FAILURE OF THE HASSE PRINCIPLE FOR ENRIQUES SURFACES

ANTHONY VARILLY-ALVARADO AND BIANCA VIRAY

Abstract. We construct an Enriques surface X over Q with empty etale-Brauer set (andhence no rational points) for which there is no algebraic Brauer-Manin obstruction to theHasse principle. In addition, if there is a transcendental obstruction on X, then we obtaina K3 surface that has a transcendental obstruction to the Hasse principle.

1. Introduction

Let X be a smooth, projective, geometrically integral scheme over a number field k. We saythat X satisfies the Hasse principle if the set X(k) of k-rational points is nonempty wheneverthe set of adelic points X(A

k

) is also nonempty. Manin and Skorobogatov have definedintermediate “obstruction sets” that fit between X(k) and X(A

k

) (cf. §2 or [Man71,Sko99]):

X(k) ✓ X(Ak

)et,Br ✓ X(Ak

)Br ✓ X(Ak

)Br1 ✓ X(Ak

). (1.1)Lind and Reichardt, Harari, Skorobogatov, and Poonen constructed the first schemes thatshow, respectively, that each above containment (from right to left) can be strict [Lin40,Rei42,Har96,Sko99,Poo10].

A wide open area of research considers the finer question: to what extent do the setsin (1.1) give distinct obstructions to the Hasse principle after fixing some numerical in-variants, like dimension, of X? For curves, Scharaschkin and Skorobogatov independentlyasked if X(A

k

)Br1 6= ; implies that X(k) 6= ;, i.e., if the algebraic Brauer-Manin obstructionexplains all counterexamples to the Hasse principle [Sch99,Sko01]. (This question has sincebeen upgraded to a conjecture [Fly04,Poo06, Sto07].) In the case of geometrically rationalsurfaces, Colliot-Thelene and Sansuc conjectured that the same implication holds [CTS80].In contrast, for most other Enriques-Kodaira classes of surfaces we expect that this is nolonger the case.

However, there are strikingly few examples of surfaces that corroborate this expectation.In a pioneering paper, Skorobogatov constructs the first surface for which the failure of theHasse principle is not explained by an algebraic Brauer-Manin obstruction (or, for that mat-ter, a transcendental Brauer-Manin obstruction) [Sko99]; the other known example, due toBasile and Skorobogatov, is of a similar nature [BS03]. In both cases, the surfaces consideredare bi-elliptic and the failure is caused by an etale-Brauer obstruction, i.e. X(A

k

)et,Br = ;.We show that Enriques surfaces give rise to a similar insu�ciency phenomenon. More

precisely, our main result is as follows.

Theorem 1.1. There exists an Enriques surface X/Q such that

X(AQ)et,Br = ; and X(AQ)Br1 6= ;.Moreover, if X(AQ)Br = ;, then Y (AQ)Br Y \Br1 Y = ;, where Y is a K3 double cover of X.

1991 Mathematics Subject Classification. Primary 11G35; Secondary 14G05, 14G25, 14G40.The second author was partially supported by NSF Grant DMS-0841321 and a Ford Foundation Disser-

tation Fellowship. This collaboration was partially supported by Rice University.1

Note the rather curious dichotomy: we obtain either a K3 surface with a transcendentalBrauer-Manin obstruction to the Hasse principle, or an Enriques surface whose failure ofthe Hasse principle is unaccounted for by a Brauer-Manin obstruction.

It is important to remark that Cunnane already showed that Enriques surfaces neednot satisfy the Hasse principle [Cun07]; however, his counterexamples are explained by analgebraic Brauer-Manin obstruction. On the other hand, Harari and Skorobogatov, and laterCunnane, showed that the Brauer-Manin obstruction is insu�cient to explain all failures ofweak approximation on Enriques surfaces [HS05,Cun07], thereby opening up the analogousquestion for the Hasse principle. Theorem 1.1 represents a key step towards a completeanswer to this question.

1.1. Outline of proof. The proof of Theorem 1.1 is constructive. Let a := (a, b, c) 2 Z3,and consider the intersection Ya of the three quadrics

xy + 5z

2 = s

2

(x + y)(x + 2y) = s

2 � 5t

2

ax

2 + by

2 + cz

2 = u

2.

in P5 = Proj Q[s, t, u, x, y, w]. Suppose that

abc(5a + 5b + c)(20a + 5b + 2c)(4a

2 + b

2)(c2 � 100ab)(c2 + 5bc + 10ac + 25ab) 6= 0.

Then Ya is smooth and thus defines a K3 surface. The involution

� : P5 ! P5, (s : t : u : x : y : z) 7! (�s : �t : �u : x : y : z)

has no fixed points when restricted to Ya, so Xa := Ya/� is an Enriques surface.

Theorem 1.2. Let a = (a, b, c) 2 Z3>0 satisfy the following conditions:

(1) for all prime numbers p | (5a + 5b + c), 5 is not a square modulo p,(2) for all prime numbers p | (20a + 5b + 2c), 10 is not a square modulo p,(3) the quadratic form ax

2 + by

2 + cz

2 + u

2 is anisotropic over Q3,(4) the integer �bc is not a square modulo 5,(5) the triplet (a, b, c) is congruent to (5, 6, 6) modulo 7,(6) the triplet (a, b, c) is congruent to (1, 1, 2) modulo 11,(7) Ya(AQ) 6= ;, and(8) the triplet (a, b, c) is Galois general, meaning that a certain number field defined in

terms of a, b, c is as large as possible. A precise definition is given in §4.1.2.Then

Xa(AQ)et,Br = ; and Xa(AQ)Br1 6= ;.Moreover, if Xa(AQ)Br = ;, then Ya(AQ)Br \Br1 = ;.

Theorem 1.1 follows almost at once from Theorem 1.2: we show that the triplet a =(12, 111, 13) satisfies conditions (1)–(8).

Remarks 1.3.(i) In lieu of a paper outline, let us explain the role that the assumptions of Theorem 1.2

play in our constructions. In §3, we show that conditions (1)–(4) imply the etale-Brauer set of Xa is empty. In §4, we use conditions (5), (6), and (8) to describeexplicit generators for the Picard groups of Xa and Ya, as well as to compute the lowdegree Galois cohomology of these groups. In §5, we use the results from §4 together

2

with conditions (5)–(8) to determine both the Brauer set and algebraic Brauer setof Xa. Finally, in §6, we prove Theorems 1.2 and 1.1.

(ii) Our construction relies heavily on an example due to Birch and Swinnerton-Dyer ofa del Pezzo surface of degree 4 that violates the Hasse principle [BSD75]. While weexpect the construction will work with other such surfaces, there are a few subtletieswhich are not readily apparent in the argument. We elaborate on this point in §6.2.

1.2. Notation. Throughout k denotes a perfect field, k is a fixed algebraic closure and G

k

denotes the absolute Galois group Gal(k/k). For any k-scheme X, we write X for the basechange X⇥

k

k and k(X) for the function field of X. For a smooth, projective, geometricallyintegral variety X we identify Pic X with the Weil class group; in particular, we use additivenotation for the group law on Pic X. In addition, we write K

X

for the class of the canonicalsheaf in the group Pic X. Finally, we denote by ⇢(X) the geometric Picard number of X,i.e., the rank of the Neron-Severi group of X.

For a homogeneous ideal I in a graded ring R, we write V (I) for ProjR/I; if explicitgenerators of I = hf

i

: i 2 Si are given, then we write V (fi

: i 2 S) instead of V (I).Henceforth, “condition(s)” refers to items (1)� (8) in Theorem 1.2.

Acknowledgements. We thank Bjorn Poonen and Brendan Hassett for many helpful con-versations. We are indebted to Michael Stoll and Damiano Testa for providing the ideasbehind the proof of Prop 4.2. We also thank Jean-Louis Colliot-Thelene, Daniel Erman,Matthias Schutt, Damiano Testa, Yuri Tschinkel, and Olivier Wittenberg for several com-ments. All computations were done using Magma [BCP97].

2. Background

2.1. K3 surfaces and Enriques surfaces. Assume that the characteristic of k is not 2. AK3 surface is a smooth projective k-surface X of Kodaira dimension 0 with trivial canonicaldivisor and h

1(X,OX

) = 0. The geometric Picard group of a K3 surface is a free abeliangroup of rank at most 22 (in characteristic 0, the rank is at most 20). In addition, theintersection lattice of a K3 surface can be embedded primitively in U

�3 � E8(�1)�2, theunique even unimodular lattice of signature (3, 19).

An Enriques surface is a smooth projective k-surface X of Kodaira dimension 0 withnumerically trivial canonical divisor and second Betti number b2 = 10. These assumptionsimply that K

X

6= 0 and 2K

X

= 0. Enriques surfaces are also characterized in terms of K3surfaces as follows [CD89, Proposition 1.3.1] [Bea96, Proposition VIII.17].

Theorem 2.1. Let X be an Enriques surface and let f : Y ! X be any etale double coverassociated to K

X

2 (Pic X) [2]. Then Y is a K3 surface. Conversely, the quotient of a K3surface by a fixed-point free involution is an Enriques surface.

2.1.1. Geometric birational models of Enriques surfaces. Assume that k is algebraicallyclosed. A generic Enriques surface can be constructed as follows. Consider a K3 surface Y

of degree 8 in P5k

= Proj k[s, t, u, x, y, z] given by

V

�Q

i

(s, t, u)� eQ

i

(x, y, z) : i = 1, 2, 3�, (2.1)

where Q

i

2 k[s, t, u] and eQ

i

2 k[x, y, z] are quadratic polynomials for i = 1, 2, 3. Forgeneric Q

i

and eQ

i

, the involution � of §1.1 has no fixed points when restricted to Y , soX := Y/� is an Enriques surface. Cossec and Verra show that a stronger statement holds:

3

Let f : Y ! X be the double cover of an Enriques surface X by a K3 surface Y . Then thereexists a birational map between Y and a surface of the form (2.1) that identifies � with theunique fixed-point free involution ◆ such that f � ◆ = f [Cos83,Ver83].

2.1.2. The Picard group. For any Enriques surface X, we have the following exact sequenceof Galois modules [CD89, Theorem 1.2.1 and Proposition 1.2.1]

0 ! hKX

i ! Pic X ! Num X ! 0. (2.2)

Additionally, Num X

⇠= U �E8(�1), the unique even unimodular lattice of rank 10 and sig-nature (1, 9) [CD89, Theorem 2.5.1]. We also have an exact sequence relating the geometricPicard group of X and that of any K3 double cover Y of X [HS05, Equation (3.15)]

0 ! Z/2Z ! Pic X ! �Pic Y

�� ! 0. (2.3)

Combining (2.2) and (2.3) we obtain an isomorphism of Galois modules NumX

⇠= �Pic Y

��.

2.2. The Brauer group and the Brauer-Manin obstruction. For the remainder of §2,we restrict to the case where k is a number field. Let BrX := H2

et(X, Gm

) be the Brauergroup of a variety X. An element A 2 BrX is algebraic if it belongs to the subgroupBr1 X := ker

�BrX ! BrX

�; otherwise A is called transcendental.

Functoriality of the Brauer group yields an evaluation pairing

h · , · i : X(Ak

)⇥ BrX�!Q/Z.

For any set S ✓ BrX (S need not be a subgroup), this pairing is used to define the set

X(Ak

)S = {(Pv

)v

2 X(Ak

) : h(Pv

),Ai = 0 for all A 2 S} .

Class field theory guarantees that X(k) ✓ X(Ak

)S, for any S. When S = BrX and Br1 X,respectively, we obtain the Brauer set of X and the algebraic Brauer set of X; we denotethese sets by X(A

k

)Br and X(Ak

)Br1 . In summary, we have

X(k) ✓ X(Ak

)Br ✓ X(Ak

)Br1 ✓ X(Ak

).

We say there is a Brauer-Manin obstruction to the Hasse principle if X(Ak

)Br = ; andX(A

k

) 6= ;. The obstruction is algebraic if in addition X(Ak

)Br1 = ;. See [Sko01, §5.2] formore details.

2.3. Torsors under finite etale groups and the etale-Brauer obstruction. Let G bean fppf group scheme over a scheme X. Recall that a (right) G-torsor over X is an fppf X-scheme Y equipped with a right G-action such that the morphism Y ⇥

X

G ! Y ⇥X

Y givenby (y, g) 7! (y, yg) is an isomorphism. A detailed account of torsors is given in [Sko01, PartI]. A torsor f : Y ! X under a finite etale k-group scheme G determines a partition

X(k) =[

⌧2H1(k,G)

f

⌧ (Y ⌧ (k)) ,

where Y

⌧ is the twisted torsor associated to ⌧ (see [Sko01, Lemma 2.2.3]). Running over allpossible G-torsors of this form, we assemble the etale-Brauer set

X(Ak

)et,Br :=\

f : Y!X

torsor underfinite etale G

0

@[

⌧2H1(k,G)

f

⌧

�Y

⌧ (Ak

)Br�1

A.

4

By construction, we have X(k) ✓ X(Ak

)et,Br; we say there is an etale-Brauer obstruction tothe Hasse principle if X(A

k

) 6= ; and X(Ak

)et,Br = ;. Note that X(Ak

)et,Br ✓ X(Ak

)Br, sothe etale-Brauer obstruction is at least as strong as the Brauer-Manin obstruction.

2.3.1. K3 double covers as torsors. Any K3 double cover f : Y ! X of an Enriques surfaceX is a Z/2Z-torsor over X. By Kummer theory, we have H1(k, Z/2Z) = k

⇥/(k⇥)2, so a

class ⌧ may be represented by an element d 2 k

⇥, up to squares. If Y has the form (2.1),then the twisted torsor Y

⌧ is given explicitly as

V

�dQ

i

(s, t, u)� eQ

i

(x, y, z) : i = 1, 2, 3�. (2.4)

Over Q, a class ⌧ is represented uniquely by a squarefree integer d; we write Y

(d)a instead of

Y

⌧

a .

3. Absence of Q-points

Lemma 3.1. Let a 2 Z3>0 satisfy conditions (1) and (2). If d is a squarefree integer divisible

by a prime p di↵erent from 2 and 5, then Y

(d)a (Z/p

2Z) = ;.Proof. Let (s : t : u : x : y : z) be a primitive Z/p

2Z-point of Y

(d)a . The equations of Y

(d)a

imply thatxy + 5z

2, (x + y)(x + 2y), and ax

2 + by

2 + cz

2 (3.1)are all congruent to 0 modulo p. Let us first assume that x ⌘ �y (mod p). Substitutingthis congruence into the first and third quadrics of (3.1) we obtain

y

2 ⌘ 5z

2 (mod p),

(a + b)y2 ⌘ �cz

2 (mod p).

These congruences are either linearly independent, in which case z ⌘ y ⌘ x ⌘ 0 (mod p), orelse p | (5a+5b+c). In the latter case, by condition (1) we know that 5 is not a square moduloany prime p dividing 5a+5b+c, so the congruences again imply that z ⌘ y ⌘ x ⌘ 0 (mod p).The defining equations of Y

(d)a then imply that s ⌘ t ⌘ u ⌘ 0 (mod p), a contradiction. The

argument for the case x ⌘ �2y (mod p) is similar. ⇤Proposition 3.2. Let a 2 Z3

>0 satisfy conditions (1)–(4). Then

Xa(AQ)et,Br = ;.Proof. By definition of the etale-Brauer set, it su�ces to show that

Y

(d)a (AQ)Br = ; (3.2)

for all squarefree integers d (including 1). Lemma 3.1 establishes (3.2) for all such d exceptthose in h�1, 2, 5i. Since a, b and c are positive, it is easy to see that Y

(d)a (R) = ; for any

negative d. Condition (3) implies that Y

(d)a (Q3) = ; for all d ⌘ 2 (mod 3) and condition (4)

implies that Y

(10)(Q5) = ;. Thus, the only remaining case is d = 1. Birch and Swinnerton-Dyer prove that the del Pezzo surface S ⇢ P4 of degree 4 given by

xy = s

2 � 5z

2, (3.3)

(x + y)(x + 2y) = s

2 � 5t

2, (3.4)

5

has no adelic points orthogonal to the quaternion Azumaya algebra

A :=⇣5,

x + y

x

⌘2 im (BrS ! Brk(S)) ;

see [BSD75]. The existence of a map h : Ya ! S and functoriality of the evaluation pairingimply that Ya(AQ)h

⇤(A) = ;. ⇤

4. Picard Groups

We compute explicit presentations for the groups Pic Xa, Num Xa, and Pic Y a, and thuscompute their Galois cohomology. In §4.1, we describe how to obtain genus 1 fibrationson Y a and Xa. In §4.2, we prove that the fibers of these fibrations give us a finite indexsubgroup of Pic Y a, and in §4.3, we explain how to compute its saturation in Pic Y a. In§4.4, we give explicit generators for Num Xa in terms of the fibrations. Finally, in §4.5, weuse the explicit computations from the previous sections to determine the low degree Galoiscohomology of Pic Y a,Pic Xa, and Num Xa.

4.1. Genus 1 fibrations. Throughout this section, we work over an algebraically closedfield.

Let Y be a K3 surface of degree 8, given as a closed subscheme of P5 by the vanishing ofa net of quadrics. Let Z be the sextic curve in P2 that parametrizes the degeneracy locus ofthis net. Expanding on [Bea96, Example IX.4.5], we explain how an isolated singular pointP 2 Z gives rise to two distinct genus 1 fibrations on Y .

Let Q be the quadric corresponding to P in the net defining Y . Then Q has rank 4, andthere are two rulings on it, each realizing V (Q) as a P3-bundle over P1. Restricting to Y ,we obtain two maps �

P

,�

0P

: Y ! P1Q whose respective general fibers are smooth complete

intersections of two quadrics in P3, i.e., smooth genus 1 curves. We write F

P

(resp. G

P

) forthe class in Pic Y of a fiber in �

P

(resp. �0P

).For the family of K3 surfaces given by Y a, the curve Z ⇢ P2 is the union of 4 lines, each

defined over Q, and a conic. If 2a 6= b, then the conic is geometrically irreducible and thereare exactly 14 distinct singular points P1, . . . , P14 on Z. These points give rise to the 28classes F

i

:= F

Pi , Gi

:= G

Pi , i = 1, . . . , 14 in Pic Y a, as above. The class F

i

+G

i

is equivalentto a hyperplane section for all i, and we have

F

2i

= G

2i

= 0, F

i

· Gi

= 4, F

i

· Gj

= F

i

· Fj

= G

i

· Gj

= 2 for all i 6= j.

These relations imply that G1, F1, F2, . . . , F14 generate a rank 15 sublattice of Pic Y a. Wehave listed equations for representatives of these classes and the Galois action of GQ on themin Appendix A. We will use this information in later sections.

4.1.1. Fibrations that descend to Xa. We work in characteristic zero for the remainder ofthis section. Let f : Y ! X be the K3 double-cover of a generic Enriques surface X; assumethat Y is of the form (2.1). In this case, the degeneracy locus Z contains at least one singularpoint P ; let �

P

,�

0P

: Y ! P1 be the fibrations described above. If there is an involution◆ : P1 ! P1 such that � � � = ◆ � �, then each fibration descends to X, i.e., there is a genus

6

1 fibration �X

: X ! P1 such that the diagram

Y �������!f

Y/� = X

??y�P

??y�X

P1 �������! P1/◆ = P1

commutes, and similarly for �0P

.Careful inspection of the singular locus of Z allows us to determine which points P have

an associated involution ◆ as above. Indeed, since the quadrics defining Y are di↵erencesof a quadric in k[x, y, z] and a quadric in k[s, t, u], the sextic curve defining Z ⇢ P2 is theunion of two (possibly reducible) cubic curves. Generically, these two cubics intersect in ninedistinct points. It is precisely these singular points of Z that have an associated involution◆ such that � � � = ◆ � �.

The map P1 ! P1/◆ is ramified above the fixed points of ◆. Since f : Y ! X is unramified,

the morphism �

X

: X ! P1 must have non-reduced fibers above the fixed points of ◆. Wedenote by C

P

and eC

P

(resp. D

P

and eD

P

) the reduced subschemes of the nonreduced fibersF

P

(resp. G

P

) of �P

.Let us specialize to our particular Enriques surface Xa and its K3 double cover Y a. We

already know that Z contains 14 singular points P1, . . . , P14. We may renumber these pointsso that P1, . . . , P9 correspond to the fibrations that descend to Xa. This gives us 36 curvesC

i

,

eC

i

, D

i

,

eD

i

, i = 1, . . . , 9 on Xa. After possibly interchanging D

i

and eD

i

for some i, wehave the linear equivalence relations

C

i

+D

i

= eC

j

+ eD

j

, 2(Ci

� eC

i

) = 2(Di

� eD

i

) = 0, f

⇤C

i

= f

⇤ eC

i

= F

i

, f

⇤D

i

= f

⇤ eD

i

= G

i

,

for all i, j. Combining the projection formula with the intersection numbers on Y a, weobtain

C

2i

= D

2i

= 0, C

i

· Dj

= C

i

· Cj

= D

i

· Dj

= 1, C

i

· Di

= 2, for all i 6= j.

We have listed the action of the Galois group GQ on C

i

, D

i

in Appendix A.

4.1.2. Splitting field of the genus 1 fibrations. Let a, b and c be indeterminates and a =(a, b, c); consider Ya and Xa as surfaces over Q(a, b, c). The splitting field K of the fibersof all the genus 1 fibrations is a degree 218 extension (explicit generators can be found inAppendix A). We can consider K as a 218-cover of A3. We say that the triplet a0 :=(a0, b0, c0) 2 Z3

>0 is Galois general if the special fiber K(a0,b0,c0) is a field, i.e., if the splittingfield of the fibers of the genus 1 fibrations of Ya0 is a degree 218 extension over Q.

4.2. Upper bounds for ⇢(Ya) (after van Luijk). Let p 2 Z be a prime of good reductionfor Ya, and write Ya,p

for the mod p reduction of Ya. Suppose that Ya has good reductionat two distinct primes p1 and p2, that ⇢(Ya,pi) n for each i, and that the discriminantsof the Picard groups of the reductions lie in di↵erent square classes. Then ⇢(Ya) n � 1(see [vL07, Proof of Theorem 3.1]).

Let ` 6= p be a prime, and write p

(T ) for the characteristic polynomial of the action ofFrobenius on H2

et

�Ya,p

, Q`

�. Then ⇢(Ya,p

) is bounded above by the number of roots of p

(counted with multiplicity) that are of the form p⇣, where ⇣ is a root of unity [vL07, Corol-lary 2.3]. Using the Lefschetz trace formula and Newton’s identities, we may compute the

7

coe�cient of T

i in p

(T ) in terms of #Ya,p

(Fp

), . . . ,#Ya,p

(Fp

22�i). The functional equation

p

22

p

(T ) = ±T

22

p

(p2/T ), (4.1)

then allows us to compute the coe�cient of T

23�i, up to sign. In addition, for a subgroupM ✓ Pic Y a,p

, the polynomial M

(T/p) divides p

(T ), where M

(T ) is the characteristicpolynomial of Frobenius acting on M . Thus, knowing the action of Frobenius on a rank r

subgroup of Pic Y a,p

, together with #Ya,p

(Fp

), . . . ,#Ya,p

(Fp

d(22�r)/2e), allows us to computeup to two possible

p

(T )’s, each corresponding to a choice of sign in (4.1). In some cases,this is enough information to rule out one of the sign choices; for more ways to determinethe sign choice, see [EJ10].

Proposition 4.1. Let a 2 Z3>0 satisfy conditions (5) and (6). Then ⇢(Ya) 15.

Proof. Let M

p

denote the subgroup of Pic Y a,p

generated by G1, F1, . . . , F14. Let e

p

(T ) :=p

�22

p

(pT ), so that the number of roots of e

p

(T ) that are roots of unity gives an upper boundfor the geometric Picard number of the reduced surface. As described above, computing theaction of Frobenius on M

p

and computing #Ya,p

(Fp

i) for i := 1, 2, 3, 4 is enough to determine

p

(T ) for p = 7 and 11:

e 7(T ) =

17(T � 1)8(T + 1)8(7T

6 + 6T

5 + 9T

4 + 4T

3 + 9T

2 + 6T + 7),

e 11(T ) =

111

(T � 1)8(T + 1)4(T 2 + 1)2(11T

6 � 2T

5 + T

4 + 12T

3 + T

2 � 2T + 11).

In both cases, the roots of the degree 6 factor of e

p

(T ) are not integral, so they are notroots of unity. We conclude that ⇢(Ya,p

) 16 for p = 7 and 11.Next, we compute the square class of the discriminant � of each reduced Picard lattice

via the Artin-Tate conjecture, which is known to hold for K3 surfaces endowed with a genus1 fibration (see [ASD73,Mil75]):

limT!p

p

(T )(T � p)rk(Pic Ya,p)

= p

21�rk(Pic Ya,p) · # Br(Ya,p

) · |�|

Observe that #Br(Ya,p

) is always a square [LLR05]. Write⇥|�|⇤ for the class of |�| in

(Q⇤)/(Q⇤)2. We compute⇥|�(Y a,7)|

⇤= 3 and

⇥|�(Y a,11)|⇤

= 2,

and thus ⇢(Ya) 15, completing the proof. ⇤

4.3. Determining Pic Y a. The sublattice L := hG1, F1, . . . , F14i of Pic Y a has discriminant217. In this subsection we determine its saturation in Pic Y a.

Each line ` on the curve Z corresponds to a pencil of quadrics; the vanishing locus ofthis pencil S

`

defines a del Pezzo surface of degree 4, embedded in a hyperplane in P5. Theinclusion of the pencil in the net of quadrics defining Y a gives a morphism Y a ! S

`

. Recallthat Z contains four lines, so we obtain four such maps. Pulling back the exceptional curveson each of the del Pezzo surfaces, we obtain 16 order 2 elements in (Pic Y a)/L, generated

8

by12(F1 + F2 + F3 + F10 + F12),

12(F1 + G1 + F4 + F5 + F6 + F10 + F11), (4.2)

12(F1 + F4 + F7 + F13 + F14),

12(F1 + G1 + F7 + F8 + F9 + F11 + F12). (4.3)

We owe the idea behind the proof of the following proposition to Michael Stoll and Dami-ano Testa [ST10].

Proposition 4.2. Let a 2 Z3>0 satisfy condition (8). The sublattice L

0 ✓ Pic Y a spanned byG1, F1, . . . , F14 and the classes in (4.2) and (4.3) is saturated.

Proof. The lattice L

0 has discriminant 29, so it su�ces to show that the induced map

� : L

0/2L

0 ! Pic Y a/2 PicY a

is injective. The action of GQ on L

0 factors through a finite group G whose order divides218 (see §4.1.2). Consider the induced G-equivariant homomorphism

�

G :�L

0/2L

0�G ! �Pic Y a/2 PicY a

�G

,

and note that if ker� is nonzero then (ker�)G = ker�G is also nonzero, because anyrepresentation of a 2-group by a nonzero F2-vector space has a nonzero invariant sub-space [Ser77, Proposition 26]. Using condition (8), it is easy to establish that

�L

0/2L

0�G

is a 2-dimensional F2-vector space, spanned by the classes v1 := [G1 + F10] and v2 :=[F2 + F3 + F12]. If �G(v1) = 0 then G1 + F10 2 2 PicY a; however, the intersection pair-ing on Pic Y a is even, and 1

2(G1 + F10) · 12(G1 + F10) = 1. Hence �G(v1) 6= 0. Similarly,

�

G(v2),�G(v1 + v2) 6= 0, and we conclude that ker�G = 0. ⇤Corollary 4.3. Let a 2 Z3

>0 satisfy conditions (5), (6), and (8). Then L

0 = Pic Y a.

Proof. By Proposition 4.1 we have ⇢(Ya) 15. Thus L

0 has full rank inside Pic Y a, and itis saturated by Proposition 4.2. ⇤4.4. Determining Num Xa. Let M := hD1, C1, C2, . . . , C9i ✓ Num Xa. In §4.1, usingintersection numbers, we calculated that rkM = 10, and that the discriminant of the inter-section lattice is 4. Since Num Xa is a rank 10 unimodular lattice (see §2.1.2), M is an index2 subgroup of Num Xa. The following proposition shows that there are only two possiblesaturations of M .

Proposition 4.4. There exists a divisor R on Xa such that 2R is linearly equivalent toeither

C1 + C2 + · · · + C9 or D1 + C2 + · · · + C9, (4.4)and such that Num Xa = hR,D1, C1, . . . , C9i.Proof. From the discussion above, we know that Num Xa/M

⇠= Z/2Z. Let R be a divisorwhose class in Num Xa is not in M . Without loss of generality, we may assume that

2R = n1C1 + · · ·n9C9 + n10D1

in Pic Xa, where each n

i

2 {0, 1}. Since R pairs integrally with D1 and C

i

for all i andR

2 ⌘ 0 (mod 2), we must have that n2 ⌘ n3 ⌘ · · ·n9 ⌘ 1 (mod 2) and n1 + n10 ⌘ 1(mod 2), giving the desired result. ⇤

9

4.5. Galois cohomology. Throughout this section, we assume that a 2 Z3>0 is Galois

general, and we let K

a,b,c

be the splitting field described in §4.1.2. Since Pic Y a and Num Xa

are torsion-free, to compute the Galois cohomology of Pic Y a and Num Xa it su�ces tocompute the action of GQ on the curves C

i

, D

i

, F

i

, G

i

. Thus, the action of GQ on Pic Y a

and Num Xa factors through Gal(Ka,b,c

/Q). The action of Gal(Ka,b,c

/Q) is described inTable A.2. Together with the generators for Pic Y a and Num Xa given in §§4.3–4.4, itallows us to prove the following proposition.

Proposition 4.5. Let a 2 Z3>0 satisfy condition (8). Then H1(GQ,Num Xa) = {1}. ⇤

We used Magma [BCP97] for the computations involved in Proposition 4.5, but the indus-trious reader may verify the results by hand, using the following lemma.

Lemma 4.6. Let G be a (pro-)finite group and A be a (continuous) torsion-free G-module.Then

H1(G, A)[m] ⇠= (A/mA)G

A

G

/mA

G

. ⇤

Proposition 4.5 implies that the long exact sequence in group cohomology associatedto (2.2) is given by

0 ! hKXai ! Pic Xa ! Num Xa ! H1(GQ, hK

Xai) ! H1(GQ,Pic Xa) ! 0.

The explicit description of the Galois action given in Table A.2 allows us to verify thatPic Xa = K

Xa � Z · [C1 + D1] and that Num Xa = Z · [C1 + D1]. The following propositionnow follows easily.

Proposition 4.7. Let a 2 Z3>0 satisfy condition (8). Then

H1(GQ,Pic Xa) ⇠= H1(GQ, hKXai). ⇤

Remark 4.8. Table A.2 also allows us to show that if a satisfies conditions (5), (6) and (8)then H1(GQ,Pic Y a) ⇠= Z/2Z. While this result is not logically necessary for our purposes, itis worth noting that it implies that the algebra h

⇤(A) in the proof of Proposition 3.2 gives arepresentative of the nontrivial class of this cohomology group, and thus, up to adjustmentby constant Azumaya algebras arising from the base field, h

⇤(A) is the only algebra thatcan give an algebraic Brauer-Manin obstruction to the Hasse principle on Ya.

5. The Brauer-Manin obstruction

Proposition 5.1. Let a 2 Z3>0 satisfy conditions (7) and (8). Then Xa(AQ)Br1 6= ;.

Proof. Since a satisfies condition (8), by Proposition 4.7 we know that H1(GQ,Pic Xa) ⇠=H1(GQ, hK

Xai). Using this in conjunction with [Sko01, Theorem 6.1.2] and the isomorphismcoming from the Hochschild-Serre spectral sequence

Br1 X/ im Br Q ⇠�! H1(GQ,Pic X),

we obtain the following partition of the algebraic Brauer set

Xa(AQ)Br1 =[

⌧2H1(GQ,hKXai)

f

⌧ (Y ⌧

a (AQ)) . (5.1)

By condition (7) we have Ya(AQ) 6= ;, and thus Xa(AQ)Br1 6= ;. ⇤10

Proposition 5.2. Let a 2 Z3>0 satisfy conditions (5)– (8), and assume that Xa(AQ)Br = ;.

Then there exists an element A 2 BrY \ Br1 Y such that Ya(AQ)A = ;.Proof. Since BrX

⇠= Z/2Z for any Enriques surface X [HS05, p. 3223], there is at most onenontrivial class in BrX/ Br1 X. If Br Xa = Br1 Xa, then Xa(AQ)Br 6= ; (by Proposition 5.1),contradicting our hypotheses. Thus we may assume that there exists an element A0 2 BrXa\Br1 Xa. Then Xa(AQ)Br = Xa(AQ)Br1 \Xa(AQ)A0 , and (5.1), together with functoriality ofthe Brauer group imply that

Xa(AQ)Br =[

⌧2H1(GQ,hKXai)

f

⌧

⇣Y

⌧

a (AQ)(f⌧ )⇤A0

⌘.

By assumption, we have Xa(AQ)Br = ;, so it su�ces to show that f

⇤A0/2 Br1 Ya, because

then we can take A = f

⇤A0. Equivalently, we show that the map f

⇤ : Br Xa ! BrY a isinjective, using Beauville’s criterion [Bea09, Cor. 5.7]: f

⇤ is injective if and only if thereis no divisor class D 2 Pic Y a such that D

2 ⌘ 2 (mod 4) and �(D) = �D. This is astraightforward computation, using the integral basis of Pic Y a given in Corollary 4.3. ⇤

6. Proof of Theorems 1.2 and 1.1

Lemma 6.1. Y(12,111,13)(AQ) 6= ;.Proof. By the Weil conjectures, if p � 22 is a prime of good reduction, then Y(12,111,13) has F

p

-points. Hensel’s lemma then implies that Y(12,111,13)(Qp

) 6= ; for all such primes. It remainsto consider the infinite place, the primes less than 22, and the primes of bad reduction:2, 3, 5, 13, 37, 59, 151, 157, 179, 821, 881 and 1433. It is easy to check that Y(12,111,13)(R) 6= ;,and that for a finite prime p > 5 in our list, Y(12,111,13) has a smooth F

p

-point. For theremaining three primes we exhibit explicit local points:

(p

129 : 2p

21/5 :p

2113 : 1 : 4 : 5) 2 Y(12,111,13)(Q2),

(0 : 0 :p

821/5 : �2 : 1 :p

2/5) 2 Y(12,111,13)(Q3),

(1 : 2p�1 :

p136 : 1 : �4 : 1) 2 Y(12,111,13)(Q5). ⇤

Proof of Theorem 1.2. Combining Propositions 3.2 and 5.1 we obtain the first statement.The second statement follows from Proposition 5.2. ⇤Proof of Theorem 1.1. The triplet a = (12, 111, 13) satisfies the eight required conditions:(1)� (6) and (8) are easily checked, and (7) follows from Lemma 6.1. ⇤6.1. More possibilities. Although we have shown only that the triplet a = (12, 111, 13)satisfies conditions (1)–(8), there are in fact many triplets that do the job. Additionally, wecan weaken conditions (5) and (6), and thus find even more triplets.

We used conditions (5) and (6) to ensure that ⇢(Ya,7) = ⇢(Ya,11) = 16 and that thediscriminants of the respective intersection lattices did not di↵er by a square. Testingall possible isomorphism classes of Ya modulo 7 and modulo 11, we found 10 380 possiblecongruence classes of a modulo 77 that have the desired property concerning Picard numbersand discriminants.

A computer search shows that there are 202 triplets a 2 Z3>0 satisfying conditions (1)–(4)

and (7) and these weaker versions of (5) and (6) with a + b + c 500. We do not know howto verify condition (8) e�ciently on a computer. However, Hilbert’s irreducibility theorem

11

says we should expect condition (8) to hold “almost always”, so we feel this data does givesome evidence that several triplets a satisfying conditions (1)–(8) exist.

6.2. Varying the del Pezzo surface. The Enriques surface Xa was built up from thedegree 4 del Pezzo surface S given by (3.3) and (3.4). The equality H1(GQ,Num Xa) = {1}depends on the interaction of � with S in a way that is perhaps not obvious in the proof.More precisely, consider the following K3 surfaces

xy = z

2 � 5s

2xy = s

2 � 5u

2

Y1 : (x + y)(x + 2y) = z

2 � 5t

2Y2 : (x + y)(x + 2y) = s

2 � 5t

2

Q1(x, y, z) = eQ1(s, t, u) Q2(x, y, z) = e

Q2(s, t, u)

where Q

i

,

eQ

i

are any quadrics such that �|Yi has no fixed points, and let X

i

:= Y

i

/�, i = 1, 2,be the corresponding Enriques surfaces. Note that each Y

i

maps to a del Pezzo surfacewhich is Q-isomorphic to S. In this case # H1(GQ,Num X

i

) � 2 regardless of the choice ofQ

i

,

eQ

i

. In fact, the quaternion algebras⇣

5z

2

s

2 ,

x+y

x

⌘2 Brk(X1) and

�5,

x+y

x

� 2 Brk(X2) liftto non-trivial elements in Br X1 and Br X2, respectively.

Remark 6.2. Although z

2

s

2 is a square in k(Y2), it is not a square in k(X2), so⇣

5z

2

s

2 ,

x+y

x

⌘2

Brk(X1) is not necessarily equal to�5,

x+y

x

� 2 Brk(X1).

Remark 6.3. One can check that, up to a Q-automorphism of P5 that commutes with �,Y1 and Y2 are the only K3 surfaces that arise as double covers of some Enriques surface andthat map to a del Pezzo surface which is Q-isomorphic, but not equal, to S.

Appendix A. The splitting field of fibers of genus 1 fibrations

The splitting field K of the genus 1 curves C

i

,

eC

i

,

eD

i

, F

i

, G

i

is generated by

i,

p2,

p5,

pa,

pc,

pc

2 � 100ab, � :=p�c

2 � 5bc� 10ac� 25ab, (A.1)

4p

ab,

q�2 + 2

p2,

q�c� 10

pab, (A.2)

✓0 :=p

4a

2 + b

2, �1 :=

pa + b + c/5, �2 :=

pa + b/4 + c/10, (A.3)

✓

+1 :=

p20a

2 � 10ab� 2bc + (10a + 2c)✓0, ✓+2 :=

p�5a� 5/2b� 5/2✓0, (A.4)

⇠

+1 :=

p20a + 10b + 3c + 20�1�2, ⇠+

2 :=p

4a + 2b + 2/5c + 4�1�2. (A.5)

The field extension K/Q is Galois, as the following relations show:q�2� 2

p2 =

2ip�2 + 2

p2,

q�c + 10

pab =

pc

2 � 100abp�c� 10

pab

✓

�1 :=

p20a

2 � 10ab� 2bc + (10a + 2c)✓0 =4a�

✓

+1

, ✓

�2 :=

p�5a� 5/2b + 5/2✓0 =

5p

ab

✓

+2

⇠

�1 :=

p20a + 10b + 3c� 20�1�2 =

pc

2 � 100ab

⇠

+1

, ⇠

�2 :=

p4a + 2b + 2/5c� 4�1�2 =

2�5⇠+

2

Table A.1 lists two linear forms `1, `2 next to each divisor class. On the K3 surface, thevanishing of these linear forms defines a curve representing the corresponding class. On theEnriques surface, the curve f(Y a \ V (`1, `2))red represents the given divisor class.

12

Y

aX

aD

efiningequations

Y

aX

aD

efiningequations

F

1C

1s�

p5t,

x+

2y

F

4C

4p

cs�

p5u,

10ax�

(c+p

c

2�100

ab)

y

eC1

s+p

5t,

x+

y

eC4

pcs

+p

5u,

10ax�

(c�

pc

2�100

ab)

y

G

1D

1s�

p5t,

x+

yG

4D

4p

cs�

p5u,

10ax�

(c�

pc

2�100

ab)

y

eD1

s+p

5t,

x+

2y

eD4

pcs

+p

5u,

10ax�

(c+p

c

2�100

ab)

y

F

2C

2

p�

2+

2 p2s�

p5t,

x+p

2y

+p

5(1�p

2)z

F

5C

5i p

24 pabs�

u,

pax

+p

by

+p�

c�10 p

abz

eC2

p�

2+

2 p2s

+p

5t,

x+p

2y�

p5(1�

p2)

z

eC5

i p2

4 pabs

+u,

pax

+p

by�

p�

c�10 p

abz

G

2D

2

p�

2+

2 p2s�

p5t,

x+p

2y�

p5(1�

p2)

zG

5D

5i p

24 pabs�

u,

pax

+p

by�

p�

c�10 p

abz

eD2

p�

2+

2 p2s

+p

5t,

x+p

2y

+p

5(1�p

2)z

eD5

i p2

4 pabs

+u,

pax

+p

by

+p�

c�10 p

abz

F

3C

3

p�

2�2 p

2s�

p5t,

x�p

2y

+p

5(1+p

2)z

F

6C

6

p2

4 pabs

+u,

pax�

pby

+p�

c+

10 pabz

eC3

p�

2�2 p

2s

+p

5t,

x�p

2y�

p5(1

+p

2)z

eC6

p2

4 pabs�

u,

pax�

pby�

p�

c+

10 pabz

G

3D

3

p�

2�2 p

2s�

p5t,

x�p

2y�

p5(1

+p

2)z

G

6D

6

p2

4 pabs

+u,

pax�

pby�

p�

c+

10 pabz

eD3

p�

2�2 p

2s

+p

5t,

x�p

2y

+p

5(1+p

2)z

eD6

p2

4 pabs�

u,

pax�

pby

+p�

c+

10 pabz

Y

aX

aD

efiningequations

Y

aX

aD

efiningequations

F

7C

7p

ct�

u,

(5a

+c)

x+

(c+�)

yF

10

s�p

5z,

x

eC7p

ct+

u,

(5a

+c)

x+

(c�

�)y

G

10

s�p

5z,

y

G

7D

7p

ct�

u,

(5a

+c)

x+

(c�

�)y

F

11

u�p

cz,

pax

+i p

by

eD7p

ct+

u,

(5a

+c)

x+

(c+�)

yG

11

u�p

cz,

pax�

i pby

F

8C

8✓

+2t�

u,

x+

(2a

+b+✓

0 )/(

b+✓

0 )y�

✓

+1/(2

a)z

F

12

t�z,

x+

(1�i)

y

eC8

✓

+2t+

u,

x+

(2a

+b+✓

0 )/(

b+✓

0 )y

+✓

+1/(2

a)z

G

12

t�z,

x+

(1+

i)y

G

8D

8✓

+2t�

u,

x+

(2a

+b+✓

0 )/(

b+✓

0 )y

+✓

+1/(2

a)z

F

13

⇠

+2s�

⇠

+1t,

(�

1 +2�

2 )x

+(2�

1 +2�

2 )y�

u

eD8

✓

+2t+

u,

x+

(2a

+b+✓

0 )/(

b+✓

0 )y�

✓

+1/(2

a)z

G

13

⇠

+2s�

⇠

+1t,

(�

1 +2�

2 )x

+(2�

1 +2�

2 )y

+u

F

9C

9✓

�2t�

u,

x+

(2a

+b�

✓

0 )/(

b�✓

0 )y�

✓

�1/(2

a)z

F

14

⇠ �2s�

⇠ �1t,

(�

1 �2�

2 )x

+(2�

1 �2�

2 )y�

u

eC9

✓

�2t+

u,

x+

(2a

+b�

✓

0 )/(

b�✓

0 )y

+✓

�1/(2

a)z

G

14

⇠ �2s�

⇠ �1t,

(�

1 �2�

2 )x

+(2�

1 �2�

2 )y

+u

G

9D

9✓

�2t�

u,

x+

(2a

+b�

✓

0 )/(

b�✓

0 )y

+✓

�1/(2

a)z

eD9

✓

�2t+

u,

x+

(2a

+b�

✓

0 )/(

b�✓

0 )y�

✓

�1/(2

a)z

Table

A.1

.D

efining

equations

ofa

curve

representing

adivisor

class

13

Action

onA

ctionon

Action

onA

ctionon

Action

onA

ctionon

Action

onA

ctionon

Action

onsplitting

fieldP

icX

aP

icY

asplitting

fieldP

icX

aP

icY

asplitting

fieldP

icX

aP

icY

ap

57!� p

5C

1 7!D

1F

1 $G

1i7!

�i

C

3 7!eD

3F

3 7!G

34 pab7!

i

4 pab

C

5 7!C

6F

5 7!F

6

D

1 $eC

1C

5 7!eD

5F

5 7!G

5

p�

c�10 p

ab$

C

6 7!eD

5F

6 7!G

5

C

2 7!eC

2F

11 7!

G

11

p�

c+

10 pab

C

9 7!eD

9F

9 7!G

9

C

3 7!eC

3F

12 7!

G

12

F

11 7!

G

11

C

4 7!eD

4F

4 7!G

4

F

10 7!

G

10

pc7!

� pc

C

4 7!eD

4F

4 7!G

4�7!

��

C

7 7!D

7F

7 7!G

7

pc

2�100

ab7!

C

4 7!D

4F

4 7!G

4

C

7 7!eD

7F

7 7!G

7C

9 7!D

9F

9 7!G

9� p

c

2�100

ab

C

6 7!D

6F

6 7!G

6

F

11 7!

G

11

F

14 7!

G

14

F

14 7!

G

14

�

1 7!��

1F

13 7!

G

14

pa7!

� pa

C

5 7!D

5C

6 7!D

6

p27!

� p2

C

2 $C

3F

2 $F

3

⇠

+i

$⇠ �i

F

14 7!

G

13

C

6 7!D

6F

6 7!G

6

p�

2+

2 p2$

C

5 7!eD

5F

5 7!G

5p�

2�2 p

2C

6 7!eD

6F

6 7!G

6

�

2 7!��

2F

13 $

F

14

✓

0 7!�✓

0C

8 $C

9F

8 $F

9

p�

2+

2 p27!

C

2 7!eD

2F

2 7!G

2

⇠

+i

$⇠ �i

✓

+i

$✓

�i

�p�

2+

2 p2

C

3 7!eD

3F

3 7!G

3

⇠

+17!�⇠

+1F

13 7!

G

13

✓

+17!�✓

+1C

8 7!D

8F

8 7!G

8

p�

c�10 p

ab7!

C

5 7!D

5F

5 7!G

5

F

14 7!

G

14

C

9 7!D

9F

9 7!G

9�

p�

c�10 p

ab

C

6 7!D

6F

6 7!G

6

⇠

+27!�⇠

+2F

13 7!

G

13

✓

+27!�✓

+2C

8 7!eD

8F

8 7!G

8

F

14 7!

G

14

C

9 7!eD

9F

9 7!G

9

Table

A.2

.T

he

Galois

action

on

the

fibers

ofthe

genus

1fibrations.

The

action

on

the

splitting

field

is

described

by

the

action

on

the

generators

listed

in

A.1,A

.2,A

.3,A

.4,A

.5.

Ifa

generator

ofK

is

not

listed,then

we

assum

ethat

it

is

fixed.

We

use

the

sam

econvention

for

the

curves.

14

References

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Department of Mathematics, Rice University, Houston, TX 77005, USAE-mail address: varilly@rice.eduURL: http://www.math.rice.edu/~av15

Department of Mathematics, Box 1917, Brown University, Providence, RI 02912, USAE-mail address: bviray@math.brown.eduURL: http://math.brown.edu/~bviray

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