An example of a Brauer--Manin obstruction to weak approximation at a prime with good reduction
aa r X i v : . [ m a t h . AG ] F e b An example of a Brauer–Manin obstruction to weakapproximation at a prime with good reduction
Margherita Pagano
Abstract:
Following Bright and Newton, we construct an explicit K surface overthe rational numbers having good reduction at 2, and for which 2 is the only primeat which weak approximation is obstructed. Let k be a number field and A k be the ring of adèles of k , i.e. the restricted product of k ν forall places ν of k , taken with respect to the rings of integers O ν ⊆ k ν . Let X be a smooth,proper, geometrically irreducible variety over k . In order to study the rational points on X itis useful to look at the image of X ( k ) in the set of the adèlic points X ( A k ) . More precisely,Manin [Man71] has shown that there exists a pairing Br( X ) × X ( A k ) → Q / Z such that the rational points of X lie in the image of the right kernel of the pairing, denoted by X ( A k ) Br . If X ( A k ) Br is not equal to the whole X ( A k ) we say that there is a Brauer–Maninobstruction to weak approximation on X .In this paper we follow the ideas presented in [BN20] to construct an example of a K surfaceover the rational numbers with a Brauer–Manin obstruction to weak approximation arisingfrom a prime with good ordinary reduction. More precisely, there exist an element A ∈
Br( X ) and a prime p of good ordinary reduction such that the evaluation map |A| : X ( Q p ) → Br( Q p ) is non-constant.Let X ⊆ P Q be the projective K surface defined by the equation x y + y z + z w + w x + xyzw = 0 . (1) Theorem 1.
The class of the quaternion algebra A = z + w x + xyzx , − zx ! ∈ Br k ( X ) defines an element in Br( X ) . The evaluation map |A| : X ( Q ) → Br( Q ) is non-constant, andtherefore gives an obstruction to weak approximation on X . Finally, X ( Q ) is not dense in X ( Q ) , with respect to the analytic topology. V over a numberfield L such that H ( V, Ω V ) = 0 , every prime of good ordinary reduction is involved in aBrauer–Manin obstruction over an extension of the base field. The example provided in thisarticle is optimal, in the following sense: we build an element A ∈
Br( X )[2] such that theevaluation map associated to it is non-constant without the need to take an algebraic extensionof Q . Moreover, as pointed out in [BN20, Remark . ], when we are dealing with K surfacesdefined over the rational numbers, the only prime with good reduction that can play a role inthe obstruction to weak approximation is the prime .The following example and, more in general, the result proven by Bright and Newton givea negative answer to the following question, asked by Swinnerton-Dyer [CTS13, Question 1]. Question 1.
Let k be a number field and let S be a finite set of places of k containing theArchimedean places. Let V S be a smooth projective O S -scheme with geometrically integralfibres, and let V /k be the generic fibre. Assume that
Pic( V ) is finitely generated and torsion-free. Swinnerton-Dyer asks if there is an open and closed Z ⊆ Q ν ∈ S V ( k ν ) such that V ( A k ) Br = Z × Y ν / ∈ S V ( k ν ) . Roughly speaking, is it true that the Brauer–Manin obstruction involves only the places ofbad reduction and the Archimedean places?Finally, we point out that the element A defined in Theorem 1 has to be a transcendentalelement in Br( X ) . Indeed, Colliot-Thélène and Skorobogatov proved [CTS13, Lemma 2.2]that for every element in the algebraic Brauer group the associated evaluation map at a primewith good reduction has to be constant.In general, let V be a variety over a field k , k an algebraic closure of k and V the basechange of V to k , i.e. V := V × k k ; the algebraic and transcendental Brauer groups of V aredefined as the kernel and the image of the natural map Br( V ) → Br( V ) .For curves and surfaces with negative Kodaira dimension we have Br( V ) = Br ( V ) . Hence,K surfaces are the first example of varieties where the transcendental Brauer group is po-tentially non-trivial. However, this is not always the case: for example in [ISZ11] they showthat, under certain conditions, the whole Brauer group of a diagonal quartic surface over Q isalgebraic. The first example of a transcendental element in the Brauer group of a K surfacedefined over a number field was given by Wittenberg in [Wit04]. In particular, Wittenberg con-structed a -torsion transcendental element that obstructs weak approximation on the surface.Other examples of -torsion transcendental elements that obstruct weak approximation canbe found in [HVAV11] and [Ier10]. In all these articles, the obstruction to weak approximationcomes from the fact that the transcendental quaternion algebra has non-constant evaluationat the place at infinity. With a construction similar to the one used in [HVAV11], Hassett andVárilly-Alvarado [HVA13] have also built an example of a -torsion element on a K surfacethat obstructs the Hasse principle.Furthermore, there are examples of transcendental elements of order on K surfaces thatobstruct weak approximation (for example, see [Pre13], [New16] and [BVA20]). In all thesecases, the evaluation map at the place at infinity has to be trivial, since Br( R ) does not containelements of order , and the obstruction to weak approximation comes from the evaluation2ap at the prime , which in every example is a prime of bad reduction for the K surfacetaken into account. Therefore, none of the examples mentioned above can be used to give anegative answer to Question 1. Outline of the paper.
Section contains the proof of Theorem 1. In Section we showthat the K surface X has good ordinary reduction at the prime . Moreover, we explain theideas behind the construction of the quaternion algebra A of Theorem 1 and why we couldexpect that it obstructs weak approximation on X . Acknowledgements.
I am deeply grateful to Martin Bright for introducing me to the topicand for the ideas he shared with me that were very helpful in writing this paper.
In the first part of the proof we will show that the element
A ∈
Br( k ( X )) lies in Br( X ) . Next,we will exhibit two points P , P ∈ X ( Q ) such that ev P ( A ) = ev P ( A ) . Finally, we will prove that, for every place ν different from , the evaluation map |A| : X ( Q ν ) → Br( Q ν ) is constant. Proof of Theorem 1.
Let f := z + w x + xyz and C x , C z , C f be the closed subsets of X definedby the equations x = 0 , z = 0 and f = 0 respectively. The quaternion algebra A defines anelement in Br( U ) , where U := X \ ( C x ∪ C z ∪ C f ) . The purity theorem for the Brauer group[CTS, Theorem 3.7.2], assures us the existence of the exact sequence → Br( X )[2] → Br( U )[2] ⊕ ∂ D −−−→ M D H ( k ( D ) , Z / (2)where D ranges over the irreducible divisors of X with support in X \ U and k ( D ) denotesthe residue field at the generic point of D .In order to use the exact sequence (2) we need to understand what the prime divisors of X with support in X \ U = C x ∪ C z ∪ C f look like. It is possible to check the following:• C x has as irreducible components D and D , defined by the equations { x = 0 , z = 0 } and { x = 0 , y + z w = 0 } respectively;• C z has as irreducible components D and D , where D is defined by the equations { z = 0 , x y + w = 0 } ;• C f has as irreducible components D , D and D , where D and D are defined by theequations { y z − x z + y w = 0 , x + y z = 0 , xyz + z + xw = 0 } and { z + xw =0 , y = 0 } respectively. 3herefore, we can rewrite (2) in the following way: → Br( X )[2] → Br( U )[2] ⊕ ∂ Di −−−→ M i =1 H ( k ( D i ) , Z / . (3)Moreover, we have an explicit description of the residue map on quaternion algebras: for anelement ( a, b ) ∈ Br( U )[2] we have ∂ D i ( a, b ) = " ( − ν i ( a ) ν i ( b ) a ν i ( b ) b ν i ( a ) ∈ k ( D i ) × k ( D i ) × ≃ H ( k ( D i ) , Z / where ν i is the valuation associated to the prime divisor D i . This follows from the definitionof the tame symbols in Milnor K -theory together with the compatibility of the residue map ∂ D with the tame symbols given by the Galois symbols (see [GS17], Proposition . . ).We can proceed with the computation of the residue maps ∂ D i for i = 1 , . . . , :1. ν ( f ) = ν ( x ) = ν ( z ) = 1 . Hence, ∂ D (cid:18) fx , − zx (cid:19) = "(cid:18) − zx (cid:19) = 1 ∈ k ( D ) × k ( D ) × . ν ( x ) = 1 and ν ( f ) = ν ( z ) = 0 . Hence, ∂ D (cid:18) fx , − zx (cid:19) = " − (cid:18) fx (cid:19) − (cid:18) − zx (cid:19) = " z f = 1 ∈ k ( D ) × k ( D ) × where the last equality follows from the fact that x = 0 on D , thus f | D = z . ν ( x ) = ν ( f ) = 0 and ν ( z ) = 1 . Hence, ∂ D (cid:18) fx , − zx (cid:19) = "(cid:18) fx (cid:19) = "(cid:18) wx (cid:19) = 1 ∈ k ( D ) × k ( D ) × where the last equality follows from the fact that z = 0 on D , thus f | D = w x. ν ( x ) = ν ( z ) = 0 and ν ( f ) = 1 . Hence, ∂ D (cid:18) fx , − zx (cid:19) = "(cid:18) − xz (cid:19) = "(cid:18) yx (cid:19) = 1 ∈ k ( D ) × k ( D ) × where the last equality follows from the fact that x + y z = 0 on D , thus − xz = (cid:0) yx (cid:1) . ν ( x ) = ν ( z ) = 0 and ν ( f ) = 1 . Hence, ∂ D (cid:18) fx , − zx (cid:19) = "(cid:18) − xz (cid:19) = "(cid:18) wz (cid:19) = 1 ∈ k ( D ) × k ( D ) × where the last equality follows from the fact that z + w x = 0 on D , thus − xz = (cid:0) wz (cid:1) .4herefore, ∂ D i ( A ) = 0 for all i ∈ { , . . . , } , hence A ∈
Br( X ) .We now show that the element A obstructs weak approximation on X . Let X ⊆ P Z be theprojective scheme defined by the equation x y + y z + z w + w x + xyzw = 0 . (4) X is a Z -model for X and has good reduction at the prime .Let P := (1 : 0 : 1 : 0) ∈ X ( Z ) ; then P is such that ev P ( A ) = (1 , − . Therefore ev P ( A ) is the trivial class in Br( Q ) . Moreover, Hensel’s lemma assures us of the existenceof a solution P = (1 : 2 : 1 : d ) ∈ X ( Z ) whose reduction modulo is (1 : 2 : 1 : 2) . Hence,ev P ( A ) = ( f ( P ) , − with f ( P ) ≡ . Therefore, we get that ev P ( A ) defines a non-trivial element in the Brauer group of Q [Ser73,Theorem 3.1]. The existence of such points implies that there is a Brauer–Manin obstructionto weak approximation arising from A . Indeed, X ( A Q ) A = ( ( x p ) ∈ X ( A Q ) : X p inv p A ( x p ) = 0 ) ( X ( A Q ) . In order to conclude the proof of the theorem we investigate the behaviour of the evaluationmap at the other primes and at infinity. For every prime p let X p be the base change of X to Z p . We distinguish the following cases.Case p / ∈ { , , , ∞} . In this case, X has good reduction at p . Therefore, we can use[CTS13, Proposition . ] to conclude that the evaluation map |A| : X ( Z p ) → Br( Q p ) is constant. Moreover, P = (1 : 0 : 1 : 0) ∈ X ( Z p ) andev P ( A ) = (1 , − which is trivial in Br( Q p ) ; hence the evaluation map is trivial on the whole X ( Z p ) = X ( Q p ) . Case p ∈ { , , } . Under this assumption, X p / Z p is not smooth. In these three cases, wewant to show that the evaluation map is trivial on X ( Z p ) by showing that it factors through Br( Z p ) .The special fibre Y p := X p × Z p Spec( F p ) is a non-smooth F p -scheme. However, Y p is anirreducible F p -scheme, with just isolated singularities. The Z p -points of X p are all smooth. Infact, Y p contains just one singular point defined over F p that does not even lift to a Z /p Z -point. Let U be the smooth locus of X p ; because of what we have just said we have X ( Q p ) = X ( Z p ) = U ( Z p ) . Let U be the base change of U to Spec( Q p ) . The purity theorem on U [CTS13, Theorem . . ]gives us the exact sequence Br( U )[2] → Br( U )[2] ∂ Dp −−→ H ( k ( D p ) , Z / Z ) D p is the divisor associated to the special fibre ( D p is the smooth locus of Y p ). We justneed to show that ∂ D p ( A ) = 0 . Let ν p be the valuation corresponding to the prime divisor D p ;then ν p (cid:18) fx (cid:19) = 0 and ν p (cid:18) − zx (cid:19) = 0 . Indeed, the point (1 : 0 : 1 : 0) ∈ Y p ( F p ) is smooth, hence it lies in D p . Moreover fx (1 : 0 : 1 : 0) = 1 and − zx (1 : 0 : 1 : 0) = − . Therefore both fx and − zx do not vanish on D p , which implies that ∂ D p ( A ) = 0 . Therefore, A lies in Br( U ) ⊆ Br( X p ) and the evaluation map factors as U ( Z p ) Br( Q p ) . Br( Z p ) |A| Since
Br( Z p ) is trivial, the evaluation map has to be constant and trivial.Case p = ∞ . The evaluation map |A| : X ( R ) → Br( R ) is constant and equal to .We will show that it is constant on the dense open subset W := { P ∈ X ( R ) : x ( P ) , z ( P ) , f ( P ) = 0 } ⊆ X ( R ) . Indeed, from the continuity of the evaluation map it follows that it has to be constant also onthe whole of X ( R ) . Let P = ( α : β : γ : δ ) ∈ W , thus γ = 0 . First, assume that − γα > .Then ev P ( A ) = (cid:18) f ( P ) x ( P ) , − z ( P ) x ( P ) (cid:19) = (cid:18) f ( P ) α , − γα (cid:19) is trivial in Br( R ) . Now, suppose that − γα < . Without loss of generality, we can assumethat both α and γ are positive. We want to show that in this case f ( P ) has to be positive:• if δ = 0 , then P ∈ X ( R ) implies that β ( α + β γ ) = 0 . Therefore β = 0 , since α + β γ ≥ α > . Hence, f ( P ) = γ > ;• if δ = 0 then P ∈ X ( R ) implies f ( P ) = − βδ ( α + β γ ) . Hence, since α + β γ > , f ( P ) > if and only if − βδ > . β, δ do not have the same sign. Hence, we just need to show that there isno point P ∈ W with α, γ positive and β, δ with the same sign. First, we observe that β, δ can not be both positive, since otherwise α β + β γ + γ δ + δ α + αβγδ > . On the other hand β, δ cannot also be both negative. Indeed, we have that P ∈ X ( R ) ifand only if α ( − β ) + ( − β ) γ + γ ( − δ ) + ( − δ ) α = α ( − β ) γ ( − δ ) . Without loss of generality we may assume that α ≥ max {− β, γ, − δ } ; but if α, − β, γ, − δ are all positive, then α ( − β ) + ( − β ) γ + γ ( − δ ) + ( − δ ) α > α ( − β ) γ ( − δ ) . Hence, ( α : β : γ : δ ) / ∈ X ( R ) . A The aim of this section is to give a glance of the ideas behind the construction of the quaternionalgebra A .Through this section we will indicate by X the base change of the Z -model X to Z , with X the base change of X to Q and with Y the reduction of X at the prime .In the previous section we already mentioned that the K surface X has good reduction atthe prime (i.e. X is a smooth Z -scheme); actually X has good ordinary reduction at theprime , as we will show in the following lemma. Lemma 2. Y is an ordinary K surface over F .Proof. It is enough to show that the cardinality of Y ( F ) is even (see [Tae20]). Y is theprojective F -variety defined by the equation x y + y z + z w + w x + xyzw = 0 . Therefore, it is possible to compute | Y ( F ) | directly, which turns out to be equal to .Before proceeding with the actual construction, we mention a property of the K surface Y ,related to the fact that it is ordinary. For every q ≥ , let Ω qY/ F be the sheaf of q -differentialforms on Y . We define Z qY/ F := ker (cid:16) d : Ω qY/ F → Ω q +1 Y/ F (cid:17) and B qY/ F := im (cid:16) d : Ω q − Y/ F → Ω qY/ F (cid:17) . Let C Y/ F p : Z • Y/ F → Ω • Y/ F be the Cartier operator on Y (for the construction of the Cartier operator see [Ill79, § ]). Forevery q the sheaf of logarithmic differential q -forms Ω qY/ F , log is defined as the kernel of themorphism − C Y/ F : Z qY/ F → Ω qY/ F . Hence, Ω qY/ F , log fits in the following exact sequence: → Ω qY/ F , log → Z qY/ F − C Y/ F −−−−−→ Ω qY/ F . (5)7 emark 1. The Cartier operator is defined, more generally, for smooth S -schemes X incharacteristic p . The Cartier operator C X/S goes from Z • X/S to Ω • X ( p ) /S , where X ( p ) is thebase change of X by the Frobenius morphism F S : S → S . In this setting Ω qX/S, log is definedas the kernel of the morphism W ∗ − C X/S : Z qX/S → Ω qX ( p ) /S where W ∗ is the map induced on the differential forms by the natural projection W : X ( p ) → X. However, since in our case S = F , we have Y (2) = Y and W = id.The sheaf Ω qY/ F , log is a sheaf of F -vector spaces on the Zariski site of Y . Moreover, if welook at it on Y ét , then its formation is compatible with étale base change [Ill79, 2.1.8].Let k be an algebraic closure of F and Y the base change of Y to k . Bloch and Kato proved[BK86, Proposition 7.3] that Y ordinary implies that the natural map H (cid:16) Y , Ω Y /k, log (cid:17) ⊗ F k → H (cid:16) Y , Ω Y /k (cid:17) is an isomorphism. For K surfaces, H (cid:16) Y , Ω Y /k (cid:17) is a one-dimensional k -vector space. Hence H (cid:16) Y , Ω Y /k, log (cid:17) has to be a one-dimensional F -vector space. Let ω be the only non-trivialelement in H (cid:16) Y , Ω Y /k, log (cid:17) . Since C Y /k respects the Galois action, we get that the element ω is Galois fixed, hence it comes from H (cid:16) Y, Ω Y/ F (cid:17) . Therefore, the unique non-trivial elementof H (cid:16) Y, Ω Y/ F (cid:17) must be logarithmic. Lemma 3.
Let F be the function field of Y . The image of ω ∈ H (cid:16) Y, Ω Y/ F (cid:17) in Ω F/ F canbe written as dη η ∧ dη η , where η = z + w x + xyzx and η = zx . Proof.
Let ξ ∈ Y be the generic point of Y and ω ξ the image of ω in Ω F/ F under the inclusion H (cid:16) Y, Ω Y/ F (cid:17) ֒ → Ω F/ F . For convenience, to give an explicit description of a non-zero element in H (cid:16) Y, Ω Y/ F (cid:17) , wewill use the following notation: instead of the variables { x, y, z, w } , for the equation defining Y we will use the variables { x , x , x , x } and we denote by G ( x , x , x , x ) the polynomialdefining Y .For every permutation { p, q, i, j } of { , , , } we define W p,q ⊆ Y as the open subset of Y where x p · ∂G∂x q does not vanish. Moreover, we set ω p,q := d (cid:16) x i x p (cid:17) ∧ d (cid:16) x j x p (cid:17) x p · ∂G∂x q ∈ H ( W p,q , Ω Y ) . Y is smooth, the open sets { W p,q } cover it. It is easy to check that, since we are workingover the field F , for every ( p, q ) = ( p ′ , q ′ ) ω p,q | W p,q ∩ W p ′ ,q ′ = ω p ′ ,q ′ | W p,q ∩ W p ′ ,q ′ . Therefore, there exists ω ∈ H (cid:16) Y, Ω Y/ F (cid:17) such that for every p, q as above ω | W p,q = ω p,q ∈ H (cid:16) W p,q , Ω Y/ F (cid:17) . Now, going back to the usual notation with which we denote our variables by { x, y, z, w } , let G w be the partial derivative of the polynomial defining Y with respect to the variable w . Wehave ω ξ = ( ω , ) ξ = d (cid:0) yx (cid:1) ∧ d (cid:0) zx (cid:1) G w x = d (cid:16) G w x (cid:17) ∧ d (cid:0) zx (cid:1) G w x · zx . Indeed, d G w x ! ∧ d zx ! = d z + w x + xyzx ! ∧ d zx ! = d z x + w x + yzx ! ∧ d zx ! Using that d (cid:0) zx (cid:1) ∧ d (cid:0) zx (cid:1) = 0 , together with the fact that we are working over a field ofcharacteristic , we get d G w x ! ∧ d zx ! = zx · d yx ! ∧ d zx ! . We can finally proceed with the construction of the quaternion algebra A and explain whywe could expect that A gives an obstruction to weak approximation on X .Let R be the henselisation of the discrete valuation ring O X ,Y and K h be the fraction fieldof R . Bloch and Kato [BK86] introduced a decreasing filtration n U m H (cid:0) K h , µ ⊗ (cid:1)o m ≥ on H (cid:0) K h , µ ⊗ (cid:1) as follows. For a , a ∈ K h , let ( a , a ) denote the class δ ( a ) ∪ δ ( a ) ∈ H (cid:0) K h , µ ⊗ (cid:1) where δ : (cid:0) K h (cid:1) × → H (cid:0) K h , µ ⊗ (cid:1) is the connecting map coming from the Kummer sequence.Let U H (cid:0) K h , µ ⊗ (cid:1) := H (cid:0) K h , µ ⊗ (cid:1) ; for m > , let U m H (cid:0) K h , µ ⊗ (cid:1) be the subgroup of H (cid:0) K h , µ ⊗ (cid:1) generated by the elements of the form (cid:0) a m , a (cid:1) , with a ∈ R and a ∈ (cid:0) K h (cid:1) × . Moreover, for every m ≥ letgr m := U m H (cid:0) K h , µ ⊗ (cid:1) U m +1 H (cid:0) K h , µ ⊗ (cid:1) .
9n [BK86, § ], Bloch and Kato proved that the map ρ : Ω F, log ⊕ Ω F, log → gr := H (cid:0) K h , µ ⊗ (cid:1) U (cid:0) H (cid:0) K h , µ ⊗ (cid:1)(cid:1)(cid:18) dη η ∧ dη η , (cid:19) (˜ η , ˜ η ) (cid:18) , dη η (cid:19) (˜ η , is an isomorphism, where ˜ η , ˜ η are arbitrary lifts of η , η to K h . Remark 2.
We are working over Q , which contains a second primitive root of unity, hencewe have an isomorphism [GS17, Proposition . . ] H (cid:0) K h , µ ⊗ (cid:1) ≃ Br (cid:0) K h (cid:1) [2] which sends ( a, b ) to the class of the quaternion algebra ( a, b ) .Let B ∈
Br( X )[2] ; we can always look at B as an element in H (cid:0) K h , µ ⊗ (cid:1) . Hence, thereexists an m ≥ such that B gives a non-zero element in gr m . Bright and Newton proved[BN20] that knowing such an m gives information about the behaviour of the evaluation map |B| : X ( O L ) → Br( L ) P ev P ( B ) for finite field extensions L/ Q . We will make this sentence more precise in the followingparagraph. Bright and Newton define an evaluation filtration on the Brauer group of X in the followingway. Given a finite field extension L of Q , let e L/ Q be the ramification index of L , O L itsring of integers and π a uniformiser. For all positive integers r and P ∈ X , let B ( P, r ) be theset of points Q ∈ X ( O L ) such that Q has the same image as P in X ( O L /π r ) (equivalently,we will say that Q ≡ P (mod π r ) ). We defineEv n Br X := {B ∈ Br( X ) | ∀ L/ Q finite , ∀ P ∈ X ( O L ) |B| is constant on B ( P, e L/ Q n + 1) } , ( n ≥ Ev − Br X := {B ∈ Br( X ) | ∀ L/ Q finite , |B| is constant on X ( O L ) } Ev − Br X := {B ∈ Br( X ) | ∀ L/ Q finite , |B| is zero on X ( O L ) } In order to compare the evaluation filtration with the filtration U m H (cid:0) K h , µ ⊗ (cid:1) , Brightand Newton used the filtration { fil n Br( X )[2] } on Br( X )[2] given by Kato’s Swan con-ductor [BN20, § ]. Briefly, using the Swan conductor, Kato [Kat89] defined a filtration on H (cid:0) K h , Z / Z (1) (cid:1) . From the Kummer sequence, H (cid:0) K h , Z / Z (1) (cid:1) is isomorphic to Br (cid:0) K h (cid:1) [2] .Therefore, we get a filtration { fil n Br (cid:0) K h (cid:1) [2] } n ≥ on Br (cid:0) K h (cid:1) [2] ; the pullback of this filtrationgives us a filtration { fil n Br( X )[2] } n ≥ on Br( X )[2] .10oreover, since Q contains a second root of unity, we have that Z / Z (1) ≃ µ ⊗ , thus wecan identify H (cid:0) K h , Z / Z (1) (cid:1) ≃ H (cid:0) K h , µ ⊗ (cid:1) . (6)Kato [Kat89, Lemma . ], proved that in our setting the isomorphism of equation (6) inducesisomorphisms U m H (cid:0) K h , µ ⊗ (cid:1) ≃ fil − m Br (cid:0) K h (cid:1) [2] , ≤ m ≤ . (7)Moreover, for m > we have that both fil m Br (cid:0) K h (cid:1) [2] and U m H (cid:0) K h , µ ⊗ (cid:1) vanish.In particular, using the filtration U m on H (cid:0) K h , µ ⊗ (cid:1) we can describe fil Br( W )[2] as n B ∈
Br( X )[2] such that the image of B in H (cid:0) K h , µ ⊗ (cid:1) lies in U H (cid:0) K h , µ ⊗ (cid:1)o . Finally, Bright and Newton proved [BN20, Theorem A] that there is an equality Ev Br( X ) = fil Br( X ) . A Summing up, the construction of A goes through the following steps. By Lemma 3 we knowthat ω ∈ H (cid:0) Y, Ω Y/ F (cid:1) is such that ω ξ = dη η ∧ dη η ∈ Ω F, log is a non-trivial logarithmic -form of F , where η := G w x , η := zx The isomorphism ρ assuresus that for every choice of lifts ˜ η , ˜ η in K h , the element (˜ η , ˜ η ) ∈ Br (cid:0) K h (cid:1) [2] has non-trivialimage in gr , that is (˜ η , ˜ η ) lies in H (cid:0) K h , µ ⊗ (cid:1) \ U H (cid:0) K h , µ ⊗ (cid:1) . At this point, the ideabehind the construction of A is to find lifts ˜ η , ˜ η such that (˜ η , ˜ η ) defines an element in Br( X ) .Indeed, if such lifts exist, then the image of (˜ η , ˜ η ) in Br( X ) does not lie in fil Br( X ) . Rathersurprisingly, as proven in Theorem 1, it turns out that the choice of lifts ˜ η := z + w x + xyzx and ˜ η := − zx defines an element A := (˜ η , ˜ η ) ∈ Br( X ) . By construction, using equation (7), the image of A in Br (cid:0) K h (cid:1) [2] is not in fil Br (cid:0) K h (cid:1) [2] .Hence, if we look at A in Br( X ) , then it does not lie in fil Br( X ) ⊇ fil Br( X ) . In particular,by [BN20, Theorem A (3) ] we have that there exists a finite field extension L/ Q and two points P, Q ∈ X ( O L ) such that P and Q have the same image in X ( O L /π L ) and ev P ( A ) = ev Q ( A ) . We saw in the proof of Theorem 1 that there exist two points P and P defined over Z , withthe same reduction modulo and whose evaluation map is different. Namely, there is no needto take a field extension of Q in our case. Remark 3.
From the identification Br (cid:0) K h (cid:1) [2] ≃ U (cid:0) H (cid:0) K h , µ ⊗ (cid:1)(cid:1) ≃ fil Br (cid:0) K h (cid:1) [2]
11e get that
Br( X )[2] = fil Br( X )[2] . Hence, clearly,
A ∈ fil Br( X )[2] . [BN20, Theorem A (4) ] tells us that Ev Br( X )[2] = {B ∈ fil Br( X )[2] | rsw , ( A ) ∈ Ω F ⊕ } . Hence, if A is such that rsw , ( A ) = ( α, , then A lies in Ev Br( X )[2] . rsw , is the calledthe refined Swan conductor (for the definition of it see [Kat89, § ]). In particular, in our case rsw , : Br (cid:0) K h (cid:1) [2]fil Br (cid:0) K h (cid:1) [2] → Ω F ⊕ Ω F . Furthermore, the refined Swan conductor morphism is strictly related to the morphism ρ .Indeed, by Kato [Kat89, Lemma 4.3], we know that Br (cid:0) K h (cid:1) [2]fil Br (cid:0) K h (cid:1) [2] ≃ Br (cid:0) K h (cid:1) [2] U Br (cid:0) K h (cid:1) [2] = gr and rsw , ( ρ ( α, β )) = ( α, β ) . By construction, in our case, the image of A in gr is of the form ρ (cid:18) dη η ∧ dη η , (cid:19) . Hence rsw , ( A ) ∈ Ω F ⊕ ; thus A ∈ Ev Br( X )[2] . Therefore the evaluation map from X ( Z ) to Br( Q ) depends only on the reduction of the points modulo . As already mentioned in the introduction, this paper was strongly inspired by the followingresult proven by Bright and Newton.
Theorem 4. [BN20, Theorem C]
Let V be a smooth, proper variety over a number field L with H ( V, Ω V ) = 0 . Let p be a prime of L at which V has good ordinary reduction, withresidue characteristic p . Then there exists a finite field extension L ′ /L , a prime p ′ of L ′ lyingover p , and an element A ∈ Br V L ′ { p } such that the evaluation map |A| : V ( L ′ p ′ ) → Br( L ′ p ′ ) isnon-constant. In particular, if V ( A L ′ ) = ∅ then A obstructs weak approximation on V L ′ . In our example, V = X = Proj (cid:18) Q [ x, y, z, w ] x y + y z + z w + w x + xyzw (cid:19) ⊆ P Q is a smooth projective variety defined over the number field Q .Since X is a K surface, the hypotheses of Theorem 4 are satisfied. As already pointed outin the introduction, in this example we were able to find a quaternion algebra in the Brauergroup of X . Hence, in this case, we were able to construct an element A that satisfies Theorem12 which is already defined over the rational numbers, without the need of any field extensionof Q . Moreover, A does not just lie in the torsion part of the Brauer group of X , it hasorder exactly .It is still unclear whether one can hope to find, for all K surfaces over the rational numberswith ordinary good reduction at the prime , an element in the -torsion part of Brauer groupthat is defined over the rational numbers and obstructs weak approximation.13 eferences [BK86] Spencer Bloch and Kazuya Kato. p -adic étale cohomology. Inst. Hautes ÉtudesSci. Publ. Math. , (63):107–152, 1986.[BN20] Martin Bright and Rachel Newton. Evaluating the wild Brauer group, 2020. https://arxiv.org/pdf/2009.03282.pdf .[BVA20] Jennifer Berg and Anthony Várilly-Alvarado. Odd order obstructions to the Hasseprinciple on general K3 surfaces.
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