Châtelet Surfaces and Non-Invariance of the Brauer-Manin Obstruction for 3-Folds3

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A 6= = 6= ( Let K S K K. K ∅ ∅ ∅ ∅ S ) ) . . , . . 2 HAN WU not explained by the Brauer-Manin obstruction, was constructed firstly by Skoroboga- tov [Sko99], subsequently by Poonen [Poo10], Harpaz and Skorobogatov [HS14], Colliot- Thélène, Pál and Skorobogatov [CTPS16] and so on. We say that X satisfies weak approximation with Brauer-Manin obstruction (respectively with Brauer-Manin obstruction off S) if Br S Br X(K) is dense in X(AK) (respectively in pr (X(AK ) )). For an elliptic curve defined over Q, if its analytic rank equals one, then this elliptic curve satisfies weak approximation with Brauer-Manin obstruction, cf. [Wan96]. For an abelian variety defined over K, if its Tate-Shafarevich group is finite, then this abelian variety satisfies weak approximation with Brauer-Manin obstruction off K , cf. [Sko01, Proposition 6.2.4]. For any smooth, proper and rationally connected variety∞ X defined over an number field, it is conjectured by J.-L. Colliot-Thélène that X satisfies weak approximation with Brauer-Manin obstruction. The Colliot-Thélène’s conjecture holds for Châtelet surfaces, cf. [CTSSD87a, CTSSD87b]. For any smooth, projective, and geometrically connected curve C defined over an number field K, it is conjectured by Stoll [Sto07] that C satisfies weak approximation with Brauer-Manin obstruction off K : see Conjecture 4.0.1 for more details. Before the Stoll’s conjecture, if this curve C is∞ a counterexample to the Hasse principle, Skorobogatov [Sko01, Chapter 6.3] asked a weaker open question: is the failure of the Hasse principle of C explained by the Brauer-Manin obstruction?

1.2. Questions. Given a nontrivial extension of number fields L/K, and a finite subset S Ω , let S Ω be the subset of all places above S. Let X be a smooth, projective, ⊂ K L ⊂ L and geometrically connected variety defined over K. Let XL = X Spec K Spec L be the base change of X by L. In this paper, we consider the following questions.× Question 1.2.1. If the variety X has a K-rational point, and satisfies weak approximation with Brauer-Manin obstruction off S, must XL also satisfy weak approximation with Brauer-Manin obstruction off SL?

Question 1.2.2. Assume that the varieties X and XL are counterexamples to the Hasse principle. If the failure of the Hasse principle of X is explained by the Brauer-Manin obstruction, must the failure of the Hasse principle of XL also be explained by the Brauer- Manin obstruction?

1.3. Main results for Châtelet surfaces. A Châtelet surface defined over Q, which is a counterexample to the Hasse principle, was constructed by Iskovskikh [Isk71]. Poonen [Poo09] generalized it to any given number field. For any number field K, Liang [Lia18] constructed a Châtelet surface defined over K, which has a K-rational point and does not satisfy weak approximation off K . By using weak approximation and strong approximation off all 2-adic places for A1∞(cf. Lemma 2.0.1) to choose elements in K, and using Čebotarev’s density theorem (cf. Theorem 2.0.2) to add some splitting conditions, we construct three kinds of Châtelet surfaces to generalize their arguments. Proposition 1.3.1. For any extension of number fields L/K, and any finite subset S Ω all complex and 2-adic places splitting completely in L, there exist Châtelet surfaces⊂ K \{ } V1, V2, V3 defined over K, which have the following properties.

S SL ′ The subset V1(AK ) V1(AL ) is nonempty, but V1(Kv) = V1(Lv ) = for all • v S and all v′ S ⊂, cf. Proposition 3.1.1. ∅ ∈ ∈ L The Brauer group Br(V2)/Br(K) ∼= Br(V2L)/Br(L) ∼= Z/2Z, is generated by an • element A Br(V ). The subset V (K) V (L) is nonempty. ∈ 2 2 ⊂ 2 For any v S, there exist Pv and Qv in V2(Kv) such that the local invariants inv (A(P ))∈ = 0 and inv (A(Q )) = 1/2. For any other v / S, and any P v v v v ∈ v ∈ V2(Kv), the local invariant invv(A(Pv))=0. For any v′ S , there exist P ′ and Q ′ in V (L ′ ) such that the local invariants ∈ L v v 2 v inv ′ (A(P ′ )) = 0 and inv ′ (A(Q ′ )) = 1/2. For any other v′ / S , and any v v v v ∈ L P ′ V (L ′ ), the local invariant inv ′ (A(P ′ ))=0, cf. Proposition 3.2.1. v ∈ 2 v v v The Brauer group Br(V )/Br(K) = Br(V )/Br(L) = Z/2Z, is generated by an • 3 ∼ 3L ∼ element A Br(V3). The subset V3(AK ) V3(AL) is nonempty. For any v ∈ Ω , and any P V (K )⊂, the local invariant inv (A(P )) = 0 if ∈ K v ∈ 3 v v v CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS3

v / S; the local invariant inv (A(P ))=1/2 if v S. ∈ v v ∈ For any v′ Ω , and any P ′ V (L ′ ), the local invariant inv ′ (A(P ′ )) = 0 if ∈ L v ∈ 3 v v v v′ / S ; the local invariant inv ′ (A(P ′ ))=1/2 if v′ S , cf. Proposition 3.3.1. ∈ L v v ∈ L Combining our construction method with the global reciprocity law, we have the following results for Châtelet surfaces. Corollary 1.3.2 (Corollary 3.2.4). For any extension of number fields L/K, and any finite nonempty subset S ΩK all complex and 2-adic places splitting completely in L, there exists a Châtelet surface⊂ V\{defined over K such that V (K} ) = . For any subfield L′ L ′ 6 ∅ ⊂ over K, the Brauer group Br(V )/Br(K) ∼= Br(VL′ )/Br(L ) ∼= Z/2Z. And the surface VL′ has the following properties.

′ ′ For any finite subset T ΩL′ such that T SL′ = , the surface VL′ satisfies • weak approximation off T⊂′. ∩ 6 ∅ ′ ′ For any finite subset T ΩL′ such that T SL′ = , the surface VL′ does not • ⊂ ′ ∩ ∅ satisfy weak approximation off T . In particular, the surface VL′ does not satisfy weak approximation. Corollary 1.3.3 (Corollary 3.3.3). For any extension of number fields L/K, there exists ′ a Châtelet surface V defined over K such that V (AK ) = . For any subfield L L over ′ 6 ∅ ⊂ K, the Brauer group Br(V )/Br(K) ∼= Br(VL′ )/Br(L ) ∼= Z/2Z. And the surface VL′ has the following properties.

′ If the degree [L : K] is odd, then the surface VL′ is a counterexample to the Hasse • principle. In particular, the surface V is a counterexample to the Hasse principle. ′ If the degree [L : K] is even, then the surface VL′ satisﬁes weak approximation. • In particular, in this case, the set V (L′) = . 6 ∅ 1.4. Main results for Châtelet surface bundles over curves. We will apply our results for Châtelet surfaces to construct Châtelet surface bundles over curves to give negative answers to Questions 1.2.

1.4.1. A negative answer to Question 1.2.1. For any quadratic extension of number fields L/K, and assuming the Stoll’s conjecture, a Châtelet surface bundle over a curve was constructed by Liang[Lia18] to give a negative answer to Question 1.2.1. Also an unconditional example with explicit equations was given for L = Q(√5) and K = Q in loc. cit. His method only works for quadratic extensions. In this paper, we generalize it to any nontrivial extension of number fields. For any nontrivial extension of number fields L/K, assuming the Stoll’s conjecture, we have the following theorem to give a negative answer to Question 1.2.1. Theorem 1.4.1.1 (Theorem 6.2.1). For any nontrivial extension of number fields L/K, and any finite subset T ΩL, assuming the Stoll’s conjecture, there exists a Châtelet surface bundle over a curve:⊂ X C defined over K such that → X has a K-rational point, and satisfies weak approximation with Brauer-Manin • obstruction off K , X does not satisfy∞ weak approximation with Brauer-Manin obstruction off T. • L For K = Q and L = Q(√3), based on the method given in Theorem 6.2.1, we give an explicit unconditional example in Subsection 7.1.

1.4.2. Negative answers to Question 1.2.2. To the best knowledge of the author, Question 1.2.2 has not yet been seriously discussed in the literature. For any number ﬁeld K, and any nontrivial ﬁeld extension L of odd degree over K, assuming the Stoll’s conjecture, we have the following theorem to give a negative answer to Question 1.2.2. 4 HAN WU

Theorem 1.4.2.1 (Theorem 6.3.1). For any number field K, and any nontrivial field extension L of odd degree over K, assuming the Stoll’s conjecture, there exists a Châtelet surface bundle over a curve: X C defined over K such that → X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction, XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

−1 Let ζ7 be a primitive 7-th root of unity. For K = Q and L = Q(ζ7 + ζ7 ), based on the method given in Theorem 6.3.1, we give an explicit unconditional example in Subsection 7.2. The 3-fold X is a smooth compactification of the following 3-dimensional affine subvariety given by equations y2 377z2 = 14(x4 89726)y′2 + (x2 878755181)(5x2 4393775906) − − − − y′2 = x′3 343x′ 2401 ( − − in (x,y,z,x′,y′) A5. ∈ For any number field K having a real place, and any nontrivial field extension L/K having a real place, assuming the Stoll’s conjecture, we have the following theorem to give a negative answer to Question 1.2.2. Theorem 1.4.2.2 (Theorem 6.3.2). For any number field K having a real place, and any nontrivial field extension L/K having a real place, assuming the Stoll’s conjecture, there exists a Châtelet surface bundle over a curve: X C defined over K such that → X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction, XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

For K = Q and L = Q(√3), based on the method given in Theorem 6.3.2, we give an explicit unconditional example in Subsection 7.3. The 3-fold X is a smooth compactification of the following 3-dimensional affine subvariety given by equations y2 + 23z2 = 5(x4 + 805)(x′ 4)2 5(x4 + 115) − − y′2 = x′3 16 ( − in (x,y,z,x′,y′) A5. ∈ Exceptions 1.4.2.3 (Subsection 7.4). For Question 1.2.2, besides cases of Theorem 6.3.1 and Theorem 6.3.2, there are some exceptions. When the degree [L : K] is even and L has no real place, we can give some unconditional examples, case by case, to give negative answers to Question 1.2.2, although we do not have a uniform way to construct them. In Subsection 7.4, we give an example to explain how it works for the case that K = Q and L = Q(i). The 3-fold X, is a smooth compactification of the following 3-dimensional affine subvariety given by equations y2 + 15z2 = (x4 10x2 + 15)(y′2 + 32)/8 (5x4 39x2 + 75)y′/2 − − − y′2 = x′3 16 ( − in (x,y,z,x′,y′) A5. ∈ 1.4.3. Main ideas behind our constructions in the proof of theorems. Given a nontrivial extension of number fields L/K, we start with a curve C such that C(K) and C(L) are finite, nonempty and C(K) = C(L). Using our results for Châtelet surfaces and a construction method from Poonen6 [Poo10], we construct a Châtelet surface bundle over this curve: β : X C such that the fiber of each C(K) point is isomorphic to V , and that the → ∞ fiber of each C(L) C(K) point is isomorphic to V0. Assuming the Stoll’s conjecture, by a main proposition\ from [Poo10, Proposition 5.4] and the functoriality of Brauer-Manin pairing, the Brauer-Manin sets of X, roughly speaking, is the union of adelic points sets CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS5 of rational fibers. Using some fibration arguments, the arithmetic properties of V∞ and V0 will determined the arithmetic properties of X. For Theorem 6.2.1 and 6.3.1, these ideas will be enough. For Theorems 6.3.2, we choose the curve C with an additional connected condition at one real place, and make full use of this real place information.

2. Notation and preliminaries

Given a number field K, let be the ring of its integers, and let Ω be the set of all its OK K nontrivial places. Let K ΩK be the subset of all archimedean places, and let 2K ΩK be the subset of all 2-adic∞ ⊂ places. Let r be the subset of all real places, and⊂ let ∞K ⊂ ∞K c be the subset of all complex places. Let Ωf =Ω be the set of all finite ∞K ⊂ ∞K K K \∞K places of K. Let Kv be the completion of K at v ΩK . For v K , let τv : K ֒ Kv be ∈f ∈ ∞ → the embedding of K into its completion. For v Ω , let be its valuation ring, and let ∈ K OKv Fv be its residue field. Let S ΩK be a finite subset, and let S = v∈Ωf \S(K Kv ) be ⊂ O K ∩O S the ring of S-integers. Let AK , AK be the ring of adèles and adèlesT without S components of K. A strong approximation theorem [CF67, Chapter II §15] states that K is dense in S AK for any nonempty S. In this paper, we only use the following special case:

2K Lemma 2.0.1. Let K a number ﬁeld. The set K is dense in AK .

In this paper, we always assume that a field L is a finite extension of K. Let SL ΩL be the subset of all places above S. ⊂ It is not difficult to generalize [Neu99, Theorem 13.4] to the following version of Čebotarev’s density theorem. Theorem 2.0.2 (Čebotarev). Let L/K be an extension of number fields. Then the set of places of K splitting completely in L, has positive density.

× 2.1. Hilbert symbol. We use Hilbert symbol (a,b)v 1 , for a,b Kv and v ΩK . By deﬁnition, (a,b) = 1 if and only if x2 ax2 bx2∈= {± 0 has} a K -solution∈ in P2∈with v 0 − 1 − 2 v homogeneous coordinates (x0 : x1 : x2), which equivalently means that the curve deﬁned 2 2 2 2 1 over Kv by the equation x0 ax1 bx2 =0 in P , is isomorphic to P . The Hilbert symbol − − × ×2 gives a symmetric bilinear form on Kv /Kv with value in Z/2Z, cf. [Ser79, Chapter XIV, Proposition 7]. And this bilinear form is nondegenerate, cf. [Ser79, Chapter XIV, Corollary 7].

2.2. Preparation lemmas. We state the following lemmas for later use. Lemma 2.2.1. Let K be a number ﬁeld, and let v be an odd place of K. Let a,b K× ∈ v such that v(a), v(b) are even. Then (a,b)v =1.

−v(a) −v(b) Proof. Choose a prime element πv Kv. Let a1 = aπv and b1 = bπv . Since the ∈ −v(a) −v(b) ×2 valuations v(a) and v(b) are even, the elements πv and πv are in Kv . So (a,b)v = × (a1,b1)v and a1,b1 Kv . By Chevalley-Warning theorem (cf. [Ser73, Chapter I §2, ∈ O 2 2 ¯ 2 Corollary 2]), the equation x0 a¯1x1 b1x2 = 0 has a nontrivial solution in Fv. For v is odd, by Hensel’s lemma, this solution− − can be lifted to a nontrivial solution in . Hence OKv (a,b)v = (a1,b1)v =1.

Lemma 2.2.2. Let K be a number ﬁeld, and let v be an odd place of K. Let a,b,c K× ∈ v such that v(b) < v(c). Then (a,b + c)v = (a,b)v.

−1 −1 ×2 Proof. For v(b) < v(c), we have v(b c) > 0. By Hensel’s lemma, we have 1+ b c Kv . −1 ∈ So (a,b + c)v = (a,b(1 + b c))v = (a,b)v.

Lemma 2.2.3. Let K be a number ﬁeld, and let v Ω . Then K×2 is an open subgroup ∈ K v of K×. If v Ωf , then × is also an open subgroup of K×. So, they are nonempty open v ∈ K OKv v subset of Kv. 6 HAN WU

f Proof. It is obvious for v K . Consider v ΩK . Let p be the prime number such that ∈ ∞ ∈ ×2 × 3 v p in K. Then by Hensel’s lemma, the group K contains the set 1+p K , which | v ∩OKv O v is an open subgroup of K×. Hence K×2 and × are open subgroups of K×. v v OKv v f Lemma 2.2.4. Let K be a number ﬁeld, and let v ΩK . For any n Z, the set x K v(x)= n is a nonempty open subset of K . ∈ ∈ { ∈ v| } v Proof. By Lemma 2.2.3, the set × is an open subgroup of × Choose a prime element Kv Kv . ×O n × πv Kv. Then the set x K v(x) = n = π , so it is a nonempty open subset of ∈ { ∈ v | } v OKv Kv. f × Lemma 2.2.5. Let K be a number ﬁeld, and let v ΩK . For any a Kv , the sets × × ∈ × ∈ x K (a, x)v = 1 , x K (a, x)v = 1 K and x (a, x)v = 1 are { ∈ v | } { ∈ v | } ∩ O v { ∈ OKv | } nonempty open subsets of Kv.

Proof. For the unit 1 belongs to these sets, they are nonempty. By Lemma 2.2.3, the sets ×2 × × Kv and Kv are nonempty open subsets of Kv. The set x Kv (a, x)v =1 is a union O ×2 × { ∈ | } of cosets of Kv in the group Kv . So the sets are open in Kv. f × Lemma 2.2.6. Let K be a number ﬁeld, and let v ΩK . For any a Kv , the sets × × ∈ ∈ x Kv (a, x)v = 1 and x Kv (a, x)v = 1 Kv are open subsets of Kv. {Furthermore,∈ | if a / K−×2}, then{ they∈ are nonempty.| − } ∩ O ∈ v × ×2 Proof. If the set x Kv (a, x)v = 1 = , then it is a union of cosets of Kv in the × { ∈ | − } 6 ∅ group Kv . By Lemma 2.2.3, it is an open subset of Kv. For Kv is open in Kv, the sets are × O open subsets of Kv . Nonemptiness is from the nondegeneracy of the bilinear form given by the Hilbert symbol, and from multiplying a square element in Kv to denominate an × O element in Kv . f × Lemma 2.2.7. Let K be a number ﬁeld, and let v ΩK . For any a Kv with v(a) odd, × ∈ ∈ the set x (a, x)v = 1 is a nonempty open subset of Kv. { ∈ OKv | − }

Proof. By Lemmas 2.2.3 and 2.2.6, the set is open in Kv. We need to show that it is 2 nonempty. For a / Kv , by the nondegeneracy of the bilinear form given by the Hilbert ∈ × ′ symbol, there exists an element b Kv such that (a,b)v = 1. If v(b) is odd, let b = ab. ′ ∈ − ′ − Then (a,b )v = (a, ab)v = (a, a)v(a,b)v = 1. Replacing b by b if necessary, we can − − − −v(b) ×2 assume that v(b) is even. Choose a prime element πv Kv. Then πv Kv , so the −v(b) ∈ ∈ element bπv is in this set. Lemma 2.2.8. Let L/K be an extension of number ﬁelds, and let v Ω c . We ∈ K \∞K assume that v′ Ω splits over v, i.e. L ′ = K . Given an element a K, if v is a ﬁnite ∈ L v v ∈ place, we assume that v(a) is odd; if v is an archimedean place, we assume τv(a) < 0. Then a / L×2. ∈

Proof. The condition that v(a) is odd for the ﬁnite place v, or that τv(a) < 0 for the ×2 ×2 ×2 archimedean place v, implies that a / K . For L ′ = K , we have a / L ′ , so a / L . ∈ v v v ∈ v ∈ 3. Main results for Châtelet surfaces

In this section, we will construct three kinds of Châtelet surfaces. Each kind in each following subsection has the arithmetic properties mentioned in Subsection 1.3. Let K be a number field. Châtelet surfaces are smooth projective models of conic bundle surfaces defined by the equation (1) y2 az2 = P (x) − in K[x,y,z] such that a K×, and that P (x) is a separable degree-4 polynomial in K[x]. ∈ 0 3 Given an equation (1), let V be the affine surface in AK defined by this equation. The natural smooth compactification V of V 0 given in [Sko01, Section 7.1] is called the Châtelet surface given by this equation, cf. [Poo09, Section 5]. CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS7

Remark 3.0.1. For any local ﬁeld K , if a K×2, then V is birational equivalent to P2 v ∈ v over Kv. By the implicit function theorem, there exists a Kv-point on V.

Remark 3.0.2. For any local field Kv, by smoothness of V, the implicit function theorem 0 implies that the nonemptiness of V (Kv) is equivalent to the nonemptiness of V (Kv), and that V 0(K ) is open dense in V (K ). Given an element A Br(V ), the evaluation of v v ∈ A on V (Kv) is locally constant. By the properness of V, the space V (Kv) is compact. So the set of all possible values of the evaluation of A on V (Kv) is finite. Indeed, by [Sko01, Proposition 7.1.2], there only exist two possible values. It is determined by the 0 0 evaluation of A on V (Kv). In particular, if the evaluation of A on V (Kv) is constant, then it is constant on V (Kv). Remark 3.0.3. In [CTPS16, Proposition 6.1], it is shown that whether the Brauer-Manin set of a smooth, projective, and geometrically connected variety is empty, is determined by its birational equivalence class. Here birational equivalence means birational equivalence between smooth, projective, and geometrically connected varieties. It is proved in the papers [CTSSD87a, Theorem B; CTSSD87b] (also explained in the book [Sko01, Theorem 7.2.1]), that the Brauer-Manin obstruction to the Hasse principle and weak approximation is the only one for Châtelet surfaces. Hence, all smooth projective models of a given equation (1) are the same as to the discussion of the Hasse principle, weak approximation, the failure of the Hasse principle explained by the Brauer-Manin obstruction, and weak approximation with Brauer-Manin obstruction. Remark 3.0.4. If the polynomial P (x) has a factor x2 a, i.e. there exists a degree-2 − xy+az y+xz polynomial f(x) such that P (x) = f(x)(x2 a), then Y = and Z = give − x2−a x2−a a birational equivalence between V and a quadratic surface given by Y 2 aZ2 = f(x) with affine coordinates (x,Y,Z). By the Hasse-Minkowski theorem and Remark− 3.0.3, the surface V satisfies weak approximation.

In this section, we always use the following way to choose an element for the parameter a in the equation (1).

3.0.1. Choosing an element for the parameter a in the equation (1). Given an extension c of number fields L/K, and a finite subset S ΩK ( K 2K ), we will choose an element a K2 with respect to these L/K and ⊂S in the\ ∞ following∪ way. ∈ OK \ f If S = , by Theorem 2.0.2, we can take a place v0 ΩK 2K splitting completely in L. Then replace∅ S by v to continue the following step.∈ \ { 0} ×2 Now, suppose that S = . For v ΩK , by Lemma 2.2.3, the set Kv is a nonempty open 6 ∅f ∈ subset of Kv. For v ΩK , by Lemma 2.2.4, the set a Kv v(a) is odd is a nonempty ∈ { ∈ | 1 } open subset of Kv. Using weak approximation for the affine line A , we can choose an element a K× satisfying the following conditions: ∈ τv(a) < 0 for all v S K , • ×2 ∈ ∩ ∞ a Kv for all v 2K, • v(∈a) is odd for all∈v S . • ∈ \∞K These conditions do not change by multiplying an element in K×2, so we can assume a K . The conditions that v(a) is odd for all v S K, and that τv(a) < 0 for all v ∈S O , imply a K2 for all v S. So a ∈ \∞K2. ∈ ∩ ∞K ∈ OK \ v ∈ ∈ OK \ Remark 3.0.5. Let S′ = v r τ (a) < 0 v Ωf 2 v(a) is odd , then S′ is a { ∈ ∞K | v }∪{ ∈ K \ K| } finite set. By the conditions that τv(a) < 0 for all v S K , and that v(a) is odd for all v S , we have S′ S. Then S′ = . ∈ ∩ ∞ ∈ \∞K ⊃ 6 ∅ Remark 3.0.6. If there exists one place in S splitting completely in L or S = , then by the choice of a above and Lemma 2.2.8, the element a L2. ∅ ∈ OK \ Remark 3.0.7. For the choice of a, we can enlarge S in Ω ( c 2 ) if necessary. K \ ∞K ∪ K 3.1. Châtelet surfaces without Kv point for any v ∈ S. In this subsection, we will construct a Châtelet surface of the first kind mentioned in Subsection 1.3. 8 HAN WU

3.1.1. Choice of parameters for the equation (1). Given an extension of number fields L/K, c 2 and a finite subset S ΩK ( K 2K), we choose an element a K K as in Subsub- section 3.0.1. ⊂ \ ∞ ∪ ∈ O \ 4 2 2 4 If S = , then let P (x)=1 x . Then the Châtelet surface V1 given by y az =1 x , has a rational∅ point (x,y,z)=(0− , 1, 0). − − Now, suppose that S = . We will choose an element b K× with respect to the chosen a in the following way. 6 ∅ ∈ ′ r f Let S = v K τv(a) < 0 v ΩK 2K v(a) is odd be as in Remark 3.0.5, then ′ { ∈ ∞ | }∪{ ∈ \ | } ×2 S S is a finite set. If v S K, then v(a) is odd, which implies a / Kv . Then ⊃ ∈ \∞× ∈ by Lemma 2.2.6, the set b Kv (a,b)v = 1 is a nonempty open subset of Kv. If ′ { ∈ | − } × v S (S K ), then by Lemma 2.2.5, the set b Kv (a,b)v = 1 is a nonempty ∈ \ ∪ ∞ { ∈ |1 } open subset of Kv. Using weak approximation for affine line A , we can choose an element b K× satisfying the following conditions: ∈

τv(b) < 0 for all v S K , • τ (b) > 0 for all v ∈ (S∩′ ∞S) , • v ∈ \ ∩ ∞K (a,b)v = 1 for all v S K, • (a,b) =1−for all v ∈S′ (\∞S ). • v ∈ \ ∪ ∞K We will choose an element c K× with respect to the chosen a,b in the following way. ∈ ′′ f ′′ Let S = v ΩK 2K v(b) is odd , then S is a finite set. The same argument as in the previous paragraph,{ ∈ \ we| can choose} an element c K× satisfying the following conditions: ∈ ×2 c Kv for all v S, • v(∈c) is odd for all∈v S′′ S′. • ∈ \ These conditions do not change by multiplying an element in K×2, so we can assume b,c . ∈ OK Let P (x)= b(x4 ac), and let V be the Châtelet surface given by y2 az2 = b(x4 ac). − 1 − − Proposition 3.1.1. For any extension of number fields L/K, and any finite subset S Ω ( c 2 ) splitting completely in L, there exists a Châtelet surface V defined over⊂ K \ ∞K ∪ K 1 S SL ′ K such that V1(AK ) V1(AL ) is nonempty, but that V1(Kv)= V1(Lv )= for all v S and all v′ S . ⊂ ∅ ∈ ∈ L

Proof. For the extension L/K, and the ﬁnite set S, we will check that the Châtelet surface V1 chosen as in Subsubsection 3.1.1, has the properties. For S = , it is clear. ∅ Now, suppose that S = . We will check these properties by local computation. 6 ∅ Suppose that v ( S′) 2 . By the choice of a, we have a K×2. By Remark 3.0.1, ∈ ∞K \ ∪ K ∈ v the surface V1 admits a Kv-point. ′ 4 Suppose that v (S S) K . Take x0 K such that τv(x0 ac) > 0. By the choice of b, we have τ (∈b(x4 \ ac∩)) ∞> 0. So (a,b(x∈4 ac)) = 1, which− implies that V 0 admits a v 0 − 0 − v 1 Kv-point with x = x0. ′ Suppose that v S (S K ), then by the choice of b, we have (a,b)v =1. Take x0 Kv ∈ \ ∪ ∞ 4 4∈ such that the valuation v(x0) < 0, then by Lemma 2.2.2, we have (a, x0 ac)v = (a, x0)v = 4 0 − 1. So (a,b(x0 ac))v = (a,b)v =1, which implies that V1 admits a Kv-point with x = x0. Suppose that −v S′′ S′. By the choice of a, b, c, the valuations v(a) and v( abc) are even. ∈ \ 0 − By Lemma 2.2.1, we have (a, abc)v = 1, which implies that V1 admits a Kv-point with x =0. − f ′ ′′ Suppose that v ΩK (S S 2K). Take x0 Kv such that the valuation v(x0) < 0, ∈ \ ∪ ∪ 4 ∈ 4 then by Lemma 2.2.2, we have (a, x0 ac)v = (a, x0)v =1. For v(a) and v(b) are even, by Lemma 2.2.1), we have (a,b) =1. So−(a,b(x4 ac)) =1, which implies that V 0 admits v 0 − v 1 a Kv-point with x = x0. So, the subset V (AS ) V (ASL ) is nonempty. 1 K ⊂ 1 L CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS9

Suppose that v S K . Then by the choice of a, b, c, we have τv(a), τv(b) and τv(ac) are ∈ 4∩ ∞ 0 negative. So (a,b(x ac))v = 1 for all x Kv, which implies that V1 has no Kv-point. By Remark 3.0.2, we− have V (K−)= . ∈ 1 v ∅ Suppose that v S K. Then by the choice of b, we have (a,b)v = 1. Let x Kv. If ∈ \∞ 4 4 − ∈ 4v(x) < v(ac), then by Lemma 2.2.2, we have (a, x ac)v = (a, x )v =1. If 4v(x) > v(ac), 4 − then by Lemma 2.2.2, we have (a, x ac)v = (a, ac)v = (a,c)v =1 (by the choice of c, the ×2 − − last equality holds). For c Kv , the valuation v(ac) is odd. So, the equality 4v(x)= v(ac) ∈ 4 4 cannot happen. In each case, we have (a,b(x ac))v = (a,b)v(a, x ac)v = 1, which 0 − − − implies that V1 has no Kv-point. By Remark 3.0.2, we have V1(Kv)= . So, the set V (K )= for all v S. ∅ 1 v ∅ ∈ Take a place v′ S . Let v S be the restriction of v′ on K. By the assumption that v 0 ∈ L 0 ∈ 0 0 splits completely in L, we have K = L ′ . Hence V (K )= V (L ′ ). v0 v0 1 v0 1 v0 So, the set V (L ′ )= for all v′ S . 1 v ∅ ∈ L Remark 3.1.2. By the choice of elements a,c in Subsubsection 3.1.1, if v S , then ∈ \∞K by comparing the valuation, the polynomial P (x) is irreducible over Kv. In this case, since the choice of c is not unique, we choose another one to get a new polynomial P ′(X), so the polynomials P (x) and P ′(x) are coprime in K. For v S , the polynomial P (x) is a ∈ ∩ ∞K product of two irreducible factors over Kv.

Using the construction method in Subsubsection 3.1.1, we have the following examples, which are special cases of Proposition 3.1.1. They will be used for further discussion.

−1 Example 3.1.3. Let K = Q, and let ζ7 be a primitive 7-th root of unity. Let α = ζ7 +ζ7 with the minimal polynomial x3 + x2 2x 1. Let L = Q(α). Then the degree [L : K]=3. − − Let S = 29 . For 29 1 mod7, the place 29 splits completely in L, indeed in Q(ζ7). Using the{ construction} ≡ method in Subsubsection 3.1.1, we choose data: S = 29 , S′ = 13, 29 , S′′ = 7 , a = 377, b = 14, c = 238 and P (x) = 14(x4 89726){. Then} the Châtelet{ } surface{ given} by y2 377z2 = P (x), has the properties of Proposition− 3.1.1. − Example 3.1.4. Let K = Q, and let L = Q(√3). Using the construction method in ′ ′′ Subsubsection 3.1.1, we choose data: S = K , S = K 23 , S = 5 , a = 23, b = 5, c = 5 and P (x) = 5(x4 +{∞ 115)}. Then the∞ Châtelet∪{ } surface{ given} by −y2 + 23z2 =−P (x), has the properties− of Proposition 3.1.1. Example 3.1.5. Let K = Q, and let L = Q(√3). Then the place 23 splits completely in L. Using the construction method in Subsubsection 3.1.1, we choose data: S = 23 , S′ = ′′ 4 { } K 23 , S = 5 , a = 23, b =5, c = 35 and P (x)=5(x +805). Then the Châtelet ∞surface∪{ given} by y2{+} 23z2 =−P (x), has the properties of Proposition 3.1.1.

3.2. Châtelet surfaces with rational points and not satisfying weak approximation. Given an number field K, Liang [Lia18, Proposition 3.4] constructed a Châtelet surface over K, which has a K-rational point and does not satisfy weak approximation off K . Using the same method as in [Poo09, Section 5] to choose the parameters for the ∞equation (1), he constructed a Châtelet surface, and there exists an element in the Brauer group of this surface, which has two different local invariants on a given finite place, i.e. this element gives an obstruction to weak approximation for this surface. In this subsection, we generalize it. Next, we will construct a Châtelet surface of the second kind mentioned in Subsection 1.3.

3.2.1. Choice of parameters for the equation (1). Given an extension of number fields L/K, c 2 and a finite subset S ΩK ( K 2K), we choose an element a K K as in Subsub- section 3.0.1. ⊂ \ ∞ ∪ ∈ O \ We will choose an element b K× with respect to the chosen a in the following way. ∈ ′ r f Let S = v K τv(a) < 0 v ΩK 2K v(a) is odd be as in Remark 3.0.5, then ′ { ∈ ∞ | }∪{ ∈ \ | } S S is a finite set. By Lemma 2.2.4, for v S K, the set b Kv v(b) = v(a) is a⊃ nonempty open subset of K ; for v S′ ∈(S \∞ ), the set{ b∈ K| v(b) =−v(a)} v ∈ \ ∪ ∞K { ∈ v| } 10 HAN WU

is a nonempty open subset of Kv . By Lemma 2.0.1, we can choose a nonzero element b [1/2] satisfying the followingO conditions: ∈ OS v(b)= v(a) for all v S K, • v(b)= v−(a) for all v ∈S′ \∞(S ). • ∈ \ ∪ ∞K We will choose an element c K× with respect to the chosen a,b in the following way. ∈ Let S′′ = v Ωf 2 v(b) = 0 , then S′′ is a finite set and S′ S′′. By Theorem { ∈ K \ K| 6 } \∞K ⊂ 2.0.2, we can take two different finite places v , v Ωf S′′ splitting completely in L. If 1 2 ∈ K \ v S , then v(a) is odd. In this case, by Lemma 2.2.7, the set c × (a,c) = 1 K Kv v ∈ \∞ × { ∈ O | − } is a nonempty open subset of K . If v v1, v2 , then b . In this case, by Lemma O v ∈{ } ∈ OKv 2.2.4, the sets c K v(c)=1 and c K v(1 + cb2)=1 are nonempty open subsets { ∈ v| } { ∈ v| } of Kv . Also by Lemma 2.0.1, we can choose a nonzero element c K [1/2] satisfying the followingO conditions: ∈ O

2 τv(1 + cb ) < 0 for all v S K , • τ (c) > 0 for all v (S′ ∈S) ∩ ∞ , • v ∈ \ ∩ ∞K (a,c)v = 1 and v(c)=0 for all v S K , • v (c)=1−and v (1 + cb2)=1 for the∈ chosen\∞ v , v above. • 1 2 1 2 2 2 2 2 Let P (x) = (cx + 1)((1 + cb )x + b ), and let V2 be the Châtelet surface given by y2 az2 = (cx2 + 1)((1 + cb2)x2 + b2). − Proposition 3.2.1. For any extension of number fields L/K, and any finite subset S c ⊂ ΩK ( K 2K) splitting completely in L, there exists a Châtelet surface V2 defined over K, which\ ∞ has∪ the following properties.

The Brauer group Br(V2)/Br(K) ∼= Br(V2L)/Br(L) ∼= Z/2Z, is generated by an • element A Br(V ). The subset V (K) V (L) is nonempty. ∈ 2 2 ⊂ 2 For any v S, there exist Pv and Qv in V2(Kv) such that the local invariants • inv (A(P ))∈ = 0 and inv (A(Q )) = 1/2. For any other v / S, and any P v v v v ∈ v ∈ V2(Kv), the local invariant invv(A(Pv))=0. For any v′ S , there exist P ′ and Q ′ in V (L ′ ) such that the local invariants • ∈ L v v 2 v inv ′ (A(P ′ )) = 0 and inv ′ (A(Q ′ )) = 1/2. For any other v′ / S , and any v v v v ∈ L P ′ V (L ′ ), the local invariant inv ′ (A(P ′ ))=0. v ∈ 2 v v v Proof. For the extension L/K, and the finite set S, we will check that the Châtelet surface V2 chosen as in Subsubsection 3.2.1, has the properties. We need to prove the statement about the Brauer group, and find the element A in this proposition. 2 By the choice of the places v1, the polynomial x + c is an Eisenstein polynomial, so 2 it is irreducible over Kv1 . Since v1(a) is even, we have K(√a)Kv1 ≇ Kv1 [x]/(cx + 1). 2 So K(√a) ≇ K[x]/(cx + 1). The same argument holds for the place v2 and polynomial (1+cb2)x2 +b2. For all places of S split completely in L, then by Remark 3.0.6, we have a L2. By the splitting condition of v , v , we have L(√a) ≇ L[x]/(cx2 + 1) and L(√a) ≇∈ OK \ 1 2 L[x]/((1 + cb2)x2 + b2). So P (x) = (cx2 + 1)((1 + cb2)x2 + b2) is separable and a product of two degree-2 irreducible factors over K and L. According to [Sko01, Proposition 7.1.1], the Brauer group Br(V2)/Br(K) = Br(V2L)/Br(L) = Z/2Z. Furthermore, by Proposition ∼ ∼ 2 7.1.2 in loc. cit, we take the quaternion algebra A = (a,cx + 1) Br(V2) as a generator element of this group. Then we have the equality A = (a,cx2 +1)=(∈ a, (1 + cb2)x2 + b2) in Br(V2). 0 For (x,y,z) = (0, b, 0) is a rational point on V2 , the set V2(K) is nonempty. We denote this rational point by Q0. We need to compute the evaluation of A on V (K ) for all v Ω . 2 v ∈ K For any v ΩK , the local invariant invv(A(Q0)) = 0. By Remark 3.0.2, it suffices to ∈ 0 compute the local invariant invv(A(Pv)) for all Pv V2 (Kv). Suppose that v ( S′) 2 . Then a K×2, so∈inv (A(P ))=0 for all P V (K ). ∈ ∞K \ ∪ K ∈ v v v v ∈ 2 v CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS11

′ 2 Suppose that v (S S) K . For any x K, by the choice of c, we have τv(cx + 1) > 0. 2 ∈ \ ∩∞ ∈ Then (a,cx + 1)v =1, so invv(A(Pv))=0 for all Pv V2(Kv). ′ ∈ 0 Suppose that v S (S K ). Take an arbitrary Pv V2 (Kv). If invv(A(Pv))=1/2, then 2 ∈ \ ∪∞ 2 2 2 ∈ (a,cx + 1)v = 1 = (a, (1 + cb )x + b )v at Pv. By Lemma 2.2.2, the ﬁrst equality implies − 2 2 2 v(x) 0. For v(a)= v(b) > 0 and v(c) 0, by Lemma 2.2.2, we have (a, (1+cb )x +b )v = 2≤ ≥ (a, x )v =1, which is a contradiction. So invv(A(Pv ))=0. f ′ 0 Suppose that v ΩK (S 2K ). Take an arbitrary Pv V2 (Kv). If invv(A(Pv ))=1/2, then 2 ∈ \ ∪ 2 2 2 ∈ (a,cx +1)v = 1 = (a, (1+cb )x +b )v at Pv. For v(a) is even, by Lemma 2.2.1, the ﬁrst − 2 −2 equality implies that v(cx + 1) is odd. For c K [1/2], we have v(x) 0. So v(c + x ) ∈ O 2≤ −2 ×2 is odd and positive. For v(b) 0, by Hensel’s lemma, we have 1+ b (c + x ) Kv . 2 2 2 ≥2 2 −2 ∈ So (a, (1 + cb )x + b )v = (a, x )v(a, 1+ b (c + x ))v = 1, which is a contradiction. So invv(A(Pv))=0.

Suppose that v S K . Take Pv = Q0, then invv(A(Pv)) = 0. By the choice of b, c, 2 ∈ ∩ ∞ we have τ ( b ) > τ ( 1 ) > 0. Take x K such that τ (x ) > τ ( b2 ), then v −cb2−1 v −c 0 ∈ v 0 v −cb2−1 τ ((cx2 + 1)((1 + cb2)x2 + b2)) > 0 and τ (cx2 + 1) < 0. So there existsq a Q V 0(K ) v 0 0 v 0 v ∈ 2 v with x = x0. Then invv(A(Qv))=1/2. Suppose that v S K. Take Pv = Q0, then invv(A(Pv )) = 0. Take x0 Kv such that ∈ \∞ ∈ 2 v(x0) < 0. For v(b) = v(a) < 0 and v(c)=0, by Lemma 2.2.2, we have (a,cx0 + 1)v = 2 − 2 2 2 2 2 2 (a,cx0)v = (a,c)v and (a, (1 + cb )x0 + b )v = (a,cb x0)v = (a,c)v. So (a, (cx0 + 1)((1 + 2 2 2 0 cb )x0 + b ))v = (a,c)v(a,c)v = 1. Hence, there exists a Qv V2 (Kv) with x = x0. For (a,c) = 1, we have inv (A(Q ))=1/2. ∈ v − v v Finally, we need to compute the evaluation of A on V (L ′ ) for all v′ Ω . 2 v ∈ L ′ For any v ΩL, the local invariant invv′ (A(Q0))=0. Suppose that∈ v′ S . Let v Ω be the restriction of v′ on K. By the assumption that v ∈ L ∈ K is split completely in L, we have Kv = Lv′ . So V2(Kv)= V2(Lv′ ). By the argument already shown, there exist P ,Q V (K ) such that inv (A(P )) = 0 and inv (A(Q )) = 1/2. v v ∈ 2 v v v v v View Pv,Qv as elements in V2(Lv′ ), and let Pv′ = Pv and Qv′ = Qv. Then invv′ (A(Pv′ )) = invv(A(Pv))=0 and invv′ (A(Qv′ )) = invv(A(Qv))=1/2. Suppose that v′ Ω S . This local computation is the same as the case v Ω S. ∈ L\ L ∈ K \ Remark 3.2.2. For any v S, and any P V (K ), the local invariant of the eval- ∈ v ∈ 2 v uation of A on Pv is 0 or 1/2. Let U1 = Pv V2(Kv) invv(A(Pv )) = 0 and U2 = P V (K ) inv (A(P )) = 1/2 . Then U{ and∈U are nonempty| disjoint open} subsets { v ∈ 2 v | v v } 1 2 of V2(Kv), and V2(Kv)= U1 U2.

The following proposition statesF that the surface V2 in Proposition 3.2.1, has the following weak approximation properties. Proposition 3.2.3. Given an extension of number fields L/K, and a finite subset S Ω ( c 2 ) splitting completely in L, let V be a Châtelet surface satisfying those⊂ K \ ∞K ∪ K 2 properties of Proposition 3.2.1. If S = , then V2 and V2L satisfy weak approximation. If ∅ ′ ′ S = , then V2 satisfies weak approximation off S for any finite subset S ΩK such that S′6 ∅S = , while it fails for any finite subset S′ Ω such that S′ S =⊂ . And in the ∩ 6 ∅ ⊂ K ∩ ∅ case S = , the surface V2L satisfies weak approximation off T for any finite subset T ΩL such that6 ∅T S = , while it fails for any finite subset T Ω such that T S =⊂. ∩ L 6 ∅ ⊂ L ∩ L ∅ Proof. According to [CTSSD87a, Theorem B; CTSSD87b], the Brauer-Manin obstruction to the Hasse principle and weak approximation is the only one for Châtelet surfaces, so Br V2(K) is dense in V2(AK ) .

Suppose that S = , then for any (Pv)v∈ΩK V2(AK ), by Proposition 3.2.1, the sum inv (A(P ))∅ = 0. For Br(V )/Br(K) ∈is generated by the element A, we have v∈ΩK v v 2 V (A )Br = V (A ). So V (K) is dense in V (A )Br = V (A ), i.e. the surface V satisﬁes P2 K 2 K 2 2 K 2 K 2 weak approximation. Suppose that S′ S = . Take v S′ S. For any ﬁnite subset R Ω v , ∩ 6 ∅ 0 ∈ ∩ ⊂ K \{ 0} take a nonempty open subset M = V2(Kv ) Uv V2(Kv) V2(AK ). 0 × v∈R × v∈ /R∪{v0} ⊂ Take an element (P ) M. By Proposition 3.2.1 and v S, we can take an v v∈ΩK ∈ Q Q 0 ∈ 12 HAN WU

′ ′ element Pv0 V2(Kv0 ) such that invv0 A(Pv0 )= 1/2. By Proposition 3.2.1, the sum inv∈ (A(P )) is 0 or 1/2 in Q/Z. If it is 1/2, then we replace P by P ′ . In this v∈ΩK \{v0} v v v0 v0 way, we get a new element (Pv)v∈Ω M. And the sum invv(A(Pv ))=0 in Q/Z. P K ∈ v∈ΩK So (P ) V (A )Br M. For V (K) is dense in V (A )Br, the set V (K) M = , v v∈ΩK ∈ 2 K ∩ 2 2 PK 2 ∩ 6 ∅ which implies that V2 satisfies weak approximation off v0 . So V2 satisfies weak approximation off S′. { } Suppose that S = and S′ S = . Take v S, and let U = P V (K ) inv (A(P )) = 6 ∅ ∩ ∅ 0 ∈ v0 { v0 ∈ 2 v0 | v0 v0 1/2 . For v S v0 , let Uv = Pv V2(Kv) invv(A(Pv ))=0 . For any v S, by Remark 3.2.2,} the set∈U \{is a} nonempty open{ ∈ subset of|V (K ). Let M =} U ∈ V (K ). It v 2 v v∈S v × v∈ /S 2 v is a nonempty open subset of V2(AK ). For any (Pv)v∈ΩK M, by Proposition 3.2.1 and the ∈ Q Q Br choice of Uv, the sum invv(A(Pv))=1/2 is nonzero in Q/Z. So V2(AK ) M = , v∈ΩK ∩ ∅ which implies V (K) M = . Hence V does not satisfy weak approximation off S′. 2 ∩P ∅ 2 The same argument applies to V2L.

Applying the construction method in Subsubsection 3.2.1, we have the following weak approximation properties for Châtelet surfaces.

Corollary 3.2.4. For any extension of number fields L/K, and any finite nonempty subset S Ω ( c 2 ) splitting completely in L, there exists a Châtelet surface V defined over ⊂ K \ ∞K ∪ K K such that V (K) = . For any subfield L′ L over K, the Brauer group Br(V )/Br(K) = ′ 6 ∅ ⊂ ∼ Br(VL′ )/Br(L ) ∼= Z/2Z. And the surface VL′ has the following properties. ′ ′ For any finite subset T ΩL′ such that T SL′ = , the surface VL′ satisfies • weak approximation off T⊂′. ∩ 6 ∅ ′ ′ For any finite subset T ΩL′ such that T SL′ = , the surface VL′ does not • ⊂ ′ ∩ ∅ satisfy weak approximation off T . In particular, the surface VL′ does not satisfy weak approximation.

Proof. For the extension L/K, and S, let V be the Châtelet surface chosen as in Subsubsec- tion 3.2.1. Applying the same argument about the ﬁeld L to its subﬁeld L′, the properties that we list, are just what we have explained in Proposition 3.2.1 and Proposition 3.2.3.

Using the construction method in Subsubsection 3.2.1, we have the following example, which is a special case of Proposition 3.2.1. It will be used for further discussion.

Example 3.2.5. Let K = Q,L = Q(√3), and let S = 73 . The prime numbers 11, 23, 73 split completely in L. Using the construction method{ in Subsubsection} 3.2.1, we choose ′ ′′ data: S = S = S = 73, v1 = 11, v2 = 23, a = 73, b = 1/73, c = 99 and P (x) = (99x2 + 1)(5428x2/5329+1/5329). Then the Châtelet surface given by y2 73z2 = P (x), has the properties of Proposition 3.2.1, Proposition 3.2.3. −

3.3. Châtelet surfaces related to the Hasse principle. Iskovskikh [Isk71] gave an 4 example of the intersection of two quadratic hypersurfaces in PQ, which is a Châtelet surface over Q given by y2 + z2 = (x2 2)( x2 + 3). He showed that this Châtelet surface is a counterexample to the Hasse principle.− − Similarly, Skorobogatov [Sko01, Pages 145- 146] gave a family of Châtelet surfaces with a parameter over Q. He discussed the property of the Hasse principle for this family. Poonen [Poo09, Proposition 5.1] generalized their arguments to any number field. Given an number field K, he constructed a Châtelet surface defined over K, which is a counterexample to the Hasse principle. He used the Čebotarev’s density theorem for some ray class fields to choose the parameters for the equation (1). The Châtelet surface that he constructed, has the property of [Poo09, Lemma 5.5] (a special situation of the following Proposition 3.3.1: the case when S = v for some place v { 0} 0 associated to some large prime element in K ), which is the main ingredient in the proof of [Poo09, Proposition 5.1]. In this subsection,O we generalize them. Next, we will construct a Châtelet surface of the third kind mentioned in Subsection 1.3. CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS13

3.3.1. Choice of parameters for the equation (1). Given an extension of number fields L/K, c 2 and a finite subset S ΩK ( K 2K), we choose an element a K K as in Subsub- section 3.0.1. ⊂ \ ∞ ∪ ∈ O \ We will choose an element b K× with respect to the chosen a in the following way. ∈ ′ r f Let S = v K τv(a) < 0 v ΩK 2K v(a) is odd be as in Remark 3.0.5, then ′ { ∈ ∞ | }∪{ ∈ \ | } S S is a finite set. If v S K, then v(a) is odd. Then by Lemma 2.2.7, the set ⊃ × ∈ \∞ ′ b Kv (a,b)v = 1 is a nonempty open subset of Kv . If v S (S K ), then by { ∈ O | − } × O ∈ \ ∪ ∞ Lemma 2.2.5, the set b (a,b)v =1 is a nonempty open subset of K . By Lemma { ∈ OKv | } O v 2.0.1, we can choose a nonzero element b [1/2] satisfying the following conditions: ∈ OK τv(b) < 0 for all v S K , • τ (b) > 0 for all v ∈ (S∩′ ∞S) , • v ∈ \ ∩ ∞K (a,b)v = 1 and v(b)=0 for all v S K, • (a,b) =1−and v(b)=0 for all v ∈S′ (\∞S ). • v ∈ \ ∪ ∞K We will choose an element c K× with respect to the chosen a,b in the following way. ∈ ′′ f ′′ ′ ′′ Let S = v ΩK 2K v(b) =0 , then S is a finite set and S S = . By Theorem 2.0.2, { ∈ \ | 6 } f ′ ′′ ∩ ∅ we can take two different finite places v1, v2 ΩK (S S 2K) splitting completely in ′ ∈× \ ∪ ∪ L. If v (S K) v1, v2 , then b . In this case, by Lemma 2.2.4, the sets ∈ \∞ ∪ { } ∈ OKv c Kv v(bc +1) = v(a)+2 , c Kv v(c)=1 and c Kv v(bc +1)=1 are nonempty { ∈ | } { ′′∈ | } { ∈ | × } open subsets of K . If v S , by Lemma 2.2.5, the set c (a,c)v = 1 is a O v ∈ { ∈ OKv | } nonempty open subset of Kv . Also by Lemma 2.0.1, we can choose a nonzero element c [1/2] satisfying theO following conditions: ∈ OK 0 < τv(c) < 1/τv(b) for all v S K , • − ′ ∈ ∩ ∞ τv(bc + 1) < 0 for all v (S S) K , • ∈ \ ∩′ ∞ v(bc +1)= v(a)+2 for all v S K , • ′′ ∈ \∞ (a,c)v =1 for all v S , • v (c)=1 and v (bc∈+1)=1 for the chosen v , v above. • 1 2 1 2 2 2 2 2 Let P (x) = (x c)(bx bc 1), and let V3 be the Châtelet surface given by y az = (x2 c)(bx2 bc− 1). − − − − − − Proposition 3.3.1. For any extension of number fields L/K, and any finite subset S c ⊂ ΩK ( K 2K) splitting completely in L, there exists a Châtelet surface V3 defined over K, which\ ∞ has∪ the following properties. The Brauer group Br(V )/Br(K) = Br(V )/Br(L) = Z/2Z, is generated by an • 3 ∼ 3L ∼ element A Br(V3). The subset V3(AK ) V3(AL) is nonempty. For any v ∈Ω , and any P V (K ), ⊂ • ∈ K v ∈ 3 v 0 if v/ S, invv(A(Pv )) = ∈ 1/2 if v S. ( ∈ For any v′ Ω , and any P ′ V (L ′ ), • ∈ L v ∈ 3 v ′ 0 if v / SL, invv′ (A(Pv′ )) = ∈ 1/2 if v′ S . ( ∈ L Proof. For the extension L/K, and the finite set S, we will check that the Châtelet surface V3 chosen as in Subsubsection 3.3.1, has the properties.

Firstly, we need to check that V3 has an AK -adelic point. Suppose that v ( S′) 2 . Then a K×2. By Remark 3.0.1, the surface V admits ∈ ∞K \ ∪ K ∈ v 3 a Kv-point. ′ Suppose that v (S S) K . Let x0 = 0. For τv(b) > 0 and τv(bc + 1) < 0, we have τ (c) < 0 and τ ((∈x2 \c)(bx∩ ∞2 bc 1)) = τ (c(bc + 1)) > 0, which implies that V 0 admits v v 0 − 0 − − v 3 a Kv-point with x =0. ′ Suppose that v S (S K ). Take x0 Kv such that the valuation v(x0) < 0. For × ∈ \ ∪ ∞ ∈ 2 2 b and c K [1/2], by Lemma 2.2.2, we have (a, x c)v = (a, x )v = 1 and ∈ OKv ∈ O 0 − 0 14 HAN WU

2 2 (a,bx0 bc 1)v = (a,bx0)v = (a,b)v. By the choice of b, we have (a,b)v = 1. Hence (a, (x2 − c)(bx− 2 bc 1)) = (a,b) = 1, which implies that V 0 admits a K -point with 0 − 0 − − v v 3 v x = x0. ′′ Suppose that v S . By the choice of a, b, c, we have (a,c)v =1, v(a) even, and bc +1 × By Lemma∈ 2.2.1, we have Let Then 2 2 ∈ Kv . (a,bc+1)v =1. x0 =0. (a, (x0 c)(bx0 bc 1))v = O 0 − − − (a,c(bc+1))v = (a,c)v(a,bc+1)v =1, which implies that V3 admits a Kv-point with x =0. Suppose that v Ωf (S′ S′′ 2 ). Then v(b)=0. Take x K such that the valuation ∈ K \ ∪ ∪ K 0 ∈ v v(x ) < 0. For b × and c [1/2], by Lemma 2.2.2, we have (a, x2 c) = (a, x2) = 0 Kv K 0 v 0 v 2 ∈ O ∈ O2 − 1 and (a,bx0 bc 1)v = (a,bx0)v = (a,b)v. For v(a) and v(b) are even, by Lemma 2.2.1, we have (a,b)− =− 1. So (a, (x2 c)(bx2 bc 1)) = (a,b) = 1, which implies that V 0 v 0 − 0 − − v v 3 admits a Kv-point with x = x0. Suppose that v S K . Let x0 = 0. Then by the choice of a, b, c, we have τv(a) < 0, ∈ ∩ ∞ 2 2 τv(c) > 0 and τv(bc + 1) > 0. So (a, (x0 c)(bx0 bc 1))v = (a,c(bc + 1))v = 1, which 0 − − − implies that V3 admits a Kv-point with x =0. Suppose that v S K. Choose a prime element πv and take x0 = πv. By the choice of a, b, c, we have b,c∈ \∞× , v(bx2)=2, and v(bc +1)= v(a)+2 3. By Lemma 2.2.2, we Kv 0 2 ∈ O 2 2 ≥ have (a, x0 c)v = (a, c)v and (a,bx0 bc 1)v = (a,bx0)v. By Hensel’s lemma, we have − −×2 2 − −2 2 bc =1 (bc + 1) Kv . So (a, (x0 c)(bx0 bc 1))v = (a, bcx0)v =1, which implies − 0 − ∈ − − − − that V3 admits a Kv-point with x = πv. Secondly, we need to prove the statement about the Brauer group, and ﬁnd the element A in this proposition. 2 By the choice of the places v1, the polynomial x c is an Eisenstein polynomial, so it is irreducible over K . Since v (a) is even, we have− K(√a)K ≇ K [x]/(x2 c). v1 1 v1 v1 − So K(√a) ≇ K[x]/(x2 c). The same argument holds for the place v and polynomial − 2 bx2 bc 1. For all places of S split completely in L, then by Remark 3.0.6, we have a − −L2. By the splitting condition of v , v , we have L(√a) ≇ L[x]/(x2 c) and ∈ OK \ 1 2 − L(√a) ≇ L[x]/(bx2 bc 1). So P (x) = (x2 c)(bx2 bc 1) is separable and a product of two degree-2 irreducible− − factors over K and− L. According− − to [Sko01, Proposition 7.1.1], the Brauer group Br(V3)/Br(K) = Br(V3L)/Br(L) = Z/2Z. Furthermore, by Proposition ∼ ∼ 2 7.1.2 in loc. cit, we take the quaternion algebra A = (a, x c) Br(V3) as a generator element of this group. Then we have the equality A = (a, x−2 c∈) = (a,bx2 bc 1) in − − − Br(V3). Thirdly, We need to compute the evaluation of A on V (K ) for all v Ω . 3 v ∈ K 0 By Remark 3.0.2, it suﬃces to compute the local invariant invv(A(Pv )) for all Pv V3 (Kv) and all v Ω . ∈ ∈ K ′ ×2 Suppose that v ( K S ) 2K. Then a Kv , so invv(A(Pv))=0 for all Pv V3(Kv). ∈ ∞′ \ ∪ ∈ ∈ Suppose that v (S S) K . By the choice of b, c, we have τv(b) > 0 and τv(bc + 1) < 0. ∈ \ ∩ ∞ 2 0 So, for any x K, we have (a,bx bc 1)v =1. Hence invv(A(Pv ))=0 for all Pv V3 (Kv). ∈ ′ − − ∈ Suppose that v S (S K ). By the choice of b, we have (a,b)v =1. Take an arbitrary 0 ∈ \ ∪ ∞ 2 2 Pv V3 (Kv). If v(x) < 0 at Pv, by Lemma 2.2.2, we have (a, x c)v = (a, x )v = 1. ∈ × − If v(x) > 0 at Pv, since b,c Kv and v(bc +1) = v(a)+2 3, by Lemma 2.2.2, we 2 ∈ O ≥ ×2 have (a, x c)v = (a, c)v. By Hensel’s lemma, we have bc = 1 (bc + 1) Kv . So 2 − − − − × ∈ (a, x c)v = (a, c)v = (a, bc)v = 1. If v(x)=0 at Pv, since b Kv and v(bc +1) = − − − 2 2 ∈ O v(a)+2 3, by Lemma 2.2.2, we have (a,bx bc 1)v = (a,bx )v =1. So invv(A(Pv))=0. ≥ f ′ − − 0 Suppose that v ΩK (S 2K ). Take an arbitrary Pv V3 (Kv). If invv(A(Pv)) = 1/2, 2 ∈ \ ∪ 2 ∈ then (a,bx bc 1)v = (a, x c)v = 1 at Pv. For v(a) is even, by Lemma 2.2.1, the last equality implies− − that v(x2 c)−is odd,− so it is positive. So v(bx2 bc 1)=0. By Lemma 2.2.1, we have (a,bx2 bc − 1) =1, which is a contradiction. So−inv−(A(P ))=0. − − v v v 0 2 Suppose that v S K . Take an arbitrary Pv V3 (Kv). If A(Pv )=0, then (a,bx bc 2 ∈ ∩∞ ∈ 2 − − 1)v = (a, x c)v = 1 at Pv. The last equality implies that τv(x c) > 0. By the choice of b, we have−τ (b) < 0, so τ (bx2 bc 1) < 0, which contradicts−(a,bx2 bc 1) = 1. v v − − − − v So invv(A(Pv ))=1/2. Suppose that v S K. By the choice of b, we have (a,b)v = 1. Take an arbitrary 0 ∈ \∞ × − Pv V (Kv). If v(x) 0 at Pv, for b and v(bc +1) = v(a)+2 3, by Lemma ∈ 3 ≤ ∈ OKv ≥ CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS15

2 2 × 2.2.2, we have (a,bx bc 1)v = (a,bx )v = 1. If v(x) > 0 at Pv, for b,c Kv and − − − 2 ∈ O v(bc +1) = v(a)+2 3, by Lemma 2.2.2, we have (a, x c)v = (a, c)v. By Hensel’s lemma, we have bc =1≥ (bc + 1) K×2. So (a, x2 c) =− (a, c) =− (a, bc) = 1. − − ∈ v − v − v − − v − So invv(A(Pv ))=1/2. Finally, we need to compute the evaluation of A on V (L ′ ) for all v′ Ω . 3 v ∈ L Suppose that v′ S . Let v Ω be the restriction of v′ on K. By the assumption ∈ L ∈ K that v is split completely in L, we have Kv = Lv′ . So V3(Kv) = V3(Lv′ ). Then for any P ′ V (L ′ ), denote P ′ in V (K ) by P . Then by the argument already shown, the local v ∈ 3 v v 3 v v invariant invv′ (A(Pv′ )) = invv(A(Pv ))=1/2. Suppose that v′ Ω S . This local computation is the same as the case v Ω S. ∈ L\ L ∈ K \ Remark 3.3.2. If the surface V3 has a K-rational point Q, then by the global reciprocity law, the sum in If the number is odd, then from Proposition v∈ΩK invv(A(Q)) = 0 Q/Z. ♯S 3.3.1 that we get, this sum is ♯S/2, which is nonzero in Q/Z. So, in this case, the surface P V3 has no K-rational point, which implies that the surface V3 is a counterexample to the

Hasse principle. If the number ♯S is even, then for any (Pv)v∈ΩK V3(AK ), by Proposition 3.3.1, the sum inv (A(P )) = ♯S/2=0 in Q/Z. For Br(V∈)/Br(K) is generated by v∈ΩK v v 3 the element A, we have V (A )Br = V (A ) = . According to [CTSSD87a, Theorem B; P 3 K 3 K CTSSD87b], the Brauer-Manin obstruction to6 the∅ Hasse principle and weak approximation is the only one for Châtelet surfaces. So, in this case, the set V3(K) = , and it is dense Br 6 ∅ in V3(AK ) = V3(AK ), i.e. the surface V3 has a K-rational point and satisﬁes weak approximation. In particular, if the number ♯S = 0, i.e. S = , though the Brauer group ∅ Br(V3)/Br(K) is nontrivial, it gives no obstruction to week approximation for V3.

Applying the construction method in Subsubsection 3.3.1, we can relate the properties in Proposition 3.2.1 to the Hasse principle and weak approximation. Corollary 3.3.3. For any extension of number fields L/K, there exists a Châtelet surface ′ V defined over K such that V (AK ) = . For any subfield L L over K, the Brauer group ′ 6 ∅ ⊂ Br(V )/Br(K) ∼= Br(VL′ )/Br(L ) ∼= Z/2Z. And the surface VL′ has the following properties. ′ If the degree [L : K] is odd, then the surface VL′ is a counterexample to the Hasse • principle. In particular, the surface V is a counterexample to the Hasse principle. ′ If the degree [L : K] is even, then the surface VL′ satisfies weak approximation. • In particular, in this case, the set V (L′) = . 6 ∅ Proof. By Theorem 2.0.2, we can take a place v Ω ( c 2 ) splitting completely in L. 0 ∈ K \ ∞K ∪ K Let S = v0 . Using the construction method in Subsubsection 3.3.1, there exists a Châtelet surface V{ defined} over K having the properties of Proposition 3.3.1. By the same argument ′ as in the proof of Proposition 3.3.1, we have Br(V )/Br(K) = Br(VL′ )/Br(L ) = Z/2Z. For ′ ∼ ′ ∼ v0 splits completely in L, it also does in L . For ♯S is odd, if [L : K] is odd, then ♯SL′ is ′ odd; if [L : K] is even, then ♯SL′ is even. Applying the same argument about the field L to its subfield L′, the properties that we list, are just what we have explained in Remark 3.3.2. ′ Remark 3.3.4. Though the Brauer group Br(V )/Br(K) ∼= Br(VL′ )/Br(L ) ∼= Z/2Z in Corollary 3.3.3, is nontrivial, it gives an obstruction to the Hasse principle for V, also VL′ if ′ ′ [L : K] is odd; but no longer gives an obstruction to week approximation for VL′ if [L : K] is even.

Using the construction method in Subsubsection 3.3.1, we have the following example, which is a special case of Proposition 3.3.1. It will be used for further discussion. −1 Example 3.3.5. Let K = Q and L = Q(ζ7 + ζ7 ) be as in Example 3.1.3, and let S = 13 . For 132 1 mod7, 412 1 mod7 and 43 1 mod7, the places 13, 41, 43 split {completely} in L.≡ Using the construction≡ method in≡ Subsubsection 3.3.1, we choose data: ′ ′′ S = 13 , S = 13, 29 , S = 5 , v1 = 43, v2 = 41, a = 377, b = 5, c = 878755181 and P{ (x}) = (x2{ 878755181)(5} {x2} 4393775906). Then the Châtelet surface given by y2 377z2 = P (x)−, has the properties− of Proposition 3.3.1. − 16 HAN WU

4. Stoll’s conjecture for curves

Whether all failures of the Hasse principle of smooth, projective, and geometrically connected curves defined over a number field, are explained by the Brauer-Manin obstruction, was considered by Skorobogatov [Sko01, Chapter 6.3] and Scharaschkin [Sch99] indepen- dently. Furthermore, Stoll [Sto07, Conjecture 9.1] made the following conjecture. Given a curve defined over a number field let connected com- C K, C(AK )• = v∈∞K f { ponents of C(Kv) C(A ). The product topology of connected components } × K v∈∞KQ{ of C(K ) with discrete topology and C(Af ) with adelic topology, gives a topology for v } K Q C(AK )•. For any A Br(C), and any v K , the evaluation of A on each connected ∈ ∈ ∞ Br component of C(Kv) is constant. So, the notation C(AK )• makes sense. Conjecture 4.0.1. [Sto07, Conjecture 9.1] For any smooth, projective, and geometrically Br connected curve C defined over a number field K, the set C(K) is dense in C(AK )• . In particular, the curve C satisfies weak approximation with Brauer-Manin obstruction off . ∞K Remark 4.0.2. If Conjecture 4.0.1 holds for a given curve, which is a counterexample to the Hasse principle, then its failure of the Hasse principle is explained by the Brauer-Manin obstruction. For an elliptic curve defined over K, if its Tate-Shafarevich group is finite, then by the dual sequence of Cassels-Tate, Conjecture 4.0.1 holds for this elliptic curve. With the effort of Kolyvagin [Kol90,Kol91], Gross and Zagier [GZ86], and many others, for an elliptic curve defined over Q, if its analytic rank equals zero or one, then its Mordell-Weil rank equals its analytic rank, and its Tate-Shafarevich group is finite. So, Conjecture 4.0.1 holds for this elliptic curve. Definition 4.0.3. Given a nontrivial extension of number fields L/K, let C be a smooth, projective, and geometrically connected curve defined over K. We say that a triple (C,K,L) is of type I if C(K) and C(L) are finite nonempty sets, C(K) = C(L) and Stoll’s Conjecture 4.0.1 holds for the curve C. We say that a triple (C,K,L) is6 of type II if (C,K,L) is of type I, and there exists a real place v′ such that C(L ′ ) is connected. ∈ ∞L v Lemma 4.0.4. Given a nontrivial extension of number fields L/K, if Conjecture 4.0.1 holds for all smooth, projective, and geometrically connected curves defined over K, then there exists a curve C defined over K such that the triple (C,K,L) is of type I. Furthermore, if L has a real place, then this triple (C,K,L) is of type II.

Proof. Since L is a finite separable extension over K, there exists a θ L such that L = K(θ). Let f(x) be the minimal polynomial of θ. Let n = deg(f), then n∈= [L : K] 2. ≥ Let f˜(w0, w1) be the homogenization of f. If n is odd, we consider a curve C defined over n+2 ˜ 2 2 K by a homogeneous equation: w2 = f(w0, w1)(w1 w0) with homogeneous coordinates 2 2 − (w0 : w1 : w2) P . For the polynomials f(x) and x 1 are separable and coprime in K[x], the curve C is∈ smooth, projective, and geometrically− connected. By genus formula for a plane curve, the genus of C equals g(C)= n(n + 1)/2 > 1. By Faltings’s theorem, the sets C(K) and C(L) are finite. It is easy to check that (w0 : w1 : w2)=(1:1:0) C(K) and (θ :1:0) C(L) C(K). By the assumption that Conjecture 4.0.1 holds for∈ all smooth, projective,∈ and geometrically\ connected curves over K, we have that the triple (C,K,L) ′ is of type I. Since n +2 is odd, the space C(Kv′ ) is connected for all v L. So, if L has a real place, then this triple (C,K,L) is of type II. If n is even, we∈ replace ∞ the n+2 ˜ 2 2 n+3 ˜ 2 2 homogeneous equation w2 = f(w0, w1)(w1 w0) by w2 = f(w0, w1)w1(w1 w0). The same argument applies to this new curve and− new triple. −

Remark 4.0.5. For some nonsquare integer d, let K = Q and L = Q(√d). Consider an 2 3 (d) elliptic curve Ed defined by a Weierstraß equation: y = x + d. Let Ed be the quadratic twist of Ed by d. The curve Ed is connected over R. It is easy to check that the point (x, y) = (0, √d) C(L) C(K). If both E (Q) and E(d)(Q) are finite, then the set E (L) ∈ \ d d d is finite, cf. [Sil09, Exercise 10.16]. If additionally, the Tate-Shafarevich group X(Ed, Q) is finite, then the triple (Ed,K,L) is of type II. CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS17

5. Poonen’s proposition

For our result is base on Poonen’s proposition [Poo10, Proposition 5.4]. We recall that paper and his general result first. There exist some remarks on it in [Lia18, Section 4.1]. Colliot-Thélène [CT10, Proposition 2.1] gave another proof of that proposition. Recall 5.0.1. Let B be a smooth, projective, and geometrically connected variety over a number field K. Let be a line bundle on B, assuming the set of global sections ⊗2 L × Γ(B, ) =0. Let = B B . Let a be a constant in K , and let s be a nonzero globalL section6 in Γ(EB, O⊗2).⊕The O ⊕ zero L locus of (1, a, s) Γ(B, ⊗2) L − − ∈ OB ⊕ OB ⊕ L ⊂ Γ(B, Sym2 ) in the projective space bundle Proj( ) is a projective and geometrically in- E E tegral variety, denoted by X with the natural projection X B. Let K be an algebraic closure field of K. Denote B Spec K by B. → ×Spec K Proposition 5.0.2. [Poo10, Proposition 5.3] Given a number field K, all notations are the same as in Recall 5.0.1. Let α: X B be the natural projection. Assume that → the closed subscheme defined by s =0 in B is smooth, projective, and geometrically • connected, BrB =0 and X(A ) = . • K 6 ∅ Then X is smooth, projective, and geometrically connected. And α∗ : Br(B) Br(X) is an isomorphism. →

6. Main results for Châtelet surface bundles over curves

6.1. Preparation Lemmas. We state the following lemmas, which will be used for the proof of our theorems. Fibration methods are used to do research on weak approximation, weak approximation with Brauer-Manin obstruction between two varieties. We modify those ﬁbration methods to ﬁt into our context.

Lemma 6.1.1. Given a number field K, and a finite subset S ΩK , let f : X Y be a K-morphism of proper K-varieties X and Y . We assume that ⊂ → (1) the set Y (K) is finite, (2) the variety Y satisfies weak approximation with Brauer-Manin obstruction off S, (3) for any P Y (K), the fiber X of f over P satisfies weak approximation off S. ∈ P Then X satisfies weak approximation with Brauer-Manin obstruction off S.

′ Proof. For any finite subset S ΩK S, take an open subset N = v∈S′ Uv v∈ /S′ X(Kv) ⊂Br \ × ⊂ X(AK ) such that N X(AK ) = . Let M = v∈S′ f(Uv) v∈ /S′ f(X(Kv)), then by 6 ∅ Br × Q Q the functoriality of Brauer-Manin pairing, M Y (AK ) = . By Assumptions (1) and (2), ST Br Q S6 ∅ Q we have Y (K)= pr (Y (AK ) ). So there exists P0 pr (M) Y (K). Consider the fiber T ∈ XP . Let L = ′ [XP (Kv) Uv] ′ XP (Kv), then it is a nonempty open subset 0 v∈S 0 × v∈ /S ∪S 0 T of X (AS ). By Assumption (3), there exists Q L X (K). So Q X(K) N, which P0 K Q T Q 0 P0 0 implies that X satisfies weak approximation with∈ Brauer-Manin obstructio∈ n off S. T T Lemma 6.1.2. Given a number field K, and a finite subset S ΩK , let f : X Y be a K-morphism of proper K-varieties X and Y . We assume that ⊂ → (1) the set Y (K) is finite, (2) the morphism f ∗ : Br(Y ) Br(X) is surjective, → (3) there exists some P Y (K) such that the fiber XP of f over P does not satisfy weak approximation∈ off S, and that X (K ) = . v∈S P v 6 ∅ Then X does not satisfy weak approximationQ with Brauer-Manin obstruction off S.

Proof. By Assumption (3), take a P Y (K) such that the fiber X does not satisfy weak 0 ∈ P0 approximation off S, and that v∈S XP0 (Kv) = . Then there exist a finite nonempty sub- ′ 6 ∅ set S ΩK S and a nonempty open subset L = ′ Uv ′ XP (Kv) XP (AK ) ⊂ \ Q v∈S × v∈ /S 0 ⊂ 0 Q Q 18 HAN WU

such that L XP0 (K) = . By Assumption (1), the set Y (K) is ﬁnite, so we can take a ∅ ′ Zariski open subset VP0 Y such that VP0 (K) = P0 . For any v S , since Uv is open in X (K ) T f −1(V )(⊂K ), we can take an open{ subset} W of f −∈1(V )(K ) such that P0 v ⊂ P0 v v P0 v ′ ′ Wv XP0 (Kv)= Uv. Consider the open subset N = v∈S Wv v∈ /S X(Kv) X(AK ), then∩L N. By the functoriality of Brauer-Manin pairing and Assumptio× n (2), we⊂ have L ⊂ Br Br Q Q ⊂ N X(AK ) . So N X(AK ) = . But N X(K)= N XP0 (K)= L XP0 (K)= , which implies that X does not satisfy6 ∅ weak approximation with Brauer-Manin obstruction∅ oﬀTS. T T T T

We use the following lemma to choose a dominant morphism from a given curve to P1. Lemma 6.1.3. Given a nontrivial extension of number fields L/K, let C be a smooth, projective, and geometrically connected curve defined over K. Assume that the triple (C,K,L) is of type I (Definition 4.0.3). For any finite K-subscheme R P1 0, , there exists a dominant K-morphism γ : C P1 such that γ(C(L) C(K))⊂ = \{0 ∞}P1(K), γ(C(K)) = P1(K), and that γ is→ étale over R. \ { } ⊂ {∞} ⊂ Proof. Let K(C) be the function field of C. For C(K) and C(L) are finite nonempty sets and C(L) C(K) = , by Riemann-Roch theorem, we can choose a rational function φ K(C)× K\× such6 that∅ the set of its poles contains C(K), and that the set of its zeros ∈ \ 1 contains C(L) C(K). This rational function φ gives a dominant K-morphism γ0 : C P \ 1 1 → such that γ0(C(L) C(K)) = 0 P (K) and γ0(C(K)) = P (K). We can choose \ 1 { }⊂1 {∞} ⊂× an automorphism ϕλ0 : P P , (u : v) (λ0u : v) with λ0 K such that the branch → 7→ −1 ∈ locus of γ0 has no intersection with ϕλ0 (R). Let λ = (ϕλ0 ) γ0. Then the morphism λ is étale over R and satisfies other conditions. ◦

The following lemma is well known. Lemma 6.1.4. Let C be a curve over a field, and let B = C P1. Then BrB =0. × Proof. By [Gro68, III, Corollary 1.2], the Brauer group for a given curve over an algebraic closed field is zero. So Br(C P1) = Br(C)=0. × ∼ Definition 6.1.5. Let C be a smooth, projective, and geometrically connected curve defined over a number field. We say that a morphism β : X C is a Châtelet surface bundle over the curve C, if → X is a smooth, projective, and geometrically connected variety, • the morphism β is faithfully flat, and proper, • the generic fiber of β is a Châtelet surface over the function field of C. • Next, we construct Châtelet surface bundles over curves to give negative answers to Ques- tions 1.2.

6.2. Non-invariance of weak approximation with Brauer-Manin obstruction. For any quadratic extension of number fields L/K, and assuming Conjecture 4.0.1, Liang [Lia18, Theorem 4.5] constructed a Châtelet surface bundle over a curve to give a negative answer to Question 1.2.1. Assuming that the extension L/K is quadratic, he constructed a Châtelet surface defined over K such that the property of weak approximation is not invariant under the extension of L/K. Then choosing a higher genus curve, he combined this Châtelet surface with the construction method of Poonen [Poo10] to get the result. His method only works for quadratic extensions. In this subsection, we generalize his result to any extension L/K. Theorem 6.2.1. For any nontrivial extension of number fields L/K, and any finite subset T ΩL, assuming that Conjecture 4.0.1 holds over K, there exists a Châtelet surface bundle⊂ over a curve: X C defined over K such that → X has a K-rational point, and satisfies weak approximation with Brauer-Manin • obstruction off K , X does not satisfy∞ weak approximation with Brauer-Manin obstruction off T. • L CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS19

Proof. Firstly, we will construct two Châtelet surfaces. Let S ΩK be the set of all ⊂ f restrictions of T on K. By Theorem 2.0.2, we can take a finite place v Ω (S 2 ) 0 ∈ K \ ∪ K splitting completely in L. For the extension L/K, and S = v0 , let V0 be the Châtelet { }2 2 surface chosen as in Subsubsection 3.2.1. Then V0 defined by y az = P0(x) over K 2− 2 having the properties of Proposition 3.2.1. Let P∞(x) = (1 x )(x a), and let V∞ 2 2 − − be the Châtelet surface defined by y az = P∞(x). By the argument in the proof of Proposition 3.2.1, two degree-2 irreducible− factors of P (x) are prime to x2 a in K[x]. So, 0 − the polynomials P0(x) and P∞(x) are coprime in K[x].

Secondly, we will construct a Châtelet surface bundle over a curve. Let P˜∞(x0, x1) and P˜0(x0, x1) be the homogenizations of P∞ and P0. Let (u0 : u1) (x0 : x1) be the coordinates 1 1 ′ 2 ˜ 2 ˜ 1× 1 ⊗2 of P P , and let s = u0P∞(x0, x1)+ u1P0(x0, x1) Γ(P P , (1, 2) ). For P0(x) × ∈ × ′O ′ and P∞(x) are coprime in K[x], by Jacobian criterion, the locus Z defined by s = 0 in pr P1 P1 is smooth. Then the branch locus of the composition Z′ ֒ P1 P1 1 P1, denoted by ×R, is finite over K. By the assumption that Conjecture 4.0.1 holds→ × over K,→ and Lemma 4.0.4, we can take a curve C defined over K such that the triple (C,K,L) is of type I. By Lemma 6.1.3, we can choose a K-morphism γ : C P1 such that γ(C(L) C(K)) = 0 P1(K), γ(C(K)) = P1(K), and that γ →is étale over R. Let B =\ C P1. {Let} ⊂ = (γ,id)∗ (1, 2), and{∞} let ⊂s = (γ,id)∗(s′) Γ(B, ⊗2). For γ is étale over× the branchL locus R, Othe locus Z defined by s = 0 in∈B is smooth.L Since Z is defined by the support of the global section s, it is an effective divisor. The invertible sheaf L (Z′) on P1 P1 is isomorphic to (2, 4), which is a very ample sheaf on P1 P1. And (γ,id) is a finite× morphism, so theO pull back of this ample sheaf is again ample,× which implies that the invertible sheaf L (Z) on C P1 is ample. By [Har97, Chapter III Corollary 7.9], the curve Z is geometrically connected.× So the curve Z is smooth, projective, and geometrically connected. By Lemma 6.1.4, the Brauer group Br(B)=0. Let X be the zero ⊗2 2 locus of (1, a, s) Γ(B, B B ) Γ(B, Sym ) in the projective space bundle Proj( ) with− the− natural∈ projectionO ⊕O ⊕Lα: X ⊂B. Using PropositionE 5.0.2, the variety X is E → pr smooth, projective, and geometrically connected. Let β : X α B = C P1 1 C be the → × → composition of α and pr1. Then β is a Châtelet surface bundle over the curve C. At last, we will check that X has the properties. We will show that X has a K-rational point. For any P C(K), the fiber β−1(P ) = V . ∈ ∼ ∞ The surface V∞ has a K-rational point (x,y,z)=(0, 0, 1), so the set X(K) = . We will show that X satisfies weak approximation with Brauer-Manin obstructio6 ∅ n off . ∞K By Remark 3.0.4, the surface V∞ satisfies weak approximation. So, for the morphism β, Assumption (3) of Lemma 6.1.1 holds. Since Conjecture 4.0.1 holds for the curve C, using Lemma 6.1.1 for the morphism β, the variety X satisfies weak approximation with Brauer-Manin obstruction off . ∞K We will show that XL does not satisfy weak approximation with Brauer-Manin obstruc- ∗ tion off T. By Proposition 5.0.2, the map αL : Br(BL) Br(XL) is an isomorphism, so ∗ → βL : Br(CL) Br(XL) is an isomorphism. By the choice of the curve C and morphism β, → −1 for any Q C(L) C(K), the fiber β (Q) ∼= V0L. By Proposition 3.2.3, the surface V0L does not satisfy∈ weak\ approximation off T . For V (L) = , by Lemma 6.1.2, the va- ∪ ∞L 0 6 ∅ riety XL does not satisfy weak approximation with Brauer-Manin obstruction off T L. So it does not satisfy weak approximation with Brauer-Manin obstruction off T. ∪ ∞

6.3. Non-invariance of the failures of the Hasse principle explained by the Brauer-Manin obstruction. For extensions L/K of the following two cases, assuming Conjecture 4.0.1, we construct Châtelet surface bundles over curves to give negative answers to Question 1.2.2. Theorem 6.3.1. For any number field K, and any nontrivial field extension L of odd degree over K, assuming that Conjecture 4.0.1 holds over K, there exists a Châtelet surface bundle over a curve: X C defined over K such that → X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction, 20 HAN WU

XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

Proof. Firstly, we will construct two Châtelet surfaces. By Theorem 2.0.2, we can take two f different finite places v1, v2 ΩK 2K splitting completely in L. For the extension L/K, and S = v , v , we choose∈ an element\ a as in Subsubsection 3.0.1. For the extension { 1 2} L/K, and S1 = v1 , by Remark 3.0.7, choosing other parameters for the equation (1) as in Subsubsection{ 3.3.1,} we have a Châtelet surface V defined by y2 az2 = P (x) over 0 − 0 K having the properties of Proposition 3.3.1. For the extension L/K, and S2 = v2 , by Remark 3.0.7, choosing other parameters for the equation (1) as in Subsubsection{ 3.1.1,} we have a Châtelet surface V defined by y2 az2 = P (x) over K having the properties ∞ − ∞ of Proposition 3.1.1. By Remark 3.1.2, the polynomial P∞(x) is irreducible over K. For P0(x) is a product of two degree-2 irreducible factors over K, the polynomials P0(x) and P∞(x) are coprime in K[x].

Secondly, we will construct a Châtelet surface bundle over a curve. Let P˜∞(x0, x1) and P˜0(x0, x1) be the homogenizations of P∞ and P0. Let (u0 : u1) (x0 : x1) be the coordinates 1 1 ′ 2 ˜ 2 ˜ 1× 1 ⊗2 of P P , and let s = u0P∞(x0, x1)+ u1P0(x0, x1) Γ(P P , (1, 2) ). For P0(x) × ∈ × ′O ′ and P∞(x) are coprime in K[x], by Jacobian criterion, the locus Z defined by s = 0 in pr P1 P1 is smooth. Then the branch locus of the composition Z′ ֒ P1 P1 1 P1, denoted by ×R, is finite over K. By the assumption that Conjecture 4.0.1 holds→ × over K,→ and Lemma 4.0.4, we can take a curve C defined over K such that the triple (C,K,L) is of type I. By Lemma 6.1.3, we can choose a K-morphism γ : C P1 such that γ(C(L) C(K)) = 0 P1(K), γ(C(K)) = P1(K), and that γ →is étale over R. Let B =\ C P1. {Let} ⊂ = (γ,id)∗ (1, 2), and{∞} let ⊂s = (γ,id)∗(s′) Γ(B, ⊗2). By the same argument× as in theL proof of TheoremO 6.2.1, the locus Z defined∈ by sL= 0 in B is smooth, projective, and geometrically connected; the Brauer group Br(B)=0. Let X be the zero locus of ⊗2 2 (1, a, s) Γ(B, B B ) Γ(B, Sym ) in the projective space bundle Proj( ) with− the− natural∈ projectionO ⊕O ⊕Lα: X ⊂ B. By PropositionE 5.0.2, the variety X is smooth,E → projective, and geometrically connected. Let β : X C be the composition of α and pr1. Then β is a Châtelet surface bundle over the curve→C. At last, we will check that X has the properties. −1 We will show X(AK ) = . For any P C(K), the fiber β (P ) ∼= V∞. By Proposi- 6 {v∅2} ∈ {v2} tion 3.1.1, the set V∞(AK ) = . So X(AK ) = . For v2 splits completely in L, take ′ f 6 ∅′ 6 ∅ a place v Ω above v2, i.e. v v2 in L. Then Kv = Lv′ . By Proposition 3.3.1, the 2 ∈ L 2| 2 2 set V (A ) = . Take a point Q C(L) C(K), then the fiber β−1(Q) = V . We have 0 L 6 ∅ ∈ \ ∼ 0L X(K )= X (L ′ ) β−1(Q)(L ′ ) V ((L ′ ) = . So X(A ) = . v2 L v2 v2 = 0 v2 K ⊃Br ∼ 6 ∅ 6 ∅ We will show X(AK ) = . By Conjecture 4.0.1, the set C(K) is finite, and C(K) = ∞K Br ∅ ∞K Br pr (C(AK ) ). By the functoriality of Brauer-Manin pairing, we have pr (X(AK ) ) −1 ∞K ⊂ P ∈C(K) β (P )(AK ). But by Proposition 3.1.1, the set V∞(Kv2 ) = , so we have Br ∅ pr∞K (X(A ) ) β−1(P )(A∞K ) = V (A∞K ) C(K)= , which implies that F K ⊂ P ∈C(K) K ∼ ∞ K × ∅ X(A )Br = . K F So, the variety∅ X is a counterexample to the Hasse principle, and its failure of the Hasse principle is explained by the Brauer-Manin obstruction. Br ∗ We will show XL(AL) = . By Proposition 5.0.2, the map αL : Br(BL) Br(XL) is ∗ 6 ∅ → an isomorphism, so βL : Br(CL) Br(XL) is an isomorphism. By the functoriality of → Br Brauer-Manin pairing, the set X (A ) contains β−1(Q)(A ) = V (A ) L L Q∈C(L)\C(K) L ∼ 0 L × (C(L) C(K)), which is nonempty. F We will\ show X(L)= . By the assumption that the degree [L : K] is odd, and v splitting ∅ 1 completely in L, the number ♯S1L is odd. By Proposition 3.3.1 and the global reciprocity law explained in Remark 3.3.2, the set V0(L)= . By Proposition 3.1.1, the set V∞(AL)= . Since each L-rational fiber of β is isomorphic to∅ V or V , the set X(L)= . ∅ 0L ∞L ∅ So, the variety XL is a counterexample to the Hasse principle, but its failure of the Hasse principle cannot be explained by the Brauer-Manin obstruction. CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS21

In the case when both ﬁelds K and L have real places, making use of these real place information, and assuming the Stoll’s conjecture, we have the following theorem to give a negative answer to Question 1.2.2

Theorem 6.3.2. For any number field K having a real place, and any nontrivial field extension L/K having a real place, assuming that Conjecture 4.0.1 holds over K, there exists a Châtelet surface bundle over a curve: X C defined over K such that → X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction, XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

′ Proof. Firstly, we will construct two Châtelet surfaces. Take a real place v0 of L, and ′ let v0 K be the restriction of v0 on K. By Theorem 2.0.2, we can take a ﬁnite place ∈f ∞ v1 ΩK 2K splitting completely in L. For the extension L/K, and S = v0, v1 , we choose∈ an\ element a as in Subsubsection 3.0.1. For the trivial extension K/K (respectively{ } the extension L/K), and S1 = v0 (respectively S2 = v1 ), by Remark 3.0.7, choosing other parameters for the equation{ (1)} as in Subsubsection{ 3.1.1,} we have a Châtelet surface 2 2 2 2 V0 (respectively V∞) deﬁned by y az = P0(x) (respectively by y az = P∞(x)) over K having the properties of Proposition− 3.1.1. By Remark 3.1.2,− the polynomial

P∞(x) is irreducible over Kv1 . If the polynomials P0(x) and P∞(x) are not coprime, then P (x)= λP (x) for some λ K×. By Remark 3.1.2, we choose another P (x)′ to replace 0 ∞ ∈ ∞ P∞(x) so that the new polynomial P∞(x) is prime to P0(x).

Br {v0} We will show XL(AL) = . By Proposition 3.1.1, the set V0(AK ) = and V∞(Kv0 ) = . By Proposition 5.0.2, the6 map∅ β∗ : Br(C ) Br(X ) is an isomorphism.6 ∅ By our choice,6 the∅ L L → L space C(Kv ) = C(Lv′ ) = C(R) is connected. For any A Br(CL), since the evaluation 0 ∼ 0 ∼ ∈ of A on C(L ′ ) is locally constant for all v′ Ω , it is constant on C(K ), so it is v ∈ L v0 constant on C(L ′ ) for all v′ S . Take points P C(K) and Q C(L) C(K). By the v ∈ 1L ∈ ∈ \ 22 HAN WU

∗ functoriality of Brauer-Manin pairing and isomorphism of βL : Br(CL) Br(XL), the set Br −1 S1L −1 S1L → XL(AL) β (Q)(A ) ′ β (P )(Lv′ ) = V0(A ) ′ V∞(Lv′ ) = . ⊃ L × v ∈S1L ∼ L × v ∈S1L 6 ∅ We will show X(L) = . By Proposition 3.1.1, the set V0(Kv ) = , so the set V0(Lv′ ) = ∅ Q 0 Q∅ 0 V (K )= . By Proposition 3.1.1, the set V (A )= . Since each L-rational ﬁber of β is 0 v0 ∅ ∞ L ∅ isomorphic to V0L or V∞L, the set X(L)= . So, the variety X is a counterexample to the∅ Hasse principle, but its failure of the Hasse principle cannot be explained by the Brauer-Manin obstruction.

7. Explicit unconditional examples

Firstly, we will give explicit examples without assuming Conjecture 4.0.1 for Theorem 6.2.1, Theorem 6.3.1 and Theorem 6.3.2. Secondly, when K = Q and L = Q(i), besides cases of Theorem 6.3.1 and Theorem 6.3.2, we construct an explicit Châtelet surface bundle over a curve in Subsection 7.4 to give a negative answer to Question 1.2.2.

7.1. An explicit unconditional example for Theorem 6.2.1. In the subsection, let K = Q and L = Q(√3). We will construct an explicit Châtelet surface bundle over a curve having properties of Theorem 6.2.1.

7.1.1. Choosing an elliptic curve. Let E be an elliptic curve defined over Q by a homogeneous equation: w2w = w3 16w3 1 2 0 − 2 with homogeneous coordinates (w : w : w ) P2. This is an elliptic curve with complex 0 1 2 ∈ multiplication. Its quadratic twist E(3) is isomorphic to an elliptic curve defined by a 2 3 3 homogeneous equation: w1w2 = w0 432w2 with homogeneous coordinates (w0 : w1 : 2 −(3) w2) P . These elliptic curves E and E defined over Q, are of analytic rank 0. Then the Tate-Shafarevich∈ group X(E, Q) is finite, so E satisfies weak approximation with Brauer- Manin obstruction off . The Mordell-Weil groups E(K) and E(3)(K) are finite, so E(L) ∞K is finite. Indeed, the Mordell-Weil groups E(K) = (0:1:0) and E(L) = (4 : 4√3 : 1), (0:1:0) . So the triple (E,K,L) is of type I. { } { ± } 7.1.2. Choosing a dominant morphism. Let P2 (0 : 1 : 0) P1 be a morphism over \{ } → Q given by (w0 : w1 : w2) (w0 4w2 : w2). Composite with the natural inclusion E (0:1:0) ֒ P2 (0:1:0)7→ . We− get a morphism E (0:1:0) P1, which can be extended\{ to a} dominant→ \{ morphism} γ : E P1 of degree\{2. The morphism} → γ maps E(K) → to = (1 : 0) , and maps (4 : 4√3 : 1) to 0 point: (0 : 1). By Bézout’s Theorem [Har97,{∞} Chapter{ I.} Corollary 7.8] or Hurwitz’s± Theorem [Har97, Chapter IV. Corollary 2.4], the branch locus of γ is (1 : 0), (2√3 2 4:1), (2√3 2e2πi/3 4:1), (2√3 2e−2πi/3 4:1) . { − − − } 2 2 7.1.3. Construction of a Châtelet surface bundle. Let P∞(x) =(1 x )(x 73), and let 2 2 − − P0(x) = (99x + 1)(5428x /5329+1/5329). Notice that these polynomials P∞ and P0 are separable. Let V be the Châtelet surface given by y2 73z2 = P (x). As mentioned in ∞ − ∞ Example 3.2.5, let V be the Châtelet surface given by y2 73z2 = P (x). Let P˜ (x , x ) 0 − 0 ∞ 0 1 and P˜0(x0, x1) be the homogenizations of P∞ and P0. Let (u0 : u1) (x0 : x1) be the 1 1 ′ 2 ˜ 2 ˜ 1 × 1 ⊗2 coordinates of P P , and let s = u0P∞(x0, x1)+u1P0(x0, x1) Γ(P P , (1, 2) ). For × ∈ × ′ O ′ P0(x) and P∞(x) are coprime in K[x], by Jacobian criterion, the locus Z defined by s =0 pr in P1 P1 is smooth. Then the branch locus of the composition Z′ ֒ P1 P1 1 P1, denoted by R,×is finite, and contained in P1 (1 : 0) . Let B = E P1. Let→ =× (γ,id→)∗ (1, 2), and let s = (γ,id)∗(s′) Γ(B, ⊗2). With\{ these} notations, we× have theL following lemma.O ∈ L Lemma 7.1.1. The curve Z defined by s =0 in B is smooth, projective, and geometrically connected.

Proof. For smoothness of Z, we need to check that the branch locus R does not intersect with the branch locus of γ : E P1. For R is contained in P1 (1 : 0) , we can assume the → \{ } homogeneous coordinate u1 =1, then the point in R satisﬁes one of the following equations: 5329u2 537372 = 0, 389017u2 1=0, 27625536u4 + 157730624u2 +5329 = 0. The 0 − 0 − 0 0 CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS23 polynomials of these equations are irreducible over Q. By comparing the degree [Q(u0): Q] with the branch locus of γ, we get the conclusion that these two branch loci do not intersect. The same argument as in the proof of Theorem 6.2.1, the locus Z deﬁned by s =0 in B is geometrically connected. So it is smooth, projective, and geometrically connected.

⊗2 2 Let X be the zero locus of (1, a, s) Γ(B, B B ) Γ(B, Sym ) in the projective space bundle Proj( −) with− the∈ naturalO projection⊕ O ⊕ L α: X⊂ B. ByE the same argument as in the proof of TheoremE 6.3.1, the variety X is smooth,→ projective, and geometrically connected. Let β : X E be the composition of α and pr1. Then it is a Châtelet surface bundle over the curve→ E. For this X, we have the following proposition. Proposition 7.1.2. For K = Q and L = Q(√3), the 3-fold X has the following properties. X has a K-rational point, and satisfies weak approximation with Brauer-Manin • obstruction off K . X does not satisfy∞ weak approximation with Brauer-Manin obstruction off . • L ∞L Proof. This is the same as in the proof of Theorem 6.2.1.

The 3-fold X that we constructed, has an affine open subvariety defined by the following equations, which is a closed subvariety of A5 with affine coordinates (x,y,z,x′,y′). y2 73z2 = (1 x2)(x2 73)(x′ 4)2 + (99x2 + 1)(5428x2/5329+ 1/5329) − − − − y′2 = x′3 16 ( − 7.2. An explicit unconditional example for Theorem 6.3.1. In the subsection, let −1 K = Q, and let ζ7 be a primitive 7-th root of unity. Let α = ζ7 + ζ7 with the minimal polynomial x3 + x2 2x 1. Let L = Q(α). Then the degree [L : K]=3. We will construct an explicit Châtelet− surface− bundle over a curve having properties of Theorem 6.3.1.

7.2.1. Choosing an elliptic curve. Let E be an elliptic curve defined over Q by a homogeneous equation: 2 3 2 3 w1w2 = w0 343w0w2 2401w2 − 2 − with homogeneous coordinates (w0 : w1 : w2) P . This elliptic curve defined over Q, is of analytic rank 0, so it satisfies weak approximation∈ with Brauer-Manin obstruction off K . By computer calculation, we have the Mordell-Weil groups E(K)= (0:1:0) and ∞E(L)= (7α2 +14α 7:0:1), (7α2 7α 14:0:1), ( 14α2 7α +21:0:1){ , (0:1:0)} . So the triple{ (E,K,L−) is of type I. − − − − }

7.2.2. Choosing a dominant morphism. Let P2 (1:0:0) P1 be a morphism over Q \{ } → 2 : given by (w0 : w1 : w2) (w1 : w2). Composite with the natural inclusion E ֒ P (1 0 : 0) . We get a morphism7→ γ : E P1, which is a dominant morphism of degree→ 3\{. The morphism} γ maps E(K) to =→(1 : 0) , and maps E(L) E(K) to 0 = (0 : 1) . By Bézout’s Theorem [Har97, Chapter{∞} I.{ Corollary} 7.8] or Hurwitz’s\ Theorem{ } [Har97,{ Chapter} 4 2 IV. Corollary 2.4], the branch locus of γ is (1 : 0) (u0 : 1) 27u0 +129654u0 5764801 = 0 . { } { | − } S 4 7.2.3. Construction of a Châtelet surface bundle. Let P∞(x) = 14(x 89726), and let P (x) = (x2 878755181)(5x2 4393775906). Notice that these polynomials− P and 0 − − ∞ P0 are separable. As mentioned in Example 3.1.3 and Example 3.3.5, let V∞ be the 2 2 Châtelet surface given by y 377z = P∞(x), and let V0 be the Châtelet surface given 2 2 − by y 377z = P0(x). Let P˜∞(x0, x1) and P˜0(x0, x1) be the homogenizations of P∞ and − 1 1 ′ 2 ˜ P0. Let (u0 : u1) (x0 : x1) be the coordinates of P P , and let s = u0P∞(x0, x1)+ 2 ˜ ×1 1 ⊗2 × u1P0(x0, x1) Γ(P P , (1, 2) ). For P0(x) and P∞(x) are coprime in K[x], by Jacobian criterion, the∈ locus Z×′ deﬁnedO by s′ = 0 in P1 P1 is smooth. Then the branch locus of pr1 × . (the composition Z′ ֒ P1 P1 P1, denoted by R, is ﬁnite and contained in P1 (1 : 0 Let B = E P1. Let→ =× (γ,id→)∗ (1, 2), and let s = (γ,id)∗(s′) Γ(B, ⊗2). With\{ these} notations, we× have theL following lemma.O ∈ L 24 HAN WU

Lemma 7.2.1. The curve Z deﬁned by s =0 in B is smooth, projective, and geometrically connected.

Proof. For smoothness of Z, we need to check that the branch locus R does not intersect with the branch locus of γ : E P1. For R is contained in P1 (1 : 0) , we can → \{ } assume the homogeneous coordinate u1 = 1, then the point in R satisﬁes one of the 2 2 4 following equations: 14u0 +5=0, 44863u0 137894762198231040 = 0, 70345184u0 2 − − 216218987126801139936u0 +1=0. The polynomials of these equations are irreducible over Q. By comparing these irreducible polynomials with the branch locus of γ, we get the conclusion that these two branch loci do not intersect. The same argument as in the proof of Theorem 6.2.1, the locus Z deﬁned by s = 0 in B is geometrically connected. So it is smooth, projective, and geometrically connected.

⊗2 2 Let X be the zero locus of (1, a, s) Γ(B, B B ) Γ(B, Sym ) in the projective space bundle Proj( −) with− the∈ naturalO projection⊕ O ⊕ L α: X⊂ B. ByE the same argument as in the proof of TheoremE 6.3.1, the variety X is smooth,→ projective, and geometrically connected. Let β : X E be the composition of α and pr1. Then it is a Châtelet surface bundle over the curve→ E. For this X, we have the following proposition. −1 Proposition 7.2.2. For K = Q and L = Q(ζ7 + ζ7 ), the 3-fold X has the following properties. X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction. XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

Proof. This is the same as in the proof of Theorem 6.3.1.

The 3-fold X that we constructed, has an affine open subvariety defined by the following equations, which is closed subvariety of A5 with affine coordinates (x,y,z,x′,y′). y2 377z2 = 14(x4 89726)y′2 + (x2 878755181)(5x2 4393775906) − − − − y′2 = x′3 343x′ 2401 ( − − 7.3. An explicit unconditional example for Theorem 6.3.2. In the subsection, let K = Q and L = Q(√3). We will construct an explicit Châtelet surface bundle over a curve having properties of Theorem 6.3.2.

7.3.1. Choosing an elliptic curve and a dominant morphism. Let E and γ : E P1 be the same as in Subsection 7.1. For E(R) is connected, the triple (E,K,L) is of type→ II.

4 7.3.2. Construction of a Châtelet surface bundle. Let P∞(x)=5(x +805), and let P0(x)= 4 5(x + 115). Notice that these polynomials P∞ and P0 are irreducible. As mentioned in − 2 2 Example 3.1.4 and Example 3.1.5, let V∞ be the Châtelet surface given by y + 23z = 2 2 P∞(x), and let V0 be the Châtelet surface given by y +23z = P0(x). Let P˜∞(x0, x1) and P˜0(x0, x1) be the homogenizations of P∞ and P0. Let (u0 : u1) (x0 : x1) be the coordinates 1 1 ′ 2 ˜ 2 ˜ 1× 1 ⊗2 of P P , and let s = u0P∞(x0, x1)+ u1P0(x0, x1) Γ(P P , (1, 2) ). For P0(x) × ∈ × ′O ′ and P∞(x) are coprime in K[x], by Jacobian criterion, the locus Z defined by s = 0 in pr P1 P1 is smooth. Then the branch locus of the composition Z′ ֒ P1 P1 1 P1, denoted by ×R, is finite, and contained in P1 (1 : 0) . Let B = E P1. Let→ =× (γ,id→)∗ (1, 2), and let s = (γ,id)∗(s′) Γ(B, ⊗2). With\{ these} notations, we× have theL following lemma.O ∈ L Lemma 7.3.1. The curve Z defined by s =0 in B is smooth, projective, and geometrically connected.

Proof. For smoothness of Z, we need to check that the branch locus R does not intersect with the branch locus of γ : E P1. For R is contained in P1 (1 : 0) , we can assume the → \{ } homogeneous coordinate u1 =1, then the point in R satisﬁes one of the following equations: CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS25

2 2 u0 1=0, 7u0 1=0. By comparing this locus with the branch locus of γ, these two branch− loci do not− intersect. The same argument as in the proof of Theorem 6.2.1, the locus Z deﬁned by s =0 in B is geometrically connected. So it is smooth, projective, and geometrically connected.

⊗2 2 Let X be the zero locus of (1, a, s) Γ(B, B B ) Γ(B, Sym ) in the projective space bundle Proj( −) with− the∈ naturalO projection⊕ O ⊕ L α: X⊂ B. ByE the same argument as in the proof of TheoremE 6.3.1, the variety X is smooth,→ projective, and geometrically connected. Let β : X E be the composition of α and pr1. Then it is a Châtelet surface bundle over the curve→ E. For this X, we have the following proposition. Proposition 7.3.2. For K = Q and L = Q(√3), the 3-fold X has the following properties. X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction. XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

Proof. This is the same as in the proof of Theorem 6.3.2.

The 3-fold X that we constructed, has an affine open subvariety defined by the following equations, which is closed subvariety of A5 with affine coordinates (x,y,z,x′,y′). y2 + 23z2 = 5(x4 + 805)(x′ 4)2 5(x4 + 115) − − y′2 = x′3 16 ( − 7.4. Exceptions. For Question 1.2.2, when the degree [L : K] is even and L has no real place, besides cases of Theorem 6.3.1 and Theorem 6.3.2, we can give some unconditional examples, case by case, to give negative answers to Question 1.2.2, although we do not have a uniform way to construct them. In this subsection, we give an explicit example to explain how it works for the case that K = Q and L = Q(i).

7.4.1. Choosing an elliptic curve. Let E be an elliptic curve defined over Q by a homogeneous equation: w2w = w3 16w3 1 2 0 − 2 with homogeneous coordinates (w : w : w ) P2. This is an elliptic curve with complex 0 1 2 ∈ multiplication. Its quadratic twist E(−1) is isomorphic to an elliptic curve defined by a homogeneous equation: w2w = w3 +16w3 with homogeneous coordinates (w : w : w ) 1 2 0 2 0 1 2 ∈ P2. These elliptic curves E and E(−1) defined over Q, are of analytic rank 0. Then the Tate- Shafarevich group X(E, Q) is finite, so E satisfies weak approximation with Brauer-Manin (−1) obstruction off K . The Mordell-Weil groups E(K) and E (K) are finite, so E(L) is finite. Indeed, the∞ Mordell-Weil group E(K)= (0:1:0) and E(L)= (0 : 4i : 1), (0 : 1:0) . So the triple (E,K,L) is of type I. { } { ± } 7.4.2. Choosing a dominant morphism. Let P2 (1:0:0) P1 be a morphism over Q \{ } → 2 : given by (w0 : w1 : w2) (w1 :4w2). Composite with the natural inclusion E ֒ P (1 0 : 0) , then we get a morphism7→ γ : E P1, which is a dominant morphism of→ degree\{ 3. The dominant} morphism γ maps E(K→) to = (1 : 0) , and maps (0 : 4i : 1) to ( i : 1). By Bézout’s Theorem [Har97, Chapter{∞} I. Corollary{ } 7.8] or Hurwitz’s± Theorem [Har97,± Chapter IV. Corollary 2.4], the branch locus of γ is (1 : 0), ( i : 1) . { ± } 4 2 4 7.4.3. Properties of Châtelet surfaces. Let P∞(x)=2(x 10x + 15), P0(x)= 2(5x 2 − − − 39x +75), and let P1(x)= P∞(x)+iP0(x). Notice that all those polynomials P∞, P0, P1 are 2 2 2 separable, and P1(x)= 2[x (5+i)][( 1+5i)x 15i]. The two polynomials x (5+i) 2 − − − − − and ( 1+5i)x 15i are irreducible over Q(i) (indeed, they are irreducible over Q(i)3). − − 2 2 Let V∞ be the Châtelet surface over Q given by y + 15z = P∞(x), and let V1 be the 2 2 Châtelet surface over Q(i) given by y + 15z = P1(x). With these notations, we have the following lemmas. 26 HAN WU

Lemma 7.4.1. The Châtelet surface V given by y2 + 15z2 = 2(x4 10x2 + 15), has a ∞ − Q -point for all v ΩQ 5 , but no Q -point. v ∈ \{ } 5

Proof. Suppose that v = Q. Let x0 = 0. Then ( 15, P∞(x0))v = ( 15, 30)v = 1, which ∞0 − − implies that the surface V∞ admits a R-point with x =0. Suppose that v =2. For 15 1 mod8, by Hensel’s lemma, the element 15 is a square − ≡ − in Q2. By Remark 3.0.1, the surface V∞ admits a Q2-point. Suppose that v =3. Let x0 =2. Then ( 15, P∞(x0))v = ( 15, 18)v = ( 15, 9)v( 15, 2)v = 0 − − − − − − 1, which implies that the surface V∞ admits a Q3-point with x =2. Suppose that v Ωf 2, 3, 5 . Take x Q such that the valuation v(x ) < 0. Then by ∈ Q\{ } 0 ∈ v 0 Lemma 2.2.2, we have ( 15, P (x )) = ( 15, 2) = 1, which implies that V 0 admits a − ∞ 0 v − v ∞ Qv-point with x = x0. Suppose that v = 5. Then ( 15, 2)v = 1. Let x Q5. If v(x) 0, then by Lemma 2.2.2, we have ( 15, P (x)) −= ( 15, 2x4−) = 1. If∈v(x) > 0, then≤ by Lemma 2.2.2, we − ∞ v − v − have ( 15, P∞(x))v = ( 15, 30)v = 1. In each case, we have ( 15, P∞(x))v = 1, which implies− that V 0 has no −Q -point. By− Remark 3.0.2, we have V −(Q )= . − ∞ 5 ∞ 5 ∅ ′ 2 2 2 Lemma 7.4.2. For any v ΩQ , the Châtelet surface V given by y +15z = 2[x ∈ (i) 1 − − (5 + i)][( 1+5i)x2 15i], has a Q(i) ′ -point. − − v Proof. For the only archimedean place is complex, we only need to consider ﬁnite places. Suppose that v′ is a 2-adic place. For 15 Q×2, by Remark 3.0.1, the surface V admits − ∈ 2 1 a Q(i)v′ -point. ×2 Suppose that v′ = 3. For 2 Q , we have ( 15, 2) ′ = 1. By Lemma 2.2.2, we − ∈ 3 − − v have ( 15, ( 4 i)( 1 10i)) ′ = ( 15, (1 + i)2) ′ . Let x = 1. Then ( 15, P (x )) ′ = − − − − − v − v 0 − 1 0 v ( 15, 2( 4 i)( 1 10i)) ′ = ( 15, (1 + i)2) ′ = 1, which implies that V 0 admits a − − − − − − v − v 1 Q(i)3-point with x =1. Suppose that v′ 5, then Q(i) ′ = Q . By Lemma 2.2.2, we have ( 15, 2( 5+ i)( 10 | v ∼ 5 − − − − − 17i))v′ = ( 15, 34)v′ . Let x0 =1+ i. Then ( 15, P1(x0))v′ = ( 15, 2( 5+ i)( 10 ′ − − ′ −0 ′− − − − − 17i))v = ( 15, 34)v =1, which implies that V1 admits a Q(i)v -point with x =1+ i. − ′ − Suppose that v 13, then Q(i)v′ ∼= Q13. Let x0 =1. By Lemma 2.2.1, we have ( 15, P1(x0))v′ = | ′ 0 ′ − ( 15, 2( 4 i)( 1 10i))v =1, which implies that V1 admits a Q(i)v -point with x =1. − − − − − −f Suppose that v′ Ω 2, 3, 5, 13 . Take x Q ′ such that the valuation v′(x ) < 0. ∈ Q(i)\{ } 0 ∈ v 0 By Lemma 2.2.2, we have ( 15, P (x )) ′ = ( 15, 2( 1+5i)x4) ′ . By Lemma 2.2.1, we − ∞ 0 v − − − 0 v have ( 15, 2( 1+5i)) ′ = 1, so ( 15, P (x )) ′ = 1, which implies that V 0 admits a − − − v − ∞ 0 v 1 Q(i)v′ -point with x = x0.

For P1(x) is a product of two degree-2 irreducible factors, according to [Sko01, Proposition 7.1.1], the Brauer group Br(V1)/Br(Q(i)) = Z/2Z. Furthermore, by Proposition 7.1.2 in ∼ 2 loc. cit, we take the quaternion algebra A = ( 15, ( 1+5i)x 15i) Br(V1) as a generator element of this group. Then we have the− equality− A = ( 15−, ( 1+5∈ i)x2 15i)= ( 15, 2(x2 (5 + i))) in Br(V ). With these notations, we have− the following− lemmas.− − − − 1 ′ Lemma 7.4.3. For any v ΩQ and any P ′ V (Q(i) ′ ), ∈ (i) v ∈ 1 v 0 if v′ =3 ′ ′ invv (A(Pv )) = ′ 6 (1/2 if v =3

Proof. By Remark 3.0.2, it suﬃces to compute the local invariant invv′ (A(Pv′ )) for all P ′ V 0(Q(i) ′ ). v ∈ 1 v Suppose that v′ is an archimedean place or a 2-adic place or a 181-adic place. Then ×2 ′ ′ ′ ′ 15 Q(i)v′ , so invv (A(Pv ))=0 for all Pv V1(Q(i)v ). − ∈ ′ ′ ∈ Suppose that v 5. Let x Q(i)v′ . If v (x) 0, then by Lemma 2.2.2, we have ( 15, ( 1+ 2 | ∈2 ′ ≤ − − 2 5i)x 15i)v′ = ( 15, x )v′ =1. If v (x) > 0, then by Lemma 2.2.2, we have ( 15, 2(x − ′ − −′ ′ ′ ′ 0 ′ − − − (5 + i)))v = ( 15, 2i)v =1. So invv (A(Pv ))=0 for all Pv V1 (Q(i)v ). − f ∈ Suppose that v′ Ω 3, all 2-adic, 5-adic and 181-adic places . Take an arbitrary P ′ ∈ Q(i)\{ } v ∈ V 0(Q(i) ′ ). If inv ′ (A(P ′ ))=1/2, then ( 15, ( 1+5i)x2 15i) ′ = 1 = ( 15, 2(x2 1 v v v − − − v − − − − CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS27

′ 2 (5 + i)))v′ at Pv′ . By Lemma 2.2.1, the ﬁrst and last equalities imply that v (( 1+5i)x 15i) and v′(x2 (5 + i)) are odd, so they are positive. Hence v′(( 1+5i−)x2 15i − ( 1+5i)(x2 −(5 + i))) = v′( 10+9i) > 0. But v′ ∤ 181, which is− a contradiction.− So− − − − invv′ (A(Pv′ ))=0.

′ ′ 0 ′ ′ ′ Suppose that v = 3. Take an arbitrary Pv V1 (Q(i)v ). If invv (A(Pv )) = 1, then 2 2 ∈ ( 15, ( 1+5i)x 15i)v′ =1=( 15, 2(x (5+i)))v′ at Pv′ . For ( 15, 1+5i)v′ = 1, the− ﬁrst− equality− implies that v′(−x) >−0. Then− by Lemma 2.2.2, we− have−( 15, x2 (5− + − − i)) ′ = ( 15, (5 + i)) ′ = 1, so ( 15, 2(x2 (5 + i))) ′ = ( 15, 2) ′ ( 15, 5 i) ′ = v − − v − − − − v − − v − − − v 1, which is a contradiction. So inv ′ (A(P ′ ))=1/2. − v v Lemma 7.4.4. The Châtelet surface V1 has no Q(i)-rational point.

Proof. If there exists Q(i)-rational point P, by the global reciprocity law inv (A(P )) = v∈ΩQ(i) v 0 in Q/Z. But from Lemma 7.4.3, this sum is 1/2, which is nonzero in Q/Z. So V has no P 1 Q(i)-rational point.

7.4.4. Construction of a Châtelet surface bundle. Let P˜∞(x0, x1) and P˜0(x0, x1) be the 1 1 homogenizations of P∞ and P0. Let (u0 : u1) (x0 : x1) be the coordinates of P P , ′ 2 2 ˜ ˜ × 1 1 ⊗2 × and let s = (u0 +2u1)P∞(x0, x1)+ u0u1P0(x0, x1) Γ(P P , (1, 2) ). By Jacobian criterion, the locus Z′ defined by s′ = 0 in P1 P1 ∈is smooth.× ThenO the branch locus of pr1 × the composition Z′ ֒ P1 P1 P1 is finite, and contained in P1 (1 : 0), ( i, 1) . Let B = E P1. Let →= (γ,id× )∗ →(1, 2), and let s = (γ,id)∗(s′) Γ(\{B, ⊗2). ±With} these notations,× we haveL the followingO lemma. ∈ L Lemma 7.4.5. The curve Z defined by s =0 in B is smooth, projective, and geometrically connected.

pr Proof. For the branch locus of the composition Z′ ֒ P1 P1 1 P1 is contained in P1 (1 : 0), ( i, 1) , and the branch locus of γ : E → P1 is× (1→ : 0), ( i, 1) , they do not\{ intersect,± which} implies the smoothness of Z. The→ same argument{ as± in the} proof of Theorem 6.2.1, the locus Z deﬁned by s = 0 in B is geometrically connected. So it is smooth, projective, and geometrically connected.

⊗2 2 Let X be the zero locus of (1, a, s) Γ(B, B B ) Γ(B, Sym ) in the projective space bundle Proj( −) with− the∈ naturalO projection⊕ O ⊕ L α: X⊂ B. ByE the same argument as in the proof of TheoremE 6.3.1, the variety X is smooth,→ projective, and geometrically connected. Let β : X E be the composition of α and pr1. Then it is a Châtelet surface bundle over the curve→ E. For this X, we have the following proposition. Proposition 7.4.6. For K = Q and L = Q(i), the 3-fold X has the following properties. X is a counterexample to the Hasse principle, and its failure of the Hasse principle • is explained by the Brauer-Manin obstruction. XL is a counterexample to the Hasse principle, but its failure of the Hasse principle • cannot be explained by the Brauer-Manin obstruction.

Proof. Let σ be the generator element of Galois group Gal(L/K). We will show X(A ) = . K 6 ∅ By our construction, each K-rational ﬁber of β is isomorphic to V∞. By Lemma 7.4.1, the set V (A{5}) = . So X(A{5}) = . For 5 splits completely in L, take a place v′ Ω ∞ K 6 ∅ K 6 ∅ ∈ L above 5, i.e. v′ 5 in L. Then Q = L ′ . By Lemma 7.4.2, the set V (A ) = . Since | 5 ∼ v 1 L 6 ∅ V σ(V ) = β−1(P ) X , the set X(Q ) = X(L ′ ) V (L ′ ) = . So 1 1 ∼ P ∈E(L)\E(K) ⊂ L 5 v ⊃ 1 v 6 ∅ X(AK ) = . F 6 ∅ F Br ∞K Br We will show X(AK ) = . For E(K) = pr (C(AK ) ), the functoriality of Brauer- ∞K∅ Br −1 ∞K Manin pairing implies pr (X(AK ) ) P ∈E(K) β (P )(AK ). By Lemma 7.4.1, the Br ⊂ set V (Q ) = , so pr∞K (X(A ) ) β−1(P )(A∞K ) = V (A∞K ) = , which ∞ 5 ∅ K ⊂ PS∈E(K) K ∼ ∞ K ∅ implies that X(A )Br = . K S So, the variety X is a counterexample∅ to the Hasse principle, and its failure of the Hasse principle is explained by the Brauer-Manin obstruction. 28 HAN WU

Br ∗ We will show XL(AL) = . By Proposition 5.0.2, the map αL : Br(CL) Br(XL) is an 6 ∅ → Br isomorphism. By the functoriality of Brauer-Manin pairing, the set XL(AL) contains V1(AL), which is nonempty. We will show X(L) = . By Lemma 7.4.4, the set V1(L) = , so also σ(V1)(L) = . For 5 splits completely in L,∅ the emptiness of V (Q ) implies V∅ (A ) = . Since X(∅L) ∞ 5 ∞ L ∅ ⊂ β−1(P )(L) = V (L) V (L) σ(V )(L), the set X(L)= . P ∈E(L) ∼ ∞ 1 1 ∅ So, the variety X is a counterexample to the Hasse principle, but its failure of the Hasse F L S S principle cannot be explained by the Brauer-Manin obstruction.

The 3-fold X that we constructed, has an affine open subvariety defined by the following equations, which is closed subvariety of A5 with affine coordinates (x,y,z,x′,y′). y2 + 15z2 = (x4 10x2 + 15)(y′2 + 32)/8 (5x4 39x2 + 75)y′/2 − − − y′2 = x′3 16 ( − Acknowledgements. The author would like to thank my thesis advisor Y. Liang for proposing the related problems, papers and many fruitful discussions. The author was partially supported by NSFC Grant No. 12071448.

References

[CF67] J. Cassels and A. Fröhlich, Algebraic number theory, Academic Press, 1967. ↑2 [CT10] J.-L. Colliot-Thélène, Zéro-cycles de degré 1 sur les solides de Poonen, Bull. Soc. Math. France 138 (2010), 249–257 (French). ↑5 [CTPS16] J.-L. Colliot-Thélène, A. Pál, and A. Skorobogatov, Pathologies of the Brauer-Manin obstruction, Math. Z. 282 (2016), 799–817. ↑1.1, 3.0.3 [CTSSD87a] J.-L. Colliot-Thélène, J.-J. Sansuc, and S. Swinnerton-Dyer, Intersections of two quadrics and Châtelet surfaces I, J. Reine Angew. Math. 373 (1987), 37–107. ↑1.1, 3.0.3, 3.2.1, 3.3.2 [CTSSD87b] , Intersections of two quadrics and Châtelet surfaces II, J. Reine Angew. Math. 374 (1987), 72–168. ↑1.1, 3.0.3, 3.2.1, 3.3.2 [Gro68] A. Grothendieck, Le groupe de Brauer III: Exemples et compléments. In: Dix exposés sur la cohomologie des schémas, Advanced Studies in Pure Mathematics, vol. 3, North-Holland, 1968 (French). pp. 88-188. ↑6.1 [GZ86] B. Gross and D. Zagier, Heegner points and derivatives of L-series, Invent. Math. 84 (1986), 225–320. ↑4.0.2 [Har97] R. Hartshorne, Algebraic geometry, Graduate Texts in Mathematics, vol. 52, Springer-Verlag, 1997. ↑6.2, 7.1.2, 7.2.2, 7.4.2 [HS14] Y. Harpaz and A. Skorobogatov, Singular curves and the étale Brauer-Manin obstruction for surfaces, Ann. Sci. Éc. Norm. Supér. 47 (2014), 765–778. ↑1.1 [Isk71] V. Iskovskikh, A counterexample to the Hasse principle for systems of two quadratic forms in five variables, Mat. Zametki 10 (1971), 253–257. ↑1.3, 3.3 [Kol90] V. Kolyvagin, Euler systems. In: The Grothendieck festschrift II, Progress in Mathematics, vol. 87, Birkhäuser, 1990. pp. 435-483. ↑4.0.2 [Kol91] , On the Mordell-Weil and the Shafarevich-Tate group of modular elliptic curves. In: Proceedings of the international congress of mathematicians, Vol. I, Springer-Verlag, 1991. pp. 429-436. ↑4.0.2 [Lia18] Y. Liang, Non-invariance of weak approximation properties under extension of the ground field, Preprint, arXiv:1805.08851v1 [math.NT] (2018). ↑1.3, 1.4.1, 3.2, 5, 6.2 [Man71] Y. Manin, Le groupe de Brauer-Grothendieck en géométrie diophantienne. In: Actes du Con- grès International des Mathématiciens, Vol. 1, Gauthier-Villars, 1971 (French). pp. 401-411. ↑1.1 [Neu99] J. Neukirch, Algebraic number theory, Springer-Verlag, 1999. ↑2 [Poo09] B. Poonen, Existence of rational points on smooth projective varieties, J. Eur. Math. Soc. 11 (2009), 529–543. ↑1.3, 3, 3.2, 3.3 [Poo10] , Insufficiency of the Brauer-Manin obstruction applied to étale covers, Ann. of Math. 171 (2010), 2157–2169. ↑1.1, 1.4.3, 5, 5.0.2, 6.2 [Sch99] V. Scharaschkin, Local-global problems and the Brauer-Manin obstruction, Thesis, University of Michigan (1999). ↑1.1, 4 [Ser73] J.-P. Serre, A course in arithmetic, Graduate Texts in Mathematics, vol. 7, Springer-Verlag, 1973. ↑2.2 [Ser79] , Local fields, Graduate Texts in Mathematics, vol. 67, Springer-Verlag, 1979. ↑2.1 [Sil09] J. Silverman, The arithmetic of elliptic curves, Graduate Texts in Mathematics, vol. 106, Springer-Verlag, 2009. ↑4.0.5 [Sko01] A. Skorobogatov, Torsors and rational points, Cambridge Tracts in Mathematics, vol. 144, Cambridge University Press, 2001. ↑1.1, 3, 3.0.2, 3.0.3, 3.2.1, 3.3, 3.3.1, 4, 7.4.3 CHÂTELET SURFACES AND NON-INVARIANCE OF THE BRAUER-MANIN OBSTRUCTION FOR 3-FOLDS29

[Sko99] , Beyond the Manin obstruction, Invent. Math. 135 (1999), 399–424. ↑1.1 [Sto07] M. Stoll, Finite descent obstructions and rational points on curves, Algebra Number Theory 1 (2007), 349–391. ↑1.1, 4, 4.0.1 [Wan96] L. Wang, Brauer-Manin obstruction to weak approximation on abelian varities, Israel J. Math. 94 (1996), 189–200. ↑1.1

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