The Brauer Manin obstruction

Transcription

1 The Braer Manin obstrction Martin Bright 17 April Definitions Let X be a smooth, geometrically irredcible ariety oer a field k. Recall that the defining property of an Azmaya algebra A is that, for any field K containing k and any point P XK, we can ealate A at P to get a central simple algebra AP oer K. In particlar, sppose that k is a nmber field. Then, for each place of k, we can ealate A at points of Xk to get central simple algebras oer k. In this way we obtain a map Xk P AP Br k in Q/Z. 1 Proposition 1.1. The map 1 is continos, and hence locally constant, for the real, complex or p-adic topology, as appropriate, on Xk. Proof. I need to find a reference for this. Recall that the set of adelic points XA k of X is simply the direct prodct Xk. Adding together all the local maps 1, we obtain a continos map XA k Br k P in Q/Z. 2 By sing a direct sm instead of a direct prodct for the middle term aboe, we are implicitly stating the following reslt: Proposition 1.2. Let X be a smooth, geometrically irredcible ariety oer a nmber field k, and let A Br X be an Azmaya algebra on X. Then, for all bt finitely many places, we hae AP = 0 Br k for all P Xk. Proof. See [3, p. 101]. Proposition 1.3. If A Br X lies in the image of the natral map Br k Br X, then the associated map 2 is zero. Proof. This is a restatement of??. The following obseration is key to the definition of the Braer Manin obstrction. 1

2 Proposition 1.4. Let X be a smooth, geometrically irredcible ariety oer a nmber field k, and consider Xk as a sbset of XA k nder the diagonal embedding. Let A be an Azmaya algebra on X. Then Xk lies in the kernel of the map 2. Proof. It is straightforward to check that the following diagram commtes: Xk XA k A A Br k Br k P in Q/Z 3 where the ertical arrows are ealtion of A at points; the top horizontal arrow is the inclsion of Xk in XA k ; and the bottom line is the exact seqence??. The composite map from XA k to Q/Z is the map of 2. The proposition follows immediately from the exactness of the bottom row. With this in mind, we make the following definition. Definition 1.5. Let X be a smooth, geometrically irredcible ariety oer a nmber field k, and let A Br X be an Azmaya algebra on X. Define XA k A := {P XA k in AP = 0}. If B is a sbset of Br X, similarly define XA k B := {P XA k in AP = 0 for all A B}. One way to look at this is as follows: the maps 2 define a pairing XA k Br X Q/Z, and we hae defined XA k B to be the sbset of XA k orthogonal to the set B nder this pairing. Remark 1.6. In iew of Proposition 1.3, this pairing is actally still defined when Br X is replaced by Br X/ Br k meaning the qotient of Br X by the image of Br k. In many of the cases which interest s, Br X/ Br k will be a finite grop, and it will be possible to calclate XA k Br X explicitly. Proposition 1.4 states that Xk XA k B for any sbset B of Br X. In particlar, if XA k B is empty, then Xk is also empty. Definition 1.7. Let X be a smooth, geometrically irredcible ariety oer a nmber field k. Let B be a sbset of the Braer grop of X. If XA k is not empty bt XA k B is empty, then we say there is a Braer Manin obstrction to the Hasse principle on X coming from B. If XA k B is strictly contained in XA k, we say that there is a Braer Manin obstrction to weak approximation on X coming from B. If B = Br X, we simply say that there is a Braer Manin obstrction to the Hasse principle or to weak approximation on X. The reason that this is sch a sefl definition is that the sets XA k A are often explicitly comptable; for certain classes of arieties, we can een compte the set XA k Br X effectiely. 2

3 Proposition 1.8. Let X be a smooth, geometrically irredcible, projectie ariety oer a nmber field k, and let A Br 1 X. Then there is an effectie procedre to compte XA k A. Proof. See [2, Section 9]. In some cases we can go mch frther: Proposition 1.9. Let X be a smooth, geometrically irredcible, projectie ariety oer a nmber field k. Sppose that Pic X is free and finitely generated, and that we are explicitly gien a finite set of diisors which generate Pic X, together with the relations between them and the Galois action on them. Then there is an effectie procedre to compte XA k Br1 X. Proof. See [2, Theorem 3.4]. 2 Examples Example 2.1 Birch Swinnerton-Dyer [1]. Let X be the non-singlar del Pezzo srface of degree 4 defined by the eqations??, and let A be the qaternion algebra A = 5, + oer κx. Then A is an Azmaya algebra on X, and there is a Braer Manin obstrction to the Hasse principle on X coming from A. Proof. It will be shown in the next chapter?? that, to proe that A is an Azmaya algebra, it is enogh to check that the principal diisor / + is the norm of a diisor defined oer Q 5. This is precisely the erification carried ot for the last part of Exercise??. Diiding both sides of the first eqation by 2, we see that / = x2 5y 2 2 = N Q 5/Q x + 5y is a norm from Q 5, and so by?? the algebra 5, / + is isomorphic to A. In a similar way, we get the following for qaterion algebras oer κx, all isomorphic: A = 5, +, 5,, + 5,, + 2 5, We will describe the map XQ Q/Z, gien by P in AP, separately for each place. For = the real place, notice that 5 is positie and hence a sqare in R; ths AP = 5, P /P +P is a triial algebra and in AP = 0, at least for all P XR where P and P +P are non-zero. Since the map P in AP is continos on XR, it follows that it is zero eerywhere. If is an odd prime p sch that 5 is a sqare in Q p, then the same argment works and shows that in p AP = 0 for all P XQ p. 3

4 Now sppose that is an odd prime p 5, sch that 5 is not a sqare in Q p and therefore not a sqare in F p. In this case, and can neer be both zero at a point of XF p, since otherwise x/y and x/z wold be sqare roots of 5 in F p. Similarly, + and + 2 are neer both zero at a point on XF p. It follows that, for each P XQ p, at least one of the isomorphic algebras 4 is clearly of the form 5, b with b Z p, and therefore in p AP = 0 by Proposition??. Next, consider the case = 2. At first glance the preios argment will not work: since P = 0 : 0 : 1 : 1 : 1 XF 2, it wold appear that and can both be een at a point of XQ 2. Bt it trns ot that P does not lift to a point of XQ 2, as can be seen, for example, by looking at the eqations modlo 16. So once again, for each P XQ 2, one of, is odd, and similarly one of +, + 2 is odd. The formla of Proposition?? shows that 5, b 2 = 1 wheneer b is odd, so once again we conclde that in 2 AP = 0 for all P XQ 2. Finally, we look at = 5. Modlo 5, the ariety X redces to a nion of for planes, meeting in a common line; two of these planes are defined oer F 5 and the other two are qadratic and conjgate. The two defined oer F 5, which therefore contain all the points of XF 5, are { = = x} and { = = x}. The line of intersection of these planes is { = = x = 0}, bt no point here lifts to a point of XQ 5. Therefore eery point of XQ 5 satisfies ±x mod 5 with,, x Z 5. This means that / + 3 mod 5, and the formla of Proposition?? gies 5, 3 5 = 1, meaning that in 5 AP = 1 2 for all P XQ 5. To smmarise, we hae proed that in AP = 0 for all P XQ where 5, and that in AP = 1 2 for all P XQ 5. It follows that in AP = 1 2 for all P XA Q = XQ. So XA Q A =, and therefore there is a Braer Manin obstrction to the Hasse principle on X. Remark 2.2. Althogh or representatie qaternion algebras 4 all had problems when = = 0, it trned ot that we neer needed to ealate them at sch a point. In fact we cold hae fond other isomorphic qaternion algebras, extending or set 4 to a set which cold be easily ealated at any point of any XQ ; this is a conseqence of the fact that A is Azmaya. We conclde by showing that the phenomenon of Example?? can also be explained by the Braer Manin obstrction. Example 2.3. Let X be the singlar cbic srface??, and let A be the qaternion algebra oer κx defined by 4z 7t A = 1,. t Then A is an Azmaya algebra on X, and there is a Braer Manin obstrction to weak approximation on X coming from A, which explains why one connected component of XR contains no rational points. 4

5 Remark 2.4. We hae not defined what an Azmaya algebra is on a singlar ariety. In this particlar case, it is straightforward to check that A is an Azmaya algebra away from the singlar points of X, and this is enogh for or prposes. Proof. The two singlar points of X are ±i : 1 : 0 : 0 where i 2 = 1. Let U denote the complement in X of these two points. Any rational points of X mst be contained in U, since neither of the singlar points are rational. As before, we start by finding alternatie ways of writing the qaternion algebra A. Firstly, note that by looking at the defining eqations of X we can see that 4z 7t = x2 + y 2 + 8zt 14t 2 t z 2 as fnctions on X, immediately giing a new way of writing A. Frthermore, since the denominator z 2 is a sqare, we can replace it with any other sqare sch as x 2, y 2 or t 2 to get new qaternion algebras isomorphic to A. Also, the defining eqations show easily that the algebra 1, z 2 2t 2 /t 2 is isomorphic to A, and again the denominator here may be replaced by any sqare. In this way we find a set of isomorphic algebras which can be ealated at any point of U. Now consider each place separately. If p 1 mod 4 then 1 is a sqare in Q p, and so in p AP = 0 for all P UQ p. If p 3 mod 4 then a similar argment to the preios example shows that, for each P XQ p, one of or algebras always ealates to 1, b with b Z p, and therefore that in p AP = 0 for all P XQ p. For = 2, notice that 4z 7t/t 1 mod 4, and that 1, b 2 = 1 wheneer b 1 mod 4, so again in 2 AP = 0 for all P XQ 2. Finally, for =, we see that 4z 7t/t is non-negatie on one component of XR, and strictly positie on a dense open sbset of that component, so in A = 0 on that component. On the other component, howeer, 4z 7t/t is strictly negatie on a dense open sbset, and therefore in A = 1 2 on that component. We dedce that rational points can only be fond in the component where 4z 7t/t 0, giing a Braer Manin obstrction to weak approximation on X. References [1] B. J. Birch and H. P. F. Swinnerton-Dyer. The Hasse problem for rational srfaces. J. Reine Angew. Math., 274/275: , Collection of articles dedicated to Helmt Hasse on his seenty-fifth birthday, III. [2] A. Kresch and Y. Tschinkel. Effectiity of Braer Manin obstrctions. arxi:math/ [Math.AG], [3] A. Skorobogato. Torsors and rational points, olme 144 of Cambridge Tracts in Mathematics. Cambridge Uniersity Press, Cambridge,