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 proofsObservations 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 ap-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 andνp(t3) ≥ 1 together imply that νp(t3(a1t0 + d1t3)(a2t0 + d2t3))≥ 3. On theother 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 473iii) Again, assume t0, t1, t2, t3 ∈ Zp to be coprime. Assertion ii) implies thatt3 is a p-adic unit. Hence, (a1t0 + d1t3)/t3 ∈ Zp. Further,(a1t0 + d1t3t3mod p)= a1 + d1ssfor 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 anextension 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 mapN : 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 sameapplies to the infinite prime.Second step: Preparations.It remains to consider the case that q remains prime in K. Then, anelement 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 isunramified, there exists some t ∈ Q∗q such that θ := tθ ∈ Kw is a unit.The surface X̃ given byT3(a1T0 + d1T3)(a2T0 + d2T3) =3∏i=1(T0 + θ(i)T1 + (θ(i))2T2)is isomorphic to X ×SpecQ SpecQq. Even more, the mapι : X ×SpecQ SpecQq → X̃, (t0 : t1 : t2 : t3) 7→ (t0 : t1t :t2t2 : 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. Ifν(t3(a1t0 + d1t3)(a2t0 + d2t3))= 0then (a1t0 + d1t3)/t3 is a q-adic unit, hence clearly a norm. Otherwise, wehaveν( 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 afterreduction to the residue field Fq3 . As θ(i) is reduced to a generator of theextension Fq3/Fq, this implies that ν(t0), ν(t1), ν(t2) > 0. Consequently,t3 must be a unit.474 Jörg JahnelFrom 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 that3 | 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 bea 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 ifu := (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 proveX(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 bya 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 linegiven by T0 = T3 = 0. By assumption, one of the planes appears withmultiplicity 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 theintersection X ′q of Xq with the hyperplane “T0 = 0”. This curve is given byd1d2T33 = 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 anFq-valued point by Hasse’s bound.More cubic surfaces violating the Hasse principle 475If 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∏3i=1(T + θ(i)) is a separable polynomial.b) q|d1d2. Then, X ′q := Xq ∩ “T0 = 0” is given by0 = 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 quadraticform 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 inanother Fq-valued point that is smooth.Otherwise, (F mod q) does not depend on T3. I.e., the left hand side ofthe equation of X vanishes modulo q. This means that one of the factorsmust 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 linesthat 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. ExamplesExample. For p = 19, a counterexample to the Hasse principle is given byT3(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 byT3(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 thethree 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 JahnelExample. For p = 19, consider the cubic surface X given byT3(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 = 5as its only solution. The two other solutions are conjugate to each otherin F192 . However, in F19, 6/5 = 5 is a non-cube.Example. Put p = 19 and consider the cubic surface X given byT3(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 threesolutions 8, 9, and 14. However, in F19, 10/9 = 18 is a cube while 9/8 = 13and 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 tothe coefficients a1, d1, a2, and d2, a family of surfaces arises, which are ofsimilar nature.Remark (Lattice basis reduction). The norm form in the p = 19 examplesproduces coefficients that are rather large. An equivalent form with smallercoefficients may be obtained using lattice basis reduction. In its simplestform, this means the following.For the rank 2 lattice in R3, generated by v1 := (θ(1), θ(2), θ(3)) andv2 :=((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 find3∏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 conceptuallyin 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 theMore cubic surfaces violating the Hasse principle 477norm 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 thesum 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 showsthat 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 isunique 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 toapply the norm-residue-homomorphismsQ∗ν/NK∗w → Q/Z [2, 45.2]. This isexactly what we did in the proof of Theorem 1.1.ii).More details on this approach are given in the author’s Habilitationthesis [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. LondonMath. 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. Casselsand A. Fröhlich, Academic Press and Thompson Book Co., London and Washington, 1967.Jörg JahnelFB6 MathematikWalter-Flex-Str. 3D-57068 Siegen, GermanyE-mail: jahnel@mathematik.uni-siegen.deURL: http://www.uni-math.gwdg.de/jahnel

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More cubic surfaces violating the Hasse principle

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