Monatsh MathDOI 10.1007/s00605-012-0387-8

The arithmetic of certain quartic curves

Nguyen Ngoc Dong Quan

Received: 6 May 2011 / Accepted: 28 January 2012© Springer-Verlag 2012

Abstract We give a sufficient condition using the Brauer–Manin obstruction underwhich certain quartic curves have no rational points. Using this sufficient condition,we construct two families of genus one quartic curves violating the Hasse principleexplained by the Brauer–Manin obstruction.

Keywords Azumaya algebras · Brauer groups · Brauer–Manin obstructions · Hasseprinciple · Quartic curves

Mathematics Subject Classification (2000) Primary: 14G05 · 11G35 · 11G30

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Notation and definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Non-existence of rational points on quartic curves . . . . . . . . . . . . . . . . . . . . . . . . . . .4 Diophantine equations of Lind–Mordell–Reichardt type . . . . . . . . . . . . . . . . . . . . . . . .5 The first family of genus one quartic curves violating the Hasse principle . . . . . . . . . . . . . . .6 The second family of genus one quartic curves violating the Hasse principle . . . . . . . . . . . . .References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Communicated by U. Zannier.

N. N. Dong Quan (B)Department of Mathematics, The University of Arizona, Tucson, AZ 85721, USAe-mail: dongquan@math.arizona.edu

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1 Introduction

The arithmetic of the quartic curves of the form

z2 = αx4 + βx2 + γ, (1)

has a long history dating back to Fermat and Euler. For example, the classical theoremdue to Fermat (around 1637) says that the smooth projective model of the affine curvedefined by

z2 = x4 + 1,

has exactly four rational points (0,±1) and ±∞ where ±∞ denote the points at infin-ity of the curve. There exist many works since the seventeenth century studying the(non) existence of rational points on certain quartic curves of the form (1). Notably,Pépin [14,16,18,19] announced the nonexistence of rational points on certain curvesof the form (1) which recently Lemmermeyer [7,8] has verified most of them usingthe arithmetic of ideals in quadratic number fields and the theory of binary quadraticforms. There is also a connection between the arithmetic of curves of the form (1)and the Hasse principle. For example, Lind [9] and Reichardt [23] independently con-struct the first counterexamples of genus one curves to the Hasse principle such as thesmooth projective model of the affine curve defined by

X(2,0,17,1) : 2z2 = x4 − 17.

The main purpose of this paper is to study the nonexistence of rational points oncertain quartic curves of the form (1) using the Brauer–Manin obstruction which wasfirst introduced by Manin [10]. Let us now describe the content of the paper. Section 2provides basic definitions and notation that we will use throughout the paper. In Sect. 3,we will give a sufficient condition under which certain quartic curves C of the form(1) satisfies C(AQ)Br = ∅. The sufficient condition only depends on the prime fac-torization of the discriminant � := β2 − 4αγ and the existence of certain rationalpoints on certain quadratic forms attached to the quartic curves C. In Sect. 4, we willshow that the sufficient condition for nonexistence of rational points on quartic curvesare not very restrictive in the sense that it is very easy to write down infinitely manyfamilies of quartic curves X satisfying that X (AQ)Br = ∅ for any given prime psuch that p ≡ 1 (mod 8). Hence, if we know that X is everywhere locally solvable,then X violates the Hasse principle explained by the Brauer–Manin obstruction. Forexample, the Lind–Reichardt curve X(2,0,17,1) and the Mordell curve (which will bedescribed in the same section) are among the curves X . In Sect. 5, we will show thatamong the curves X , there are infinitely many arithmetic families of curves D violatingthe Hasse principle explained by the Brauer–Manin obstruction. The infinite familiesof curves D arise from triples (d, h, p) where h is any integer, d is a square-freepositive integer satisfying certain local conditions and p is a prime such that p ≡ 1(mod 8). In the same section, we will give a concrete example of how to construct aninfinite arithmetic family of non-trivial genus one quartic curves with non-constant

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The arithmetic of certain quartic curves

j-invariant violating the Hasse principle explained by the Brauer–Manin obstructionusing the sufficient condition in Sect. 3. Finally, in Sect. 6, we will show that thereis another family of genus one quartic curves with non-constant j-invariant violatingthe Hasse principle explained by the Brauer–Manin obstruction.

2 Notation and definitions

Let V be a smooth geometrically irreducible projective variety defined over a globalfield k. We shall denote by Br(V) the Brauer group of V , that is, the group of equiva-lence classes of Azumaya algebras on V . It follows from the work of Manin [10] thatthere is a set V(Ak)

Br such that

V(k) ⊂ V(Ak)Br ⊂ V(Ak).

Here V(Ak) denotes the set of adelic points on V and V(Ak)Br is defined as follows.

V(Ak)Br =

{(Pv) ∈ V(Ak) such that

∑v

invvA(Pv) = 0 for all A ∈ Br(V)

}

where for each valuation v and each Azumaya algebra A in Br(V), invv : Br(kv) −→Q/Z is the local invariant map from class field theory and A(Pv) is defined as follows.A point Pv ∈ V(kv) gives a map Spec(kv) −→ V , and hence induces a pullback mapBr(V) −→ Br(kv); we write A(Pv) for the image of A under this map.

Definition 2.1 V is said to satisfy the Hasse principle if the following is true

V(k) �= ∅ ⇔ V(kv) �= ∅ for all v.

Definition 2.2 V is said to be a counter-example to the Hasse principle if V(k) = ∅and V(Ak) �= ∅. Further, if we also have V(Ak)

Br = ∅, we say that V is a counter-example to the Hasse principle explained by the Brauer–Manin obstruction.

3 Non-existence of rational points on quartic curves

In this section, we will prove a sufficient condition which provides us a way to verifythe nonexistence of rational points on certain quartic curves. We begin by proving themain lemma in this section.

Lemma 3.1 Let (α, β, γ ) ∈ Z3 be a triple of integers such that α �= 0 and α is not a

square in Z. Define � = β2 − 4αγ and assume that the following are true.

(A1) the conic Q ⊂ P2Q

defined by

Q : X2 − �Y 2 − 4αZ2 = 0, (2)

has a point (a, b, c) ∈ Z3 such that gcd(a, b, c) = 1 and abc �= 0.

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N. N. Dong Quan

(A2) � is not a perfect square in Z and � = 22s�1�2 where s ∈ Z≥0 and �1,�2are odd positive integers.

(A3) gcd(a,�2) = 1 and there exist an odd number of primes pi , 1 ≤ i ≤ 2k +1 forsome integer k ≥ 0 such that pi divides �2 and a is a quadratic non-residuein F×

pifor each 1 ≤ i ≤ 2k + 1 and such that for any prime l dividing �2 with

l �= pi , 1 ≤ i ≤ 2k + 1, a is a quadratic residue in F×l .

(A4) for each prime l dividing �1, vl(�1) is even and for each prime l dividing�2, vl(�2) = 1. Furthermore, gcd(�1,�2) = 1.

(A5) �2 divides β and �2 ≡ 1 (mod 8). Furthermore, for every prime l dividing�2, l is congruent to 1 mod 4.

(A6) for every odd prime l dividing �1,�2 is a square in Q×l .

Let C be the smooth projective model of the affine curve defined by

C : z2 = αx4 + βx2 + γ. (3)

Let Q(C) be the function field of C and let A be the class of the quaternion algebra inBr(Q(C)) defined by

A = (�2, 4cαz + 2aαx2 + aβ + �b). (4)

Then, A is an Azumaya algebra of C, that is, A belongs to the subgroup Br(C) ofBr(Q(C)). Furthermore, the quaternion algebras

A,B := (�2, 4cαz−2aαx2−aβ−�b), E :=(

�2,4cαz + 2aαx2 + aβ + �b

x2

),

all represent the same class in Br(Q(C)).

Proof We shall prove that there is a Zariski open covering {Ui } of C such that Aextends to an element of Br(Ui ) for each i . We see that (3) can be written in the form.

16(cαz)2 = (a(2αx2 + β) + �b)2 − �(b(2αx2 + β) + a)2.

Hence, it follows that

(4cαz − 2aαx2 − aβ − �b)(4cαz+2aαx2+aβ+�b) = −�(b(2αx2+β)+a)2.

(5)

By (A4), we know√

�1 ∈ Z and we thus have

(4cαz − 2aαx2 − aβ − �b)(4cαz + 2aαx2 + aβ + �b)

= −�2(2s√

�1(b(2αx2 + β) + a))2

= NormQ(√

�2)/Q(√

�2(2s√

�1(b(2αx2 + β) + a))).

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The arithmetic of certain quartic curves

Hence, it follows that A + B = 0. Furthermore, since x2 is a square, we have thatA − E = (p, x2) = 0. Since A,B and E belong to the 2-torsion part of Br(Q(C)),this implies that A = B = E .

Now let U1 be an open subvariety of C in which the rational function F := 4cαz +2aαx2 + aβ + �b has neither a zero nor pole and let U2 be an open subvariety of Cin which G := 4cαz − 2aαx2 − aβ − �b has neither a zero nor pole. Then, sinceA = B,A is an Azumaya algebra on U1 and also on U2. We prove that in the affinepart of C, the locus where both F and G have a zero is empty. Assume the contrary,that is, there is an point (x, z) on C such that

F(x, z) = G(x, z) = 0.

Hence, we deduce that

F − G = 2(2aαx2 + aβ + �b) = 0,

and hence,

2αx2 + β = −�b

a.

On the other hand, by (5), we have

2αx2 + β = −a

b.

Thus,

−�b

a= −a

b,

and it follows that

a2 − �b2 = 0.

By (A1), we deduce that

4αc2 = a2 − �b2 = 0,

contradiction. Hence, the system of equations

F(x, z) = G(x, z) = 0,

has no solution in C. Hence, it remains to show that A is regular and non-vanishingat the points at infinity. Define

H := 4cαz + 2aαx2 + aβ + �b

x2 ,

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N. N. Dong Quan

and let ∞ := (X∞ : Y∞ : Z∞) be a point at infinity. Then,

X∞ = 1, Y∞ = 0, Z∞ = ±√α.

Hence,

H(∞) = 4cαZ∞ + 2aαX2∞ + (aβ + �b)Y 2∞X2∞

= 2α(±2c√

α + a).

We contend that H(∞) �= 0; otherwise, a2 = 4αc2 and it follows from (A1) that�b2 = a2 − 4αc2 = 0, contradiction. Hence, H is regular and non-vanishing at thepoints at infinity of C.

Now let U3 be an open subvariety of C in which H has neither a zero nor pole.Then, since A = E , we deduce that A is an Azumaya algebra on U3.

By what we have shown, it follows that C = U1 ∪ U2 ∪ U3 and since A is anAzumaya algebra on each Ui for i = 1, 2, 3, we deduce that A belongs to Br(C),proving our contention. ��Theorem 3.2 Suppose the same notations and assumptions as in Lemma 3.1. Let Cbe the smooth projective model defined in Lemma 3.1. Then, C(AQ)Br = ∅.

Proof Let Q(C) be the function field of C and let A be the class of quaternion algebrain Br(Q(C)) defined by

A = (�2, 4cαz + 2aαx2 + aβ + �b).

It follows from Lemma 3.1 that A is an Azumaya algebra of C and A equals the classof the quaternion algebras B, E where B, E are the same as in Lemma 3.1.

We shall prove that for any Pl ∈ C(Ql),

invl(A(Pl)) ={

0 if l �= pi for all 1 ≤ i ≤ 2k + 1

1/2 if l = pi for each 1 ≤ i ≤ 2k + 1.(6)

Since C is smooth, we know that C0(Ql) is l-adically dense in C(Ql) where C0 is theaffine curve given by

C0 : z2 = αx4 + βx2 + γ.

Since invl(A(Pl)) is a continuous function on C(Ql) with the l-adic topology, it sufficesto prove (6) for Pl ∈ C0(Ql).

Suppose that l = ∞ or l = 2. Then, it follows from (A2) and (A5) that �2 isa square in Q

×l . Hence, for any t ∈ Q

×l , the Hilbert symbol (�2, t)l = 1. Hence,

invl(A(Pl)) = 0.Suppose that l is an odd prime such that gcd(l,�2) = 1 and �2 is a square in Q

×l .

Then, repeating in the same manner as above, we deduce that invl(A(Pl)) = 0.

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Suppose that l is an odd prime such that gcd(l,�2) = 1 and �2 is not a square inQ

×l . We contend that α �≡ 0 (mod l) and c �≡ 0 (mod l). If α ≡ 0 (mod l), then we

know that

22s�1�2 = β2 − 4αγ ≡ β2 (mod l).

On the other hand, we see from (A6) that �1 �≡ 0 (mod l). By (A4), it follows that√�1 ∈ Z

×l and it then follows that

�2 ≡(

β

2s√

�1

)2

(mod l),

contradiction. Hence, α �≡ 0 (mod l).If c ≡ 0 (mod l), then it follows from (A1) that

a2 ≡ 22s�1�2b2 (mod l).

since �2 is not a square in Q×l and �1 ∈ Z

×2l , we deduce that a ≡ b ≡ 0 (mod l).

Hence, l divides gcd(a, b, c) = 1, contradiction. Thus, c �≡ 0 (mod l).We consider the following cases.

Case I vl(x) ≥ 0.Hence, it follows from (3) that vl(z) ≥ 0. Define

F := 4cαz + 2aαx2 + aβ + �b,

G := 4cαz − 2aαx2 − aβ − �b,

and assume that F ≡ 0 (mod l) and G ≡ 0 (mod l). Then, we have

0 ≡ F − G ≡ 2(2aαx2 + aβ + �b) (mod l)

and hence,

a(2αx2 + β) + �b ≡ 0 (mod l).

On the other hand, by (5), we have

b(2αx2 + β) + a ≡ 0 (mod l).

Hence, we deduce from (A1) that

0 �≡ 4αc2 = a2 − b2� ≡ a(b(2αx2 + β) + a)

−b(a(2αx2 + β) + �b) ≡ 0 (mod l),

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N. N. Dong Quan

contradiction. Hence, at least one of F and G is a unit in Z×l , say U . Thus, the Hilbert

symbol (�2, U )l = 1. Therefore, since A and B represent the same class in Br(Q(C)),we deduce that invl(A(Pl)) = 0.

Case II vl(x) = −m < 0 for some positive integer m.Recall that α �≡ 0 (mod l). Hence, it follows from (3) that

vl(z) = 2vl(x) = −2m.

Let x0, z0 ∈ Z×l such that x = x0

lmand z = z0

l2m. Then, we see that F = F0

l2mand

G = G0

l2mwhere

F0 = 4cαz0 + 2aαx20 + l2maβ + l2m�b ∈ Zl ,

G0 = 4cαz0 − 2aαx20 − l2maβ − l2m�b ∈ Zl .

We contend that at least one of F0 and G0 is a unit in Z×l . Assume the contrary, that is,

F0 ≡ 0 (mod l) and G0 ≡ 0 (mod l). Then, it follows that F0 + G0 = 8cαz0 ≡ 0(mod l). Hence, since 8cα �≡ 0 (mod l), we deduce that z0 ≡ 0 (mod l), contra-diction. Thus, at least one of F0 and G0 is an l-unit in Z

×l , say U . Then, by Theo-

rem 5.2.7 (see [5, p. 296]), we see that the Hilbert symbol

(�2,

U

l2m

)l

= 1. Thus,

invl(A(Pl)) = 0.Suppose that l is an odd prime dividing �2. We see from (A1) that a2 ≡ 4αc2

(mod l). Hence, by (A3), we know that a �≡ 0 (mod l); thus, α �≡ 0 (mod l) andc �≡ 0 (mod l). By (A5), we see that l divides β. Hence, it follows that

0 ≡ −22s�1�2 = −� = 4αγ − β2 ≡ 4αγ (mod l).

Since α �≡ 0 (mod l), we deduce that γ ≡ 0 (mod l).We consider the following cases.

Case 1 vl(x) ≥ 0.We contend that x �≡ 0 (mod l). Assume the contrary, that is, x ≡ 0 (mod l).

Then, we see from (3) that vl(z) ≥ 0. Therefore, it follows from (3) that z ≡ 0(mod l). We know from the defining equation of C that

vl(γ ) ≥ min(vl(z2), vl(αx4 + βx2)) ≥ 2.

Thus, since gcd(�1,�2) = 1, it follows from (A4) that

1 = vl(�2) = vl(�) ≥ min(vl(β2), vl(−4αγ )) ≥ 2,

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The arithmetic of certain quartic curves

contradiction. Hence, x �≡ 0 (mod l). Thus, modulo l Eq. (3), we obtain that

z2 ≡ αx4 �≡ 0 (mod l),

and hence, z �≡ 0 (mod l).Therefore, since α �≡ 0 (mod l), one can verify from (5) that Pl ∈ U1 ∪ V1 where

U1 := {(x, z) ∈ C(Ql), x, z ∈ Z×l , ax2 − 2cz ≡ 0 (mod l)},

V1 := {(x, z) ∈ C(Ql), x, z ∈ Z×l , ax2 + 2cz ≡ 0 (mod l)}.

Suppose that Pl ∈ U1 and x ≡ h (mod l) at Pl with h ∈ F×l . Then, since c �≡ 0

(mod l), we see that z ≡ ah2

2c(mod l). Hence, since 4α ≡ a2

c2 (mod l), one can

check that

F = 4cαz + 2aαx2 + aβ + �b ≡ 4aαh2 ≡ a

(ah

c

)2

�≡ 0 (mod l),

and hence, F is an l-unit in Z×l . Thus, we deduce from (A3) that

(F

l

)=

{1 if l �= pi for all 1 ≤ i ≤ 2k + 1

−1 if l = pi for each 1 ≤ i ≤ 2k + 1.

Thus, using Theorem 5.2.7 in [5], we deduce that the Hilbert symbol

(�2, F)l =(

F

l

)=

{1 if l �= pi for all 1 ≤ i ≤ 2k + 1

−1 if l = pi for each 1 ≤ i ≤ 2k + 1.

Therefore, for each Pl ∈ U1, we have

invl(A(Pl)) ={

0 if l �= pi for all 1 ≤ i ≤ 2k + 1

1/2 if l = pi for each 1 ≤ i ≤ 2k + 1.(7)

Similarly, suppose that Pl ∈ V1 and x ≡ h (mod l) at Pl with h ∈ F×l . Then, we see

that z ≡ −ah2

2c(mod l). Hence, one can check that

G = 4cαz − 2aαx2 − aβ − �b ≡ −4aαh2 ≡ −a

(ah

c

)2

(mod l).

Thus, since l ≡ 1 (mod 4), it follows from (A3) that

(G

l

)=

{1 if l �= pi for all 1 ≤ i ≤ 2k + 1

−1 if l = pi for each 1 ≤ i ≤ 2k + 1.

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N. N. Dong Quan

Thus, repeating in the same manner as above, it follows that for each Pl ∈ V1, wehave

inv(A(Pl)) ={

0 if l �= pi for all 1 ≤ i ≤ 2k + 1

1/2 if l = pi for each 1 ≤ i ≤ 2k + 1.(8)

Therefore, we see from (7) and (8) that∑

l invlA(Pl) = 1/2 for any (Pl)l ∈ C(AQ)

with vl(x) ≥ 0.

Case 2 vl(x) = −ε < 0 for a positive integer ε.Recall that α �≡ 0 (mod l) and c �≡ 0 (mod l). Then, it follows that vl(z) < 0.

One can check using (3) that vl(z) = −2ε. Hence, there exist elements x0, z0 ∈ Z×l

such that x = x0

lεand z = z0

l2ε. Repeating the same arguments as above, we deduce

that Pl ∈ U2 ∪ V2 where

U2 := {(x, z) ∈ C(Ql) : x = x0lvl (x), z = z0l2vl (x) such that vl(x) < 0,

x0, z0 ∈ Z×l , ax2

0 − 2cz0 ≡ 0 (mod l)},V2 := {(x, z) ∈ C(Ql) : x = x0lvl (x), z = z0l2vl (x) such that vl(x) < 0,

x0, z0 ∈ Z×l , ax2

0 + 2cz0 ≡ 0 (mod l)}.

For each Pl ∈ U2, one can check using the same arguments as in Case 1 that F = F0

l2ε

where

F0 := 4cαz0 + 2aαx20 + l2εaβ + l2ε�b ≡ 4aαx2

0 ≡ a(ax0

c

)2 �≡ 0 (mod l).

Hence, F0 is an l-unit in Z×l . Thus, by Theorem 5.2.7 (see [5, p. 296]), we deduce that

the Hilbert symbol

(�2, F)l =(

F0

l

)=

{1 if l �= pi for all 1 ≤ i ≤ 2k + 1

−1 if l = pi for each 1 ≤ i ≤ 2k + 1.

Therefore, for each Pl ∈ U2, we have

invl(A(Pl)) ={

0 if l �= pi for all 1 ≤ i ≤ 2k + 1

1/2 if l = pi for each 1 ≤ i ≤ 2k + 1.(9)

Similarly, one can check that for each Pl ∈ V2, we have

invl(A(Pl)) ={

0 if l �= pi for all 1 ≤ i ≤ 2k + 1

1/2 if l = pi for each 1 ≤ i ≤ 2k + 1.(10)

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The arithmetic of certain quartic curves

Therefore, it follows from (9) and (10) that

∑l

invlA(Pl) =2k+1∑i=1

(1/2) = 1/2 mod Z,

for any (Pl)l ∈ C(AQ) with vl(x) < 0.Thus, in any event, we have

∑l invlA(Pl) = 1/2 for any (Pl)l ∈ C(AQ). Hence,

C(AQ)Br = ∅. ��Example 3.3 Let

α = 1273, β = 1292, γ = 323,

and let C(17,19,2) be the smooth projective model of the affine curve defined by

C(17,19,2) : z2 = 1273x4 + 1292x2 + 323.

Then, � = 12922 − 4 · 1273 · 323 = 22 · 192 · 17. Hence, �1 = 192 and �2 = 17.We see that the conic defined by

X2 − 22 · 192 · 17Y 2 − 4 · 1273Z2 = 0,

has a point (a, b, c) = (8170, 39, 76). Since 8170 is a quadratic non-residue in F×17,

(A1) and (A3) in Theorem 3.2 are true. One can verify that (A2), (A4), (A5) and(A6) are also true. Hence, by Theorem 3.2, we deduce that C(17,19,2)(AQ)Br = ∅.Furthermore, one can check that C(17,19,2) is everywhere locally solvable and hence,it violates the Hasse principle explained by the Brauer–Manin obstruction.

4 Diophantine equations of Lind–Mordell–Reichardt type

In this section, we shall study certain Diophantine equations of Lind–Mordell–Reichardt type.

Theorem 4.1 Let p be a prime such that p ≡ 1 (mod 8) and let

� := 22sm∏

j=1

q2nii .

Here m is a positive integer, ni , s ∈ Z≥0 and qi are odd primes for 1 ≤ i ≤ m. Let dbe a non-zero integer and assume that the following are true.

(B1) d = 2r d0 where r is a nonnegative integer and d0 is a squarefree odd integer.(B2) gcd(d, p) = 1 and d is a quadratic residue but not a quartic residue in F

×p .

(B3) gcd(p,�) = 1 and p is a quadratic residue in F×qi

for each 1 ≤ i ≤ m.

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N. N. Dong Quan

Let h be an integer and let X be the smooth projective model of the affine curve definedby

X : dz2 = x4 + (2hp�)x2 + p�(h2 p� − 1). (11)

Then X (AQ)Br = ∅.

Proof Throughout the proof, we will use the same notations as in Theorem 3.2. Thedefining equation of X can be written in the form

X : z2 = dx4 + (2dhp�)x2 + dp�(h2 p� − 1).

Hence,

α = d, β = 2dhp�, γ = dp�(h2 p� − 1),

and hence,

� = β2 − 4αγ = 4d2 p� = 22(1+r+s)d20 p

⎛⎝ m∏

j=1

q2nii

⎞⎠ .

Thus,

�1 = d20

⎛⎝ m∏

j=1

q2nii

⎞⎠ and �2 = p.

We prove that (Ai) for 1 ≤ i ≤ 6 in Theorem 3.2 are true. Indeed, one can eas-ily verify that (A2), (A4), (A5) hold. Hence, it remains to show that (A1), (A3) and(A6) are true. Let us first prove (A6). With no loss of generality, we can assume thatX (AQ) �= ∅; otherwise, it follows trivially that X (AQ)Br = ∅. Hence, for any oddprime l dividing d0 with gcd(l,�) = 1, we know that X is locally solvable at l. Thus,X (Ql) is non-empty. Furthermore, it is not difficult to show that if Pl ∈ X (Ql), thenPl = (x, z) ∈ X (Ql) for some x, z ∈ Ql . Assume first that vl(x) < 0. Hence, itfollows from the defining equation of X and (B1) that

1 + 2vl(z) = vl(dz2) = vl(x4) = 4vl(x),

contradiction since the left-hand side is odd whereas the right-hand side is even. Hence,vl(x) ≥ 0 and it follows from (11) that vl(z) ≥ 0. Modulo l the defining equation ofX , we obtain that

(x2 + hp�)2 ≡ p� (mod l).

Since gcd(l,�) = 1 and � is a square in Z, it follows from the last equation that p isa quadratic residue in F

×l . Thus, it follows from (B3) that (A6) in Theorem 3.2 holds.

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Now we define the conic Q in the same way as in (2)

Q : X2 − 22(1+r+s)d20 p

⎛⎝ m∏

j=1

q2nii

⎞⎠ Y 2 − 4d Z2 = 0. (12)

Note that if X is not locally solvable at some prime l, then X (AQ) = ∅ and it thenfollows immediately that X (AQ)Br = ∅. Thus, with no loss of generality, we canassume that X is everywhere locally solvable. It follows from the defining equationof X that

(x2 + hp�)2 − p� − dz2 = 0,

is everywhere locally solvable. Hence, we see that the conic defined by

X2 − p�Y 2 − d Z2 = 0,

is everywhere locally solvable. Multiplying the last equation by 4d2 and noting that� = 4d2 p�, we deduce that the conic Q is everywhere locally solvable. Hence, bythe Hasse–Minkowski theorem, it follows that Q has a point (a, b, c) ∈ Z

3>0 such that

abc �= 0 and gcd(a, b, c) = 1. Hence, (A1) in Theorem 3.2 is true. We contend that ais a quadratic nonresidue in F

×p . Indeed, let c = 2εc0 such that c0 is a positive odd inte-

ger and ε ∈ Z≥0. It follows from equation (12) that gcd(c0, p) = 1. Hence, since (A6)in Theorem 3.2 is true, it follows from the quadratic reciprocity law and Eq. (12) that

1 =(

p

c0

)=

(c0

p

)=

(c

p

).

Hence, there exists an integer c̄ such that c ≡ c̄2 �≡ 0 (mod p). Thus, it follows fromthe equation of Q that

dc̄4 ≡(a

2

)2(mod p).

Therefore, by (B2), we deduce that a/2 is a quadratic nonresidue in F×p . Hence, since

2 is a quadratic residue in F×p , it follows that a is a quadratic nonresidue in F

×p . Hence,

(A3) in Theorem 3.2 holds. Thus, it follows from Theorem 3.2 that X (AQ)Br = ∅.��

Remark 4.2 Lind [9] and Reichardt (1942) independently proved that for d = 2, h =0, p = 17 and � = 1, the smooth projective model X(2,0,17,1) of the affine curvedefined by

X(2,0,17,1) : 2z2 = x4 − 17,

satisfies X(2,0,17,1)(Q) = ∅ and X(2,0,17,1)(AQ) �= ∅. This is the first counterexampleof genus one curves to the Hasse principle. In 1969, Mordell [11] generalized the

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N. N. Dong Quan

Lind–Reichardt counterexample to every prime p such that p ≡ 1 (mod 8) and 2is not a quartic residue in F

×p . More precisely, if p is such a prime, then the smooth

projective model X(2,0,p,1) of the affine curve defined by

X(2,0,p,1) : 2z2 = x4 − p,

satisfies X(2,0,p,1)(Q) = ∅ and X(2,p,1)(AQ) �= ∅. This is the weaker version ofTheorem 4.1 for the special case when d = 2, h = 0, � = 1. In fact, Theorem 4.1 con-cludes that the Lind–Reichardt curve and the Mordell curve violate the Hasse principleexplained by the Brauer–Manin obstruction. Very recently, Aitken and Lemmermeyer[1] prove that for a prime p ≡ 1 (mod 8) and a squarefree integer d such thatgcd(p, d) = 1 and d is not a biquadratic residue in F

×p , the smooth projective model

X(d,0,p,1) of the affine curve defined by

X(d,0,p,1) : dz2 = x4 − p,

satisfies X(d,0,p,1)(Q) = ∅. This follows immediately from Theorem 4.1 when lettingh = 0 and � = 1.

5 The first family of genus one quartic curves violating the Hasse principle

In this section, we shall construct an infinite family of non-trivial genus one quarticcurves with non-constant j-invariant violating the Hasse principle explained by theBrauer–Manin obstruction.

Theorem 5.1 Let p be a prime such that p ≡ 1 (mod 8) and let d be a squarefreeodd integer. Assume that h is an integer such that the following are true.

(C1) d ≡ −1 (mod 8) and d = ±∏mi=1 li . Here m ∈ Z≥1 and li are odd primes

such that li �= 3 and li ≡ 3 (mod 4).(C2) for each 1 ≤ i ≤ m, li is a quadratic residue in F

×p . Furthermore, d is not a

quartic residue in F×p .

(C3) h = −8a2 for some integer a such that a ≡ 0 (mod d).

Let D be the smooth projective model of the affine curve defined by

D : dz2 = x4 + (2hp)x2 + p(h2 p − 1). (13)

Then, D violates the Hasse principle explained by the Brauer–Manin obstruction.

Proof The conclusion that D(AQ)Br = ∅ is a special case of Theorem 4.1 when � = 1.So, it remains to prove that D is everywhere locally solvable. The discriminant of Dis

Disc(D) = 212 p3d6(h2 p − 1).

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The arithmetic of certain quartic curves

Hence, D is singular only at the primes l = 2, p, li for 1 ≤ i ≤ m and the primes ldividing (h2 p − 1); so, the Weil’s inequality mandates a smooth l-adic point on D forprimes l ≥ 5 such that l �= p, l �= li for 1 ≤ i ≤ m and gcd(l, h2 p − 1) = 1. Hence,it suffices to verify that D is locally solvable at the primes l = ∞, 2, 3, p and li for1 ≤ i ≤ m and at the primes l dividing (h2 p − 1). We consider the following cases.

Case 1 l is an odd prime dividing (h2 p − 1).Define

F(x, z) = x4 + (2ph)x2 + p(ph2 − 1) − dz2 ∈ Z[x, z].

We consider the system of equations

F(x, z) ≡ 0 (mod l) (14)

and

∂ F

∂x(x, z) = 4x3 + 4phx ≡ 0 (mod l). (15)

By assumption, we know that h2 p ≡ 1 (mod l) and hence, p ≡ 1

h2 (mod l). In

particular, we know that h �≡ 0 (mod l) and hence, a �≡ 0 (mod l). Thus, by (C3),we have

2hp ≡ 2

h≡ − 1

4a2 (mod l).

Therefore, it follows that

F

(1

2a, 0

)≡ 1

16a4 + 2hp1

4a2 ≡ 1

16a4 − 1

16a4 ≡ 0 (mod l),

and

∂ F

∂x

(1

2a, 0

)= 4

1

8a3 + 4hp1

2a≡ 1

2a3 − 1

4a3 = 1

4a3 �≡ 0 (mod l).

Thus, (x, z) =(

1

2a, 0

)is a solution to the system of Eqs. (14) and (15). Hence, by

Hensel’s lemma, there exists a point of D(Ql) whose reduction mod l is

(1

2a, 0

).

Case 2 l = p.We consider the following system of equations.

F(x, z) ≡ 0 (mod p) and∂ F

∂x(x, z) �≡ 0 (mod p). (16)

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N. N. Dong Quan

By (C2), there is an integer ε such that gcd(ε, p) = 1 and d ≡ ε2 (mod p). Hence,one can check that

F

(1,

1

ε

)≡ 0 (mod p),

∂ F

∂x

(1,

1

ε

)≡ 4 �≡ 0 (mod p).

Thus, (x, z) =(

1,1

ε

)is a solution to the system (16). Therefore, by Hensel’s lemma,

we deduce that D is locally solvable at p.

Case 3 l = li for 1 ≤ i ≤ m.We consider the following system of equations for the prime li .

F(x, z) ≡ 0 (mod li ) and∂ F

∂x(x, z) �≡ 0 (mod li ). (17)

By (C2) and the quadratic reciprocity law, we know that p is a quadratic residue inF

×li

. Since li ≡ 3 (mod 4) [by (C1)], it then follows that p is a quartic residue modulo

li . Thus, there exists an element ε ∈ F×li

such that p ≡ ε4 (mod li ). Hence, by (C3),we see that h ≡ 0 (mod li ). Thus, we deduce that

F(ε, 0) ≡ ε4 − p ≡ 0 (mod li ),∂ F

∂x(ε, 0) ≡ 4ε3 �≡ 0 (mod li ).

Thus, (x, z) = (ε, 0) is a solution to the system (17). Therefore, by Hensel’s lemma,we deduce that D is locally solvable at li .

Case 4 l = 2.By (C1) and (C3), we see that

dp(h2 p − 1) ≡ −dp ≡ 1 (mod 8).

Hence,√

dp(h2 p − 1) ∈ Q×2 . Thus, the point

P2 := (x, z) =(

0,

√dp(h2 p − 1)

d

),

belongs to D(Q2), proving that D is locally solvable at 2.

Case 5 l = 3.We consider the following two systems of equations

F(x, z) ≡ 0 (mod 3) and∂ F

∂x(x, z) �≡ 0 (mod 3) (18)

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The arithmetic of certain quartic curves

Table 1 3-adic points on D

h mod 3 0 0 0 0 1 1 1 1

p mod 3 1 1 2 2 1 1 2 2

d mod 3 1 2 1 2 1 2 1 2

(x, z) (1, 0) (1, 0) (0, 1) (1, 1) (1, 0) (1, 0) (1, 1) (0, 1)

and

F(x, z) ≡ 0 (mod 3) and∂ F

∂z(x, z) �≡ 0 (mod 3). (19)

By (C1) and (C3), we know that

h ≡ 0, 1 (mod 3), p ≡ 1, 2 (mod 3) and d ≡ 1, 2 (mod 3).

Table 1 verifies that at least one of the systems (18) and (19) is solvable and hence itfollows from Hensel’s lemma that D is locally solvable at 3.

Case 6 l = ∞.By (C3), we know that −h = 8a2 ≥ 0. Hence,

√−hp ∈ R. Assume first that

d < 0. Then,√−dp ∈ R. Hence, we see that the point (x, z) =

(√−hp,

√−dp

d

)belongs to D(R).

Now assume that d > 0. Then, we see that the point (x, z) =(√

p − hp,

√d(p2−p)

d

)belongs to D(R).

Therefore, in any event, D is everywhere locally solvable, proving our contention.��

Remark 5.2 Let p be a prime such that p ≡ 1 (mod 8) and let q be an odd primesuch that q ≡ −1 (mod 8) and q = a2 − pb2 for certain integers a and b where a is aquadratic non-residue in F

×p . Then, Lemma 2.5 in [13] says that the smooth projective

model D of the affine curve defined by

D : qz2 = x4 − p,

violates the Hasse principle explained by the Brauer–Manin obstruction. This is aspecial case of Theorem 5.1 when h = 0. Furthermore, Theorem 5.1 assures that theglobal condition q = a2 − pb2 can be replaced by a weaker local condition whichonly requires that q is a quadratic residue but not a quartic residue in F

×p . Hence, using

Theorem 5.1, it is very easy to write down infinitely many counterexamples to theHasse principle explained by the Brauer–Manin obstruction. For example, take anyprime p such that p ≡ 1 (mod 8) and take any integer ε such that ε is a quadraticnon-residue in F

×p . By the Chinese remainder theorem and the Dirichlet’s theorem on

123

N. N. Dong Quan

arithmetic progressions, there are infinitely many primes q such that q ≡ −1 (mod 8)

and q ≡ ε2 (mod p). Then, the couple (q, p) satisfies (C1) and (C2) in Theorem 5.1.Now take any integer h of the form

h = −8q2a2, for any a ∈ Z,

and define the smooth projective model D as in Theorem 5.1. Thus, h satisfies (C3)and it follows from Theorem 5.1 that we have constructed infinitely many genus onecurves D violating the Hasse principle explained by the Brauer–Manin obstruction.Furthermore, one can show that among the infinite family of genus one curves we havejust constructed, there are infinitely many non-isomorphic genus one curves violatingthe Hasse principle explained by the Brauer–Manin obstruction. This can proved bywriting down the j-invariant of each genus one curve in the infinite family.

Remark 5.3 In Remark 5.2, we have shown how to construct infinitely many non-isomorphic genus one curves violating the Hasse principle explained by the Brauer–Manin obstruction for a given couple (q, p). Here, (q, p) is a couple of odd primessatisfying (C1) and (C2). Following the same methods described in Remark 5.2, wecan construct infinitely many non-isomorphic genus one curves violating the Hasseprinciple explained by the Brauer–Manin obstruction for a given couple (d, p). Here, dis a squarefree integer such that the couple (d, p) satisfies (C1) and (C2). Example 5.5below illustrates the method of constructing genus one quartic curves violating theHasse principle explained by the Brauer–Manin obstruction using a couple of primes(q, p). For a given couple (d, p), the reader can write down infinitely many genus onecurves violating the Hasse principle attached to the couple (d, p), following the sametechniques and it is left as an exercise.

Remark 5.4 Lemma 2.5 in [13] plays a crucial role in constructing an arithmetic fam-ily of projective quartic curves of genus 3 violating the Hasse principle explained bythe Brauer–Manin obstruction. Using Theorem 5.1 and following similar arguments asin [13], we expect that there are many infinite arithmetic families of projective quar-tic curves of genus 3 violating the Hasse principle explained by the Brauer–Maninobstruction arising from the curves D in Theorem 5.1.

Example 5.5 Let p = 17 and q = 383. Then, we see that q ≡ 32 (mod 17). Since 3is a quadratic nonresidue in F

×17, it follows that q = 383 is a quadratic residue but not

a quartic residue in F×17. Thus, the pair (p, q) = (17, 383) satisfies (C1) and (C2) in

Theorem 5.1. For any integer a ∈ Z, define

h := −8q2a2 = −1173512a2.

Let D(17,383,a) be the smooth projective model of the affine curve defined by

D(17,383,a) : 383z2 = x4 − (39899408a2)x2 + 397990689687616a4 − 17.

Hence, by Theorem 5.1, we deduce that the infinite family (D(17,383,a))a∈Z violatesthe Hasse principle explained by the Brauer–Manin obstruction.

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The arithmetic of certain quartic curves

Furthermore, note that the j-invariant of D(17,383,a) is

j (D(17,383,a))

= 23411217040448(35077445332157619972354605056a12−3371215253824512a8+108a4)−27

27(23411217040448a4−1).

Thus, (D(17,383,a))a∈Z is a family of nontrivial genus one quartic curves with non-constant j-invariant violating the Hasse principle explained by the Brauer–Maninobstruction.

Remark 5.6 In 2000, Colliot-Thélène and Poonen (see [6]) show how to produce oneparameter families of genus one cubic curves with non-constant j-invariant violatingthe Hasse principle. The general method developed in [6] was later made explicit in alater paper of Poonen (see [20]) who constructs an explicit family of genus one cubiccurves with non-constant j-invariant violating the Hasse principle.

6 The second family of genus one quartic curves violating the Hasse principle

In this section, we shall construct another family of non-trivial genus one quarticcurves with non-constant j-invariant violating the Hasse principle explained by theBrauer–Manin obstruction.

Theorem 6.1 Let p be a prime such that p ≡ 1 (mod 8) and let q be an odd primesuch that −q is a quadratic residue but not a quartic residue in F

×p . Assume that h is

an integer such that the following are true.

(D1) assume that1

p≡ σ 2 (mod q) for some integer σ such that 1 − 2σ �≡ 0

(mod q). Then, h ≡ σ − 1 (mod q).(D2) h = −72a2 for some integer a.(D3) q ≡ 1 (mod 8).

Let M be the smooth projective model of the affine curve defined by

M : qz2 = (ph2 − 1)x4 + 2phx2 + p. (20)

Then, M violates the Hasse principle explained by the Brauer–Manin obstruction.

Remark 6.2 Since p ≡ 1 (mod 8), we see that −q is a quadratic residue in F×p if and

only if q is a quadratic residue in F×p .

Proof We prove that M is everywhere locally solvable. The discriminant of M is

Disc(M) = 212 p3q6(ph2 − 1).

Hence, M is singular only at the primes l = 2, p, q and at the primes l dividing(ph2 − 1); so, the Weil’s inequality mandates a smooth l-adic point on M for primesl ≥ 5 such that l �= p, q and gcd(l, h2 p − 1) = 1. Hence, it suffices to verify that

123

N. N. Dong Quan

M is locally solvable at the primes l = ∞, 2, 3, p, q and at the primes l dividing(h2 p − 1). We consider the following cases.

Case 1 l = p.Define

F(x, z) = (ph2 − 1)x4 + 2phx2 + p − qz2 ∈ Z[x, z].

We consider the system of equations

F(x, z) ≡ 0 (mod p) and∂ F

∂x(x, z) = 4(ph2 − 1)x3 + 4phx �≡ 0 (mod p).

(21)

Since −q is a quadratic residue in F×p , there is an integer ε such that −q ≡ ε2 �≡ 0

(mod p). We see that

F(1, 1/ε) ≡ −1 + ε2 1

ε2 ≡ 0 (mod p)

and

∂ F

∂x(1, 1/ε) ≡ −4 �≡ 0 (mod p).

Hence, (1, 1/ε) is a solution to the system (21). Thus, by Hensel’s lemma, there existsa point on M(Qp) whose reduction mod p is (1, 1/ε).

Case 2 l = q.We consider the system of equations

F(x, z) ≡ 0 (mod q) and∂ F

∂x(x, z) = 4(ph2 − 1)x3 + 4phx �≡ 0 (mod q).

(22)

By (D1), we see that

F(1, 1) = (ph2 − 1) + 2ph + p = p(h + 1)2 − 1 ≡ 1

σ 2 σ 2 − 1 ≡ 0 (mod q),

and

∂ F

∂x(1, 1) = 4(ph2 − 1) + 4ph ≡ 4

(1

σ 2 (σ − 1)2 − 1

)+ 4

1

σ 2 (σ − 1)

≡ − 4

σ�≡ 0 (mod q) [by (D1)].

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The arithmetic of certain quartic curves

Table 2 3-adic points on M

h mod 3 0 0 0 0

p mod 3 1 1 2 2

q mod 3 1 2 1 2

(x, z) (1, 0) (1, 0) (1, 1) (0, 1)

Hence, (1, 1) is a solution to the system (22). Thus, by Hensel’s lemma, there is apoint on M(Qq) whose reduction mod q is (1, 1).

Case 3 l is an odd prime such that l �= p and l divides (ph2 − 1).By assumption, we know that ph2 ≡ 1 (mod l). In particular, we deduce that

−72a2 = h �≡ 0 (mod l). Hence, l �= 2, 3 and a �≡ 0 (mod l). We know from (D2)that

F

(1

12a, 0

)≡ 2ph

1

144a2 + p ≡ p

(h + 72a2

72a2

)≡ 0 (mod l),

∂ F

∂x

(1

12a, 0

)≡ 4ph

1

12a≡ −24ap �≡ 0 (mod l).

Hence, by Hensel’s lemma, there is a point on M(Ql) whose reduction mod l is(1

12a, 0

).

Case 4 l = 2.By (D3), we see that pq ≡ 1 (mod 8). Hence,

√pq ∈ Q

×2 . Thus, the point

P2 := (x, z) =(

0,

√pq

q

)belongs to M(Q2).

Case 5 l = 3.We consider the following two system of equations.

F(x, z) ≡ 0 (mod 3) and∂ F

∂x(x, z) �≡ 0 (mod 3) (23)

and

F(x, z) ≡ 0 (mod 3) and∂ F

∂z(x, z) �≡ 0 (mod 3). (24)

By (D2) and (D3), we know that

h ≡ 0 (mod 3), p ≡ 1, 2 (mod 3) and q ≡ 1, 2 (mod 3).

Table 2 verifies that at least one of the systems (23) and (24) is solvable and hence itfollows from Hensel’s lemma that M is locally solvable at 3.

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N. N. Dong Quan

Case 6 l = ∞.

We see that the point (x, z) =(

0,

√pq

q

)belongs to M(R).

Therefore, by what we have shown, M is everywhere locally solvable.Now we prove that M(AQ)Br = ∅. Throughout the rest of the proof, we will use

the same notations as in Theorem 3.2. The defining equation of M can be written inthe form

M : z2 = q(ph2 − 1)x4 + 2hpqx2 + pq.

Hence, we have

α = q(ph2 − 1),

β = 2hpq,

γ = pq.

Hence, � = β2 − 4αγ = 4q2 p2h2 − 4q2(ph2 − 1)p = 4pq2. Thus, �1 = q2 and�2 = p. Let Q0 be the conic defined by

Q0 : X2 − pY 2 − q(ph2 − 1)Z2 = 0.

The defining equation of M can be written in the form

((ph2 − 1)x2 + ph)2 − p − q(ph2 − 1)z2 = 0. (25)

Since M is everywhere locally solvable, it follows from (25) that Q0 is everywherelocally solvable. Hence, by the Hasse-Minkowski theorem, we deduce that Q0 con-tains a rational point (a1, b1, c1) ∈ Z

3>0 such that gcd(a1, b1, c1) = 1 and a1b1c1 �= 0.

We contend that a1 is a quadratic non-residue in F×p . Indeed, assume that c1 = 2kc0

with c0 a positive odd integer. We see from the equation of Q0 that gcd(c0, p) = 1and thus,

1 =(

p

c0

)=

(c0

p

)=

(c1

p

).

Hence, there is an integer c2 such that c2 �≡ 0 (mod p) and c1 ≡ c22 (mod p). Thus,

modulo p the defining equation of Q0, we deduce that

a21 ≡ −qc4

2 (mod p).

By assumption, we know that −q is not a quartic residue in F×p ; hence it follows that a1

is a quadratic non-residue in F×p . One can verify that the rational point (2a1q, b1, c1q)

belongs to the conic Q defined by

Q : X2 − �Y 2 − 4αZ2 = 0.

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The arithmetic of certain quartic curves

Thus, since 2a1q is a quadratic nonresidue in F×p , we deduce from (D1) that (A1) and

(A3) in Theorem 3.2 are true. On the other hand, one can verify that (A2), (A4), (A5)and (A6) are true. We see from (D1) that

ph2 − 1 ≡ (σ − 1)2

σ 2 − 1 ≡ −2σ + 1

σ 2 �≡ 0 (mod q).

Hence, it follows that vq(α) = 1, proving that α is not a perfect square in Z. Therefore,by Theorem 3.2, we deduce that M(AQ)Br = ∅. Thus, our contention follows. ��Example 6.3 Let p = 137, q = 17, σ = 1 and define

h = −72q2a2 = −20808a2,

for any a ∈ Z. Then, one can check that (D1), (D2) and (D3) are true. Furthermore,we see that

−17 ≡ 422 (mod 137).

Since 42 is a quadratic non-residue in F×137, we deduce that −17 is a quadratic residue

but not a quartic residue in F×137. Let M(137,17,1,a) be the smooth projective model of

the affine curve defined by

M(137,17,1,a) : 17z2 = (59317282368a4 − 1)x4 − 5701392a2x2 + 137, a ∈ Z.

Hence, by Theorem 6.1, we deduce that the infinite family (M(137,17,1,a))a∈Z violatesthe Hasse principle explained by the Brauer–Manin obstruction.

Acknowledgments The author would like to thank his parents Nguyen Ngoc Quang and Phan Thi ThienHuong for their constant support and encouragement. He is very grateful to Romyar Sharifi for fundinghim from his NSF Grant DMS-0901526 during the time when the paper was prepared.

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