Pathologies of the Brauer-Manin obstruction

We construct a conic bundle over an elliptic curve over a real quadratic field that is a counterexample to the Hasse principle not explained by the \'etale Brauer-Manin obstruction. We also give simple examples of threefolds with the same property that are families of 2-dimensional quadrics, and discuss some other examples and general properties of the Brauer-Manin obstruction.


Introduction
It had long been suspected that varieties X over a number field k without k-points but with a non-empty Brauer-Manin set X (A k ) Br are fairly common. The first examples were found in [27] and then in [2]. An earlier, conditional example is given in [21]. One should also expect that there are many varieties X without k-points for which the étale Brauer-Manin set X (A k )´e t,Br ⊂ X (A k ) Br is non-empty (we refer to [19] or [29] for the definition of these subsets of the space X (A k ) of adèles of X ). Different methods to construct such varieties have been found recently. In [19] Poonen constructs a threefold X with a surjective morphism to a curve C that has exactly one k-point P and the fibre X P has points everywhere locally but not globally. In Poonen's example X P is a smooth Châtelet surface. The trick with a curve with just one rational point was also used in [11] where the fibres of X → C are curves of high genus and X P is a singular curve which geometrically is a union of projective lines. In retrospect one could note that the examples in [27] and [2] are families of genus 1 curves parameterised by elliptic curves of Mordell-Weil rank 0.
In this paper we propose more flexible methods to construct such examples. We show that the varieties X such that X (k) = ∅ and X (A k )´e t,Br = ∅ include the following: a conic bundle surface X → E over a real quadratic field k, where E is an elliptic curve such that E(k) = {0}, see Sect. 5.2; a threefold over an arbitrary real number field k ⊂ R, which is a family X → C of 2-dimensional quadrics parameterised by a curve C with exactly one k-point (one can choose C to be an elliptic curve), see Sect. 3.1; a threefold over an arbitrary number field k, which is a family X → C of geometrically rational surfaces parameterised by a curve C with exactly one k-point, the fibre above which is singular, see Sect. 3.2.
In the first and second examples, in contrast to those previously known, the smooth fibres satisfy the Hasse principle and weak approximation. To put this into a historical perspective let us note that soon after Manin [17] introduced the obstruction now bearing his name, Iskovskikh [13] constructed a counterexample to the Hasse principle on a conic bundle over the projective line over Q. His intention was, as he pointed out to one of us, to give a counterexample to the Hasse principle that could not be explained by the Brauer-Manin obstruction. It is well known nowadays that Iskovskikh's counterexample can be explained by the Brauer-Manin obstruction, and conjecturally the same should be true for all counterexamples to the Hasse principle on geometrically rational surfaces, see [4,5].
The examples we construct in this paper show that this is no longer the case for conic or quadric bundles over curves of genus at least 1.
In a nutshell, the idea is this. Let k be a number field. Following Poonen we use a base variety B such that B(k) = {P}. By a continuous deformation of the adèle attached to P at an archimedean component we see that B(k) is not dense in B(A k ) Br . Density may also fail due to places of k that need not be archimedean. Suppose B contains an irreducible singular conic S so that P = S sing . If a place v of k splits in the quadratic extension given by the discriminant of the binary quadratic form that defines S, then B × k k v contains two copies of P 1 k v meeting at P. Since Br(P 1 k v ) = Br(k v ), we can modify the adèle of P at v while staying inside B(A k ) Br . However, the k-point P cannot be moved in B, so B(k) is not dense in B(A k ) Br .
Next, one constructs a surjective morphism X → B for which the fibre X P has local points in all but one or two completions of k, and ensures that X has k v -points for missing places v such the resulting adelic point of X projects to B(A k ) Br . Now, if the natural map Br(B) → Br(X ) is surjective we have found an adelic point in X (A k ) Br . But sinceX (k) ⊂ X P we have X (k) = ∅. With more work one can find examples such that X (A k )´e t,Br is non-empty, too.
In this paper we have nothing to say about the important open question whether the implication holds if X is a surface with finite geometric fundamental group, e.g. a K3 surface or an Enriques surface.
The paper is organised as follows. After some preparations in Sect. 2 we realise the aforementioned programme for threefolds in Sect. 3. Making it work for surfaces requires rather more effort. For this purpose in Sect. 4 we establish some Bashmakov-style properties of elliptic curves with a large Galois image on torsion points. These properties are used in the proof of our main result in the case of surfaces in Sect. 5. Some general observations on the Brauer-Manin set are collected in Sect. 6.
The authors are grateful to Nathan Jones and Chris Wuthrich for their helpful advice on Serre curves. We used sage [30] in our calculations with elliptic curves. This work started in April 2013 when the authors were guests of the Hausdorff Institut für Mathematik (Bonn) during the special programme "Arithmetic and geometry".

Brauer groups and torsors on quadric bundles
For the convenience of the reader we recall the following well known lemma. Lemma 2.1 Let k be a field of characteristic zero. Let X be a smooth projective quadric over k of dimension at least 1. Then the natural map Br(k) → Br(X ) is surjective. If dim(X ) ≥ 3, then this map is an isomorphism.
Proof Letk be an algebraic closure of k, and let k = Gal(k/k). For any smooth, projective and geometrically integral variety X over k there is a well known exact sequence where X = X × kk . If X is a quadric of dimension at least 1, then Pic(X ) is a permutation k -module and Br(X ) = 0 (since X is rational and the Brauer group is a birational invariant of smooth projective varieties). By Shapiro's lemma we have H 1 (k, Pic(X )) = 0, so the exact sequence implies the surjectivity of the map Br(k) → Br(X ). When dim(X ) ≥ 3, the map Pic(X ) → Pic(X ) is an isomorphism, because both groups are generated by the hyperplane section class, so in this case Br(k) → Br(X ) is an isomorphism.
In this paper a quadric bundle is a surjective flat morphism f : X → B of smooth, projective, geometrically integral varieties over a field k, the generic fibre of which is a smooth quadric of dimension at least 1, and all geometric fibres are reduced.
We denote by k(B) the function field of B, and by X k(B) the generic fibre of f : X → B. If dim(X k(B) ) = 1, then f : X → B is called a conic bundle.
The following proposition is essentially well known, at least when B = P 1 k , see [ (i) dim(X k(B) ) = 1 and there is a point P ∈ B of codimension 1 such that for each point Q = P of codimension 1 in B the fibre X Q contains a geometrically integral component of multiplicity 1; (ii) dim(X k(B) ) = 2 and for each point Q ∈ B of codimension 1 the fibre X Q contains a geometrically integral component of multiplicity 1; Proof (i) Let γ ∈ Br(k(B)) be the class of the conic X k(B) . Since γ is in the kernel of the natural map f * : Br(k(B)) → Br(X k(B) ), the assumption of (i) implies that the residue res Q (γ ) ∈ H 1 (k(Q), Q/Z) is zero if Q = P. Take any α ∈ Br(X ). By Lemma 2.1 the map f * : Br(k(B)) → Br(X k(B) ) is surjective, so the image of α in Br(X k(B) ) comes from some β ∈ Br(k(B)). Again, by the assumption of (i) we have res Q (β) = 0 if Q = P. Moreover, res P (β) = 0 or res P (β) = res P (γ ). By the purity theorem for the Brauer group [10, III, Thm. 6.1, p. 134] we conclude that β ∈ Br(B) or β − γ ∈ Br(B). Since f * (β) = f * (β − γ ) = α in Br(X k(B) ), and the natural map Br(X ) → Br(X k(B) ) is injective, we have proved (i).
The proof of (ii) uses Lemma 2.1 and the arguments from the proof of (i). In case (iii) it is well known that each fibre of f at a point of B of codimension 1 contains a geometrically integral component of multiplicity 1. Then (iii) follows from the last statement of Lemma 2.1.

Proposition 2.3
Let f : X → B be a quadric bundle over a field k of characteristic zero. Then any torsor X → X of a finite k-group scheme G is the inverse image under f of a torsor B → B of G.
Proof By our definition of quadric bundles, the morphism f is flat and all its geometric fibres are connected and reduced. The generic geometric fibre of f is simply connected. By [25,X,Cor. 2.4] this implies that each geometric fibre of such a fibration is simply connected. The result then follows from [25, IX, Cor. 6.8].

Example based on real deformation
Let k be a number field with a real place. We fix a real place v, so we can think of k as a subfield of k v = R.
Let C be a smooth, projective, geometrically integral curve over k such that C(k) consists of just one point, C(k) = {P}. By [20] such a curve exists for any number field k, and by [18, Thm. 1.1] we can take C to be an elliptic curve over k. Let ⊂ C(R) be an open interval containing P. Let f : C → P 1 k be a surjective morphism that is unramified at P. Choose a coordinate function t on A 1 k = P 1 k \ f (P) such that f is unramified above t = 0. We have f (P) = ∞. Take any a > 0 in k such that a is an interior point of the interval f ( ) and f is unramified above t = a.
Let w be a finite place of k. There exists a quadratic form Q(x 0 , x 1 , x 2 ) of rank 3 that represents zero in all completions of k other than k v and k w , but not in k v or k w . We can assume that Q is positive definite over k v = R. Choose n ∈ k with n > 0 in k v and −n Q (1, We glue Y 1 and Y 2 by identifying T = t −1 , X 3 = t x 3 , and X i = x i for i = 0, 1, 2. This produces a quadric bundle Y → P 1 k with exactly two degenerate fibres (over t = a and t = 0), each given by the quadratic form Q(x 0 , x 1 , x 2 ) of rank 3. Define X = Y × P 1 k C. This is a quadric bundle X → C with geometrically integral fibres.
For example, if k = Q, we can take k w = Q 2 and consider Y defined by Proof Since C(k) = {P} we have X (k) ⊂ X P . The fibre X P is the smooth quadric Q(x 0 , x 1 , x 2 ) + nx 2 3 = 0. This quadratic form is positive definite thus X P has no points in k v = R and so X (k) = ∅. By assumption X P has local points in all completions of k other than k v and k w . The condition −n Q(1, 0, 0) ∈ k * 2 w implies that X P contains k w -points, so X P has local points in all completions of k but one. Choose N u ∈ X P (k u ) for each place u = v. Consider a small real ε > 0 such that a − ε ∈ f ( ) and ε < a. Let M ∈ be such that f (M) = a−ε. Then the smooth real fibre X M is given by an indefinite quadratic form and so X M (k v ) = ∅. Choose any N v ∈ X M (k v ). We now have an adelic point (N u ), where we allow u = v.
We claim that (N u ) ∈ X (A k )´e t,Br . Let G be a finite k-group scheme. Proposition 2.3 implies that any torsor X / X of G comes from a torsor C /C of G, in the sense that X × C C → X and X → X are isomorphic as X -torsors with the structure group G. Let σ ∈ Z 1 (k, G) be a 1-cocycle defining the k-torsor which is the fibre of C → C at P. Twisting X / X and C /C by σ and replacing the group G by the twisted group G σ and changing notation, we can assume that C contains a k-point P that maps to P in C. The irreducible component C of C that contains P is a geometrically integral curve over k. Let X ⊂ X denote the inverse image of C in X . The fibres of the morphism X → C are geometrically integral, hence such are also the fibres of X → C and X → C . Thus X is a geometrically integral variety over k.
There are natural isomorphisms X P ∼ = X P ∼ = X P , so we can define N u ∈ X (k u ) as the point that maps to N u ∈ X (k u ) for each u = v. The map C → C is finite and étale.
By the definition of the étale Brauer-Manin obstruction, to prove that (N u ) is contained in X (A k )´e t,Br it suffices to show that (N u ) is orthogonal to Br(X ). For this it is enough to show that (N u ) is orthogonal to Br(X ). By Proposition 2.2 (ii) applied to X → C we know that the natural map Br(C ) → Br(X ) is surjective. Thus it is enough to show that the adèle on C such that its u-adic component is P when u = v and and its v-component is M , is orthogonal to Br(C ). The real point M is path-connected to P , so this adèle is in the connected component of the diagonal image of the k-point P in C (A k ). But the latter adèle is certainly in C (A k ) Br , and the proposition follows. that is, there exist field extensions k 1 , . . . , k r of k whose degrees have no common factor such that X (k i ) = ∅ for i = 1, . . . , r . Although [32,Thm. 1.3] requires the finiteness of the Shafarevich-Tate group of the Jacobian of C, Wittenberg pointed out that in the proof of his theorem this assumption is only used to ensure the existence of a suitable 0-cycle of degree 1 on C. In our case such a 0-cycle is directly provided by the k-point P, so the assumption on the Shafarevich-Tate group is not needed. For more details see Remark 5.7 below. there is a surjective morphism π : B → P 1 k with smooth and geometrically integral generic fibre such that π(S) = P 1 k .

Examples based on deformation along a rational curve defined over a completion of k
Proof According to [20] there is a smooth, projective and geometrically integral curve C over k with exactly one k-point, C(k) = {O}. Moreover, by [18,Thm 1.1] there is an elliptic curve over k with this property. Let f : Z → C be any conic bundle such that the fibre There is a closed embedding Z ⊂ P n k for some n ≥ 1. By the Bertini theorem [12, II.8.18, III.7.9], there exists a hyperplane H 1 ⊂ P n k such that Z ∩ H 1 is a smooth and geometrically integral curve. This implies that S is not a subset of H 1 .
Let d be the degree of Z in P n k . We can find a hyperplane LetP n k be the blowing-up of P n k at Thus B is a smooth, projective, geometrically integral surface with a unique k-point, equipped with a surjective morphism π : B → P 1 k with smooth and geometrically integral generic fibre. Moreover, S is contained in B, and π(S) = P 1 k .
Let P be the unique k-point of B, and let Q = π(P) ∈ P 1 k (k). Let K be the quadratic extension of k over which the components of S are defined. If w is a place of k that splits in K , then the k w -variety S × k k w is the union of two projective lines meeting at P. Let Proposition 3.4 Let w 1 and w 2 be places of k that split in K , w 1 = w 2 . Let Y → P 1 k be a conic bundle satisfying the following conditions: there exists a closed point R ∈ P 1 k , R = Q, such that the restriction Y \Y R → P 1 k \R is a smooth morphism, and the fibre of π : B → P 1 k at R is smooth; the fibre Y Q is a smooth conic that has k v -points for all completions of k except w 1 and Then for the smooth threefold X One easily checks that the projection map X → B is a quadric bundle as defined in this paper: both X and B are smooth and projective over k, the generic fibre is a geometrically integral conic, the morphism X → B is flat and all its geometric fibres are reduced. Let G be a finite k-group scheme. By Proposition 2.3 every torsor X → X of G is the pullback of a torsor B → B of G. After a twist by a k-torsor of G, as detailed in the proof of Proposition 3.1, we may assume that B has a k-point P over P. Let B be the irreducible component of B that contains P . Then B is geometrically integral. Let X be the inverse image of B under the map X → B . The k-variety X is geometrically integral, and X → B is a conic bundle.
To prove our claim it is enough to show that the adelic point (M v ) in X is orthogonal to Br(X ).
Letx ∈ R be ak-point. Since the singular loci of the morphisms Y → P 1 k and π : Since the restriction of the conic bundle X → B to the complement B \ D of the integral curve D is a smooth morphism, this conic bundle satisfies the condition of Proposition 2.2 (i). It follows that the induced map Br(B ) → Br(X ) is surjective. Thus it remains to prove that the image of ( This image is the adelic point such that for all we may replace (M v ) by the diagonal adèle (P ), which by the reciprocity law is orthogonal to Br(B ).
k be a quadric bundle all fibres of which are of dimension d ≥ 2 and contain a geometrically integral component of multiplicity one = ∅ (such quadrics exist in dimension 2 and higher, but not in dimension 1). Assume that L w has a non-empty intersection with the image of X (k w ) → B(k w ). In view of Proposition 2.2 (ii), (iii), an argument similar but shorter than the one above shows that X (k) = ∅ and X (A k )´e t,Br = ∅. In this example dim(X ) = 2 + d ≥ 4.
Example 3.6 One can construct a threefold over k = Q as in Proposition 3.4 as follows. Let E be the elliptic curve where O is the point at infinity. Let (r : s : u : v : t) be homogeneous coordinates in P 4 Q . The first of the equations Q given by the projection to (u : t) satisfies the conclusions of Lemma 3.3. The second equation of (1) defines a smooth quadric The fibre X P over P is the conic r 2 −5s 2 −17t 2 = 0 over Q, so for the places w 1 and w 2 one takes the primes 5 and 17. For p = 5, 17 choose a suitable point N p ∈ S(Q p ) with v = 1 and u = α p ∈ Q p such that α 2 p = −1. One can give a different proof of the non-emptyness of the set X (A Q )´e t,Br using the method of [11]. (By Proposition 6.1 (ii) below it does not matter which birational model is used for this). Let K = Q( √ −1). Consider the fibre X O of the composed morphism X → B → E over O ∈ E(Q). The singular surface X O is fibred into conics over the singular conic S; the inverse image of the singular point P ∈ S is a smooth conic X P ⊂ X O . Thus X O × Q K is the union of two geometrically irreducible components permuted by Gal(K /Q) that intersect transversally in X P , each of them isomorphic to Y K = Y × Q K . Since 5 and 17 split in K and the components of X O × Q K have K -points, we see that X O (A Q ) = ∅. We claim that Indeed, let G be a finite k-group scheme. The generic fibre of X → E is a geometrically integral, smooth and geometrically rational surface, so it is geometrically simply connected. One checks that the morphism X → E is flat and all its geometric fibres are connected and reduced.
, so to prove our claim it is enough to show that the natural map Br(Q) → Br(X O ) is an isomorphism. Let i : X P → X O be the closed embedding. Let ν : Y K → X O be the normalisation morphism and let C = ν −1 (X P ). Then C = X P × Q K is the intersection of the quadric Q K = Q × Q K given by the second equation in (1) with the plane u = 0. The morphism ν : C → X P is the natural projection X P × Q K → X P .
The exact sequence of étale sheaves on is similar to the exact sequence (2) in [11]. The normalisation morphism ν and the closed embedding i are finite morphisms, so ν * and i * are exact functors, hence on taking cohomology we obtain an exact sequence The discriminant of the quadratic form defining Q is not a square in K , hence Pic(Q K ) is generated by the class of the hyperplane section, and the natural map Br(K ) → Br(Q K ) is an isomorphism. By the birational invariance of the Brauer group we obtain that the natural map Br(K ) → Br(Y K ) is also an isomorphism. It is well known that Br(X P ) is the quotient of Br(Q) by the subgroup generated by the class of the conic X P , which is given by the symbol (5,17). This symbol remains non-zero in Br(K ), hence C(K ) = ∅ and so Pic(C) is also generated by the class of the hyperplane section. Since the composition of the embedding C → Y K with the birational morphism Y K → Q K is the natural embedding of C as a plane section of Q K , we see that the restriction map Pic(Y K ) → Pic(C) is surjective. Now using the fact that Br(C) is the quotient of Br(K ) by the subgroup generated by the symbol (5, 17) we easily deduce that Br(X O ) = Br(Q). To conclude, our example resembles that of Poonen [19] in that the fibres of X → E are birationally equivalent to intersections of two quadrics in P 4 . However, in our case the fibre above the unique Q-point is geometrically simply connected and satisfies Br(X O ) = Br(Q) and X O (A Q ) = ∅, so we see that the étale Brauer-Manin set of X is non-empty without studying X any further. Note that it is essential that the fibres of X → E have dimension at least 2. Indeed, each everywhere locally solvable geometrically connected and simply connected curve over a number field k has a k-point [

Elliptic curves with a large Galois image
Let k be a field of characteristic zero, with an algebraic closurek. For a field K such that k ⊂ K ⊂k we denote by K the Galois group Gal(k/K ). Let k cyc ⊂k be the cyclotomic extension of k, i.e. the abelian extension of k obtained by adjoining all roots of unity.
Let E be an elliptic curve over k, and let ρ : k → GL 2 (Ẑ) be the Galois representation in the torsion subgroup of E. The group SL 2 (Z/2) is isomorphic to the symmetric group S 3 . Let us define SL + 2 (Ẑ) as the kernel of the composition of the reduction modulo 2 map SL 2 (Ẑ) → GL 2 (Z/2) with the unique non-trivial homomorphism ε : GL 2 (Z/2) → {±1}. (The finite group GL 2 (Z/2) can be identified with the symmetric group S 3 , and then ε is the signature character.) In this section we prove the following theorem using methods that go back to Bashmakov [1, Ch. 5].

Theorem 4.1 Let E be an elliptic curve over a field k of characteristic zero such that SL
. Let K be a field such that k ⊂ K ⊂ k cyc , and let ϕ : E → E × k K be an isogeny of elliptic curves over K . Then for any point P ∈ E(K ) that cannot be written as P = m Q with m > 1 and Q ∈ E(K ), the scheme ϕ −1 (P) is integral.
In the assumption of the theorem, Lemma 4.7 (b) below shows that the isogeny ϕ : E → E can be identified with the multiplication by n map E → E for some integer n. The theorem then follows from Proposition 4.8.
Serre [24,Prop. 22] proved that for any elliptic curve E over Q the image ρ( Q ) is contained in a certain subgroup H which only depends on the discriminant of E. The group H has index 2 in GL 2 (Ẑ) and contains SL + 2 (Ẑ). N. Jones [14] showed that almost all elliptic curves over Q are Serre curves which means that ρ( Q ) = H . Theorem 4.1 thus applies to almost all elliptic curves E over Q.

Group cohomology
For an integer n > 1 we define SL + 2 (Z/n) as SL 2 (Z/n) if n is odd, and as the kernel of the following composite map if n is even: where the first arrow is reduction modulo 2, and the second arrow is the signature SL 2 (Z/2) S 3 → {±1}.

Proposition 4.2 Let n be a positive integer, and let G ⊂ GL 2 (Z/n) be a subgroup containing
Proof It is enough to prove that To fix ideas, assume i = 1. We have an inflation-restriction exact sequence which implies the lemma. Proof Let C = {±Id} ⊂ GL 2 (Z/ p r ). This is a central subgroup, so we have an inflationrestriction exact sequence The order of C is 2 but p is odd, so we have H 1 (C, (Z/ p r ) 2 ) = 0. We also have ((Z/ p r ) 2 ) C = 0, and the lemma follows.

Lemma 4.6 For a positive integer r the group H
Proof When r = 1 the group SL + 2 (Z/2 r ) has order 3, so the claim is obvious. Now suppose r ≥ 2. We denote the tautological SL

Isogenies of elliptic curves
The multiplication by n map on an elliptic curve E is denoted by [n] : E → E. Let ρ n : k → GL 2 (Z/n) be the Galois representation in the n-torsion subgroup E[n].

Proposition 4.8 Let E be an elliptic curve over a field k of characteristic zero such that
For any field K such that k ⊂ K ⊂ k cyc we have the following statements.
for any positive integer n. (ii) For any point P ∈ E(K ) that cannot be written as P = m Q for m > 1 with Q ∈ E(K ), and for any integer n > 0, the scheme [n] −1 (P) is integral.
There is an inflation-restriction exact sequence The restriction of κ(P) to H 1 (K n , E[n]) is a K -equivariant homomorphism ϕ : K n → E[n] (where K acts on K n by conjugations, and on E[n] in the usual way). Its image is thus a K -submodule of E[n]. By (i) and by Lemma 4.7 (i), we have ϕ( K n ) = E[m] for some m|n. The set of K -points of [n] −1 (P) × K K n with a natural action of K n can be identified, by a choice of the base point, with the set E[n] on which g ∈ K n acts by translation by ϕ(g).
Write n = 2 r s, where r ≥ 0 and s is odd. We first deal with the case r = 0, and then proceed by induction in r . Since P is not divisible in E(K ) by assumption, and E(K ) is torsion-free, the order of κ(P) ∈ H 1 (K , E[n]) is n.
If n is odd, by Proposition 4.2 the exact sequence (2) gives rise to an embedding of H 1 (K , E[n]) into Hom( K n , E[n]), so that ϕ has order n. It follows that ϕ( K n ) = E[n] in this case. This implies that [n] −1 (P) × K K n is irreducible, hence [n] −1 (P) is irreducible. Now suppose that n = 2n and the scheme [n ] −1 (P) is irreducible. The multiplication by 2 map defines a surjective morphism [n] −1 (P) → [n ] −1 (P) which is a torsor of E [2]. We know that K acts transitively on [n ] −1 (P)(K ) and we want to show that K acts transitively on [n] −1 (P) (K ). For this we must show that each K -fibre of [n] −1 (P) → [n ] −1 (P) is contained in one K -orbit. Recall that the K n -set [n] −1 (P)(K ) is identified with E[n] so that g ∈ K n acts as the translation by ϕ(g). The K -fibres of [n] −1 (P) → [n ] −1 (P) are the E [2]-orbits in E[n]. Therefore, it is enough to show that ϕ( K n ) contains E [2]. As the order of κ(P) is n, by Proposition 4.2 the exact sequence (2) shows that the order of ϕ is divisible by 2s. Thus ϕ( K n ) contains E[2s] and hence contains E [2]. This finishes the proof.

An elliptic curve
For an elliptic curve E over a field k of characteristic zero we denote by E c the quadratic twist of E by c ∈ k * .

Lemma 5.1 Let E be an elliptic curve over a field k of characteristic zero. For a quadratic extension K
Proof Let σ be the non-zero element of Gal(K/k). We have E(k) E(K ) σ . The choice of a square root of d in K defines an isomorphism E d × k K E × k K . This gives an

Proposition 5.2
There exist the following data: a real quadratic field k = Q( √ c), where c is a square-free positive integer not congruent to 1 modulo 8; a totally real biquadratic field K = Q( √ c, √ d), where d is a square-free positive integer; an elliptic curve E over Q of discriminant < 0, such that SL + 2 (Ẑ) ⊂ ρ( k ), E(k) ={0}, and E(K ) is torsion-free of positive rank.
Proof Let E be the curve y 2 + y = x 3 + x 2 − 12x − 21 of conductor 67 and discriminant = −67, and take c = 10, d = 2. Using sage we check that E(Q) = E 10 (Q) = 0. By Lemma 5.1 we have E(k) = 0. Using sage we check that E 2 (Q) E 5 (Q) Z, so by Lemma 5.1 we conclude that E(K ) is torsion-free of positive rank. We claim that E is a Serre curve, which means that ρ( Q ) = H , where H is a subgroup of GL 2 (Ẑ) of index 2 containing SL + 2 (Ẑ). By a result of N. Jones [14,Lemma 5], for this it is enough to show that E satisfies the following conditions: ρ ( Q ) = GL 2 (Z/ ) for all primes (this is checked using sage), ρ 8 ( Q ) = GL 2 (Z/8) (this is checked using [7]), and ρ 9 ( Q ) = GL 2 (Z/9) (this is checked using [8]).

By Proposition 4.8 (i), SL
. This finishes the proof.

Remark 5.3
For c and k = Q( √ c) as above, the conic x 2 + y 2 + z 2 = 0 has points in all non-Archimedean completions of k, but no points in the two real completions of k. Indeed, this conic has Q p -points for all odd primes p. Since 2 is ramified or inert in k, this conic also has points in the completion of k at the unique prime over 2.

A conic bundle over the elliptic curve
Let k, K and E be as in Proposition 5.2. The elliptic curve E can be given by its short Weierstraß equation is the point at infinity. We denote by π : E → P 1 k the projection sending (x, y) to x. Let P ∈ E(K ) be a point not divisible in E(K ). Let σ ∈ Gal(K /k) be the generator. Since P +σ (P) ∈ E(k) = 0 we obtain that π(P) is a point of A 1 k (k) = k, say π(P) = a ∈ k. The K -point P gives rise to a solution of y 2 = r (a) in K , hence r (a) ∈ k is totally positive. We have r (a) = 0 since E(K ) is torsion-free.
Let b ∈ k, b = a. We define a central simple algebra A over k(P 1 k ) = k(x) as a tensor product of quaternion algebras The algebra A is unramified outside of the points x = a and x = b. At each of these points the residue of A is given by the class of r (b) in k * /k * 2 .
Proof The associated Albert form contains the subform = r (b), 1, 1, 1 . By Remark 5.3 the form 1, 1, 1 is isotropic over all finite completions of k. Since r (b) is totally negative, is isotropic over both real completions of k. By the Hasse-Minkowski theorem the quadratic form is isotropic over k. An application of Proposition 5.4 concludes the proof.
We are now ready to state one of the main results of this paper. (For an explanation why we do not consider the case k = Q see Proposition 6.5 below.) Theorem 5.6 There exist a real quadratic field k, an elliptic curve E and a smooth, projective and geometrically integral surface X over k with a surjective morphism f : X → E satisfying the following properties: (i) the fibres of f : X → E are conics; (ii) there exists a closed point P ∈ E such that the field k(P) is a totally real biquadratic extension of Q and the restriction X \ f −1 (P) → E\P is a smooth morphism; (iii) X (A k )´e t,Br = ∅ and X (k) = ∅.
Proof We keep the above notation: k = Q( √ 10), E is the curve of Proposition 5.2 given by its short Weierstraß equation y 2 = r (x), viewed as a curve over k, and K = k(P) = Q( √ 10, √ 2). We fix b ∈ k such that r (b) is totally negative, and define a central simple algebra A by (3). By Lemma 5.5, A is similar to a quaternion algebra B over k(P 1 k ). Let f : X → E be a relatively minimal conic bundle such that the generic fibre X k(E) is the conic over k(E) defined by the quaternion algebra B ⊗ k(P 1 ) k(E). (We call a conic bundle X → E relatively minimal if for every conic bundle Y → E any birational morphism X → Y compatible with the projections to E, is an isomorphism. See [16,Thm. 1.6] for the well known description of the fibres of X → E.) The closed point P = π −1 (a) Spec K is a solution to y 2 = r (a). The residue of B ⊗ k(P 1 ) k(E) at P is the class of r (b) in K × /K ×2 , which is non-trivial since r (b) is totally negative. We have π −1 (b) = Spec k( √ r (b)), and the residue at this closed point is the class of r (b) in k( √ r (b)) * /k( √ r (b)) * 2 which is trivial. Thus f : X → E is smooth away from P. This gives (i) and (ii).
If we go over to one of the two real completions k v of k, the point P breaks up into two real points P 1 and P 2 . The residue at each of these points is an element of R * /R * 2 given by the image of r (b), and that element is totally negative. Thus the algebra A is ramified at both P 1 and P 2 . The fibre of X → E above each point P i ∈ E(k v ) is thus a singular conic, hence has a k v -point. In particular, X (k v ) = ∅ for each of the real completions of k. The condition < 0 implies that E(k v ) is connected, thus any point in the image of X (k v ) → E(k v ) is path connected to the point 0, which is the unique point of E(k). (In fact, the image of X (R) in E(R) is the closed interval between P 1 and P 2 which does not contain 0.) Let M v be any point of X (k v ), and let I v ⊂ E(k v ) S 1 be a real interval linking f (M v ) and 0.
The value of A at ∞∈P 1 k , hence also at 0 ∈ E(k), is (−1, −1), hence the fibre X 0 = f −1 (0) is the conic x 2 + y 2 + z 2 = 0. By Remark 5.3, X 0 has points in all completions of k except the two real completions, hence X 0 (k) = ∅. Since E(k) = {0} it follows that X (A k ) = ∅, but X (k) = ∅. For each finite place v of k choose M v ∈ X 0 (k v ).
We now prove that (M v ) ∈ X (A k )´e t,Br . By Proposition 2.3 every torsor X → X of a finite k-group scheme G is the pullback X = X × E E → X of a torsor E → E of G. By twisting E and X with a k-torsor of G (and replacing G with the corresponding inner form) we can assume that E has a k-point 0 over 0 ∈ E(k). The connected component C of E containing this k-point is a smooth, projective, geometrically integral curve. The k-morphism ϕ : C → E is finite and étale, hence C has genus 1. Choosing 0 for the origin of the group law on C we make ϕ : C → E into an isogeny of elliptic curves and write 0 = 0 .
Let Y = X × E C. Then Y is a smooth, projective, geometrically integral surface over k which is an irreducible component of X . The morphism g : Y → C is a conic bundle. By Proposition 5.2 (3) we have SL + 2 (Ẑ) ⊂ ρ( k ). Since the point P ∈ E(K ) is not divisible, we can apply Theorem 4.1 to the elliptic curve E and the isogeny ϕ : C → E. It follows that the inverse image Q = ϕ −1 (P) of the closed point P of the k-curve E is a closed point of the k-curve C. We see that g : Y → C is a conic bundle such that Y \Y Q → C\Q is a smooth morphism. By Proposition 2.2 (i), the induced map g * : Br(C) → Br(Y ) is surjective.
For each finite place v of k let M v be the k v -point in the fibre Y 0 over 0 ∈ C(k) that projects to M v ∈ X 0 (k v ). Now let v be a real place of k. By Lemma 4.7 (ii) the isogeny ϕ : C → E is identified with [n] : E → E for some n. Since < 0, the induced map is a disjoint union of copies of I v exactly one of which contains 0. Let us call this interval I v . One of its ends is 0 and the other end is a point R ∈ C(k v ) such that ϕ(R) = f (M v ). Hence the real fibre Y R is naturally isomorphic to the fibre of f : This is a lifting of the adelic point (M v ), and therefore (M v ) ∈ X (A k )´e t,Br .
(N v ) ∈ Z (A k )´e t,Br . By the covariant functoriality of the étale Brauer-Manin set we have (iii) Let U ⊂ X and V ⊂ Y be non-empty Zariski open sets such that there is an isomorphism h : V− →U . Under the assumption of (ii), the group Br(Y )/Br 0 (Y ) Br(X )/Br 0 (X ) is finite. Let β 1 , . . . , β n ∈ Br(Y ) be coset representatives. There exists a finite set of places of k such that each β i vanishes on Y (k v ) for each v / ∈ . Thus every adèle (M v ) ∈ Y (A k ) Br can be approximated by an adèle Since the morphism f : X → A induces an isomorphism f * : Br(A)− →Br(X ) and f (M p ) = 0 for all p, we see that (M p ) ∈ X (A Q ) Br . However, the connected component of (M p ) in X (A Q ) contains no Q-points. Indeed, a Q-point of X is either in E or in F. In the first case it cannot approximate M q in the q-adic topology, and in the second case it cannot approximate M p in the p-adic topology where p is any prime different from q. It is easy to see that (M p ) ∈ X (A Q )´e t,Br , so the previous discussion applies to the set of connected components of the étale Brauer-Manin set as well.

Cases where the Brauer-Manin obstruction suffices
The following result shows that for the conclusion of Theorem 5.6 to hold, the conic bundle f : X → E must contain degenerate fibres. Proposition 6.3 Let E be an elliptic curve over a number field k with a finite Shafarevich-Tate group. Let f : X → E be a Severi-Brauer scheme over E. Then X (A k ) Br = ∅ implies X (k) = ∅. Moreover, X (k) is dense in X (A k ) Br • .
Proof Since f : X → E is a projective morphism with smooth geometrically integral fibres, there exists a finite set of places such that E(k v ) = f (X (k v )) for v / ∈ . We may assume that contains the archimedean places of k. At an arbitrary place v the set f (X (k v )) is open and closed in E(k v ). Let (M v ) ∈ X (A k ) Br . By functoriality we then have ( f (M v )) ∈ E(A k ) Br where E(k v ) • = E(k v ) if v is a finite place of k, and E(k v ) • = π 0 (E(k v )) if v is an archimedean place. By a theorem of Serre, the image of E(k)⊗Ẑ in that product coincides with the topological closure of E(k), see [22,23,31]. Approximating at the places of , we find a k-point M ∈ E(k) such that the fibre X M = f −1 (M) is a Severi-Brauer variety with points in all k v for v ∈ , hence also for all places v. Since X M is a Severi-Brauer variety over k, it contains a k-point, hence X (k) = ∅. For the last statement of the theorem we include into the places where we want to approximate. If k v R, each connected component of X (k v ) maps surjectively onto a connected component of E(k v ). The Severi-Brauer varieties satisfy the Hasse principle and weak approximation, so an application of the implicit function theorem finishes the proof.

Remark 6.4
The same argument works more generally for a projective morphism f : X → E such that each fibre contains a geometrically integral component of multiplicity 1, provided that the smooth k-fibres satisfy the Hasse principle. For the last statement to hold, the smooth k-fibres also need to satisfy weak approximation.
The following proposition is a complement to Theorem 5.6 which explains why a similar counterexample cannot be constructed over Q.

Proposition 6.5 Let E be an elliptic curve over a number field k such that both E(k) and
the Shafarevich-Tate group of E are finite. Let f : X → E be a conic bundle. Suppose that there exists a real place v 0 of k such that for each real place v = v 0 no singular fibre of f : X → E is over a k v -point of E. Then X (A k ) Br = ∅ implies X (k) = ∅.
Proof If a k-fibre of f is not smooth, then this fibre contains a k-point. We may thus assume that the fibres above E(k) are smooth. Let (M v ) ∈ X (A k ) Br . Then ( f (M v )) ∈ E(A k ) Br . Set N v = f (M v ) for each place v. The finiteness of the Shafarevich-Tate group of E implies the exactness of the Cassels-Tate dual sequence (4). Hence there exists N ∈ E(k) such that N = N v for each finite place v and such that N lies in the same connected component as N v for v archimedean. The fibre X N is a smooth conic with points in all finite completions of k. For an archimedean place v = v 0 , the map X (k v ) → E(k v ) sends each connected component of X (k v ) onto a connected component of E(k v ). Since N and N v are in the same connected component of E(k v ), this implies X N (k v ) = ∅. Thus the conic X N has points in all completions of k except possibly k v 0 . By the reciprocity law it has points in all completions of k and hence in k. Remark 6.6 If E(k) is finite we cannot in general expect X (k) to be dense in X (A k ) Br • or even in X (A k )´e t,Br • . Indeed, if the fibre X N over some k-point N of E is an irreducible singular conic, then the singular point P of X N is the unique k-point of X N . Let (M v ) be any adèle in X N (A k ). Note that if v splits in the quadratic extension of k over which the components of X N are defined, then X N × k k v is a union of two projective lines meeting at P. Using the fact that Br(P 1 k v ) = Br(k v ) we see that (M v ) ∈ X (A k ) Br . Furthermore, we have (M v ) ∈ X (A k )´e t,Br , cf. [11,Remark 2.4]. On the other hand, (M v ) is not in the closure of X (k) in X (A k ) • provided M v = P for at least one finite place v of k.
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