The Brauer-Manin Obstruction and Surfaces. Mckenzie West Emory University January 9, 2016

Transcription

1 The Brauer-Manin Obstruction and Surfaces Mckenzie West Emory University January 9,

2 The Brauer-Manin Obstruction and Cubic Surfaces Mckenzie West Emory University January 9,

3 History

4 Main Question 3

5 Main Question Question Given a variety X, is X (Q)? 3

6 Main Question Question Given a variety X, is X (Q)? The Homer Method 3

7 The Hasse Principle 4

8 The Hasse Principle Suppose X is a nice Q-variety, then Q p Q p X (Q) X (A Q ) = p X (Q p ). 4

9 The Hasse Principle Suppose X is a nice Q-variety, then Q p Q p X (Q) X (A Q ) = p X (Q p ). In particular, X (Q) X (A Q ). 4

10 The Hasse Principle Suppose X is a nice Q-variety, then Q p Q p X (Q) X (A Q ) = p X (Q p ). In particular, X (Q) X (A Q ). Definition A family S of Q-varieties satisfies the Hasse principle if for every X S the converse holds, that is X (A Q ) X (Q). 4

11 What We Know 5

12 What We Know Determining if X (A Q ) is a finite computation. 5

13 What We Know Determining if X (A Q ) is a finite computation. Plane quadrics satisfy the Hasse principle. 5

14 What We Know Determining if X (A Q ) is a finite computation. Plane quadrics satisfy the Hasse principle. Severi-Brauer varieties satisfy the Hasse principle. 5

15 What We Know Determining if X (A Q ) is a finite computation. Plane quadrics satisfy the Hasse principle. Severi-Brauer varieties satisfy the Hasse principle. 5

16 What We Know Determining if X (A Q ) is a finite computation. Plane quadrics satisfy the Hasse principle. Severi-Brauer varieties satisfy the Hasse principle. Conjecture (Mordell, 1949) Cubic surfaces defined over Q satisfy the Hasse principle. 5

17 First Examples 6

18 First Examples Example (Swinnerton-Dyer, 1962 ) The following surface does not satisfy the Hasse Principle X : y(y + x)(2y + x) = (x + θz + θ 2 w) where the product is taken over the roots of T 3 7T T 7 = 0. 6

19 First Examples Example (Swinnerton-Dyer, 1962 ) The following surface does not satisfy the Hasse Principle X : y(y + x)(2y + x) = (x + θz + θ 2 w) where the product is taken over the roots of T 3 7T T 7 = 0. Conjecture (Selmer ) Diagonal cubic surfaces satisfy the Hasse principle. 6

20 First Examples Example (Swinnerton-Dyer, 1962 ) The following surface does not satisfy the Hasse Principle X : y(y + x)(2y + x) = (x + θz + θ 2 w) where the product is taken over the roots of T 3 7T T 7 = 0. Conjecture (Selmer ) Diagonal cubic surfaces satisfy the Hasse principle. Example (Cassels & Guy, 1966 ) The diagonal cubic surface defined by 5x y 3 + 9z w 3 = 0 does not satisfy the Hasse Principle. 6

21 First Examples Example (Swinnerton-Dyer, 1962 ) The following surface does not satisfy the Hasse Principle X : y(y + x)(2y + x) = (x + θz + θ 2 w) where the product is taken over the roots of T 3 7T T 7 = 0. Conjecture (Selmer ) Diagonal cubic surfaces satisfy the Hasse principle. Example (Cassels & Guy, 1966 ) The diagonal cubic surface defined by 5x y 3 + 9z w 3 = 0 does not satisfy the Hasse Principle. 6

22 The Brauer-Manin Obstruction 7

23 The Brauer-Manin Obstruction For a fixed A Br X, the following diagram commutes 0 Br Q X (Q) X (A Q ) ev A ev A φ A Br Q p Q/Z 0 p inv 7

24 The Brauer-Manin Obstruction For a fixed A Br X, the following diagram commutes 0 Br Q X (Q) X (A Q ) ev A ev A φ A Br Q p Q/Z 0 p inv 7

25 The Brauer-Manin Obstruction For a fixed A Br X, the following diagram commutes 0 Br Q X (Q) X (A Q ) ev A ev A φ A Br Q p Q/Z 0 p inv Definition (Manin, 1971, 1974) X (A Q ) A = φ 1 A (0) 7

26 The Brauer-Manin Obstruction For a fixed A Br X, the following diagram commutes 0 Br Q X (Q) X (A Q ) ev A ev A φ A Br Q p Q/Z 0 p inv Definition (Manin, 1971, 1974) X (A Q ) A = φ 1 A (0) X (A Q ) Br = X (A Q ) A A Br X 7

27 The Brauer-Manin Obstruction 8

28 The Brauer-Manin Obstruction X (Q) X (A Q ) Br X (A Q ) 8

29 The Brauer-Manin Obstruction X (Q) X (A Q ) Br X (A Q ) Definition We say X has a Brauer-Manin obstruction to the Hasse Principle if X (A Q ) and X (A Q ) Br =. 8

30 The Brauer-Manin Obstruction X (Q) X (A Q ) Br X (A Q ) Definition We say X has a Brauer-Manin obstruction to the Hasse Principle if X (A Q ) and X (A Q ) Br =. Conjecture (Colliot-Thélène & Sansuc, 1979) The Brauer-Manin obstruction is the only obstruction to the Hasse principle for cubic surfaces. 8

31 The BSD Cubics

32 Setup for the Birch and Swinnerton-Dyer Cubics (1975) 10

33 Setup for the Birch and Swinnerton-Dyer Cubics (1975) Assume the following are true a, b, c, d Z, (1, φ, ψ) is a basis for a cubic extension over Q whose Galois closure is a S 3 extension L/Q, and K = Q(θ) is the unique quadratic extension contained in L. 10

34 Setup for the Birch and Swinnerton-Dyer Cubics (1975) Assume the following are true a, b, c, d Z, (1, φ, ψ) is a basis for a cubic extension over Q whose Galois closure is a S 3 extension L/Q, and K = Q(θ) is the unique quadratic extension contained in L. Then consider X : (ax + by)(x + θy)(x + θy) = m (cx + dy + φ i z + ψ i w). 10

35 A Sketch of the Sketch 11

36 A Sketch of the Sketch Assume [x : y : z : w] X (Q) and consider the factorization of (x + θy) as an ideal in O K. 11

37 A Sketch of the Sketch Assume [x : y : z : w] X (Q) and consider the factorization of (x + θy) as an ideal in O K. Things might go wrong. 11

38 A Sketch of the Sketch Assume [x : y : z : w] X (Q) and consider the factorization of (x + θy) as an ideal in O K. Things might go wrong. Suppose φ 3 = φ + 1, then 2O K = p 1 p 2. Take X : (x y)(x 2 + xy + 6y 2 ) = 2 (x + φz + φ 2 w). 11

39 A Sketch of the Sketch Assume [x : y : z : w] X (Q) and consider the factorization of (x + θy) as an ideal in O K. Things might go wrong. Suppose φ 3 = φ + 1, then 2O K = p 1 p 2. Take X : (x y)(x 2 + xy + 6y 2 ) = 2 (x + φz + φ 2 w). One finds p 3n+1 1 (x + θy). So for an A Br X, and all P X (A Q ) inv(a(p)) = inv 2 (A 2 (P 2 )) 0, and there is a Brauer-Manin obstruction to the Hasse Principle. 11

40 The Updated Version

41 The Updated Version of the BSD Cubics 13

42 The Updated Version of the BSD Cubics X : (ax + by)(x + θy)(x + θy) = m (cx + dy + φz + ψw) 13

43 The Updated Version of the BSD Cubics X : by(x + θy)(x + θy) = (x + φz + ψw) 13

44 The Updated Version of the BSD Cubics X : by(x + θy)(x + θy) = (x + φz + ψw) Theorem (W.) Suppose po L = p 1 p 2, L/K is unramified and p 1 θ while p 2 θ. Then X : py(x + θy)(x + θy) = (x + φz + ψw) has a Brauer-Manin obstruction to the Hasse principle. 13

45 Proof 14

46 Proof There is an A Br X such that A Q K = ( L/K, x + θy ). y 14

47 Proof There is an A Br X such that A Q K = Br X / Br Q H 1 (Gal(Q/Q), Pic X ) ( L/K, x + θy ). y 14

48 Proof There is an A Br X such that A Q K = Br X / Br Q H 1 (Gal(Q/Q), Pic X ) Pic X = µ, L 1, L 2, L 3, L 4, L 5, L 6 Z 7 ( L/K, x + θy ). y 14

49 Proof There is an A Br X such that A Q K = Br X / Br Q H 1 (Gal(Q/Q), Pic X ) Pic X = µ, L 1, L 2, L 3, L 4, L 5, L 6 Z 7 9 lines: { { { y = 0 x + φz + ψw = 0 x + θy = 0 x + φz + ψw = 0 x + θy = 0 x + φz + ψw = 0 ( L/K, x + θy ). y 14

50 Proof (Cont.) 15

51 Proof (Cont.) 18 more: { z = Ax + By w = Cx + Dy 15

52 Proof (Cont.) 18 more: { z = Ax + By w = Cx + Dy 1 + φ i A + ψ i C = 0 θ(1 + φ j A + ψ j C) = Bφ j + Dψ j θ(1 + φ k A + ψ k C) = Bφ k + Dψ k (Bφ + Dψ) = bθθ 15

53 Proof (Cont.) 18 more: { z = Ax + By w = Cx + Dy 1 + φ i A + ψ i C = 0 θ(1 + φ j A + ψ j C) = Bφ j + Dψ j θ(1 + φ k A + ψ k C) = Bφ k + Dψ k (Bφ + Dψ) = bθθ field of definition E/Q with [E : L] 6 15

54 Proof (Cont.) 18 more: { z = Ax + By w = Cx + Dy 1 + φ i A + ψ i C = 0 θ(1 + φ j A + ψ j C) = Bφ j + Dψ j θ(1 + φ k A + ψ k C) = Bφ k + Dψ k (Bφ + Dψ) = bθθ field of definition E/Q with [E : L] 6 [E : L] = 1 or 2: X satisfies the Hasse principle 15

55 Proof (Cont.) 18 more: { z = Ax + By w = Cx + Dy 1 + φ i A + ψ i C = 0 θ(1 + φ j A + ψ j C) = Bφ j + Dψ j θ(1 + φ k A + ψ k C) = Bφ k + Dψ k (Bφ + Dψ) = bθθ field of definition E/Q with [E : L] 6 [E : L] = 1 or 2: X satisfies the Hasse principle [E : L] = 3 or 6: H 1 (Gal(Q/Q), Pic X ) Z/3Z 15

56 Proof (Cont.) 16

57 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y 16

58 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y L/K unramified, P = [x p : y p : z p : w p ] X (K p ): inv p (B p (P)) = v p ( xp+θ py p y p ) n B [L p : K p ] mod 1 16

59 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y L/K unramified, P = [x p : y p : z p : w p ] X (K p ): [L p : K p ] = 1 or 3 inv p (B p (P)) = v p ( xp+θ py p y p ) n B [L p : K p ] mod 1 16

60 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y L/K unramified, P = [x p : y p : z p : w p ] X (K p ): [L p : K p ] = 1 or 3 inv p (B p (P)) = v p ( xp+θ py p y p ) n B [L p : K p ] p p or p = p 2 : v p ( xp+θ py p y p ) 0 (mod 3) mod 1 16

61 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y L/K unramified, P = [x p : y p : z p : w p ] X (K p ): [L p : K p ] = 1 or 3 inv p (B p (P)) = v p ( xp+θ py p y p ) n B [L p : K p ] p p or p = p 2 : v p ( xp+θ py p y p ) 0 (mod 3) p = p 1 : v p ( xp+θ py p y p ) = 2 and [L p : K p ] = 3 mod 1 16

62 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y L/K unramified, P = [x p : y p : z p : w p ] X (K p ): [L p : K p ] = 1 or 3 inv p (B p (P)) = v p ( xp+θ py p y p ) n B [L p : K p ] p p or p = p 2 : v p ( xp+θ py p y p ) 0 (mod 3) p = p 1 : v p ( xp+θ py p y p ) = 2 and [L p : K p ] = 3 Q X (A Q ): inv(a(q)) = inv p1 (B p1 (Q)) 0. mod 1 16

63 Proof (Cont.) X : py(x + θy)(x + θy) = (x + φz + ψw), B = ( L/K, x + θy ) y L/K unramified, P = [x p : y p : z p : w p ] X (K p ): [L p : K p ] = 1 or 3 inv p (B p (P)) = v p ( xp+θ py p y p ) n B [L p : K p ] p p or p = p 2 : v p ( xp+θ py p y p ) 0 (mod 3) p = p 1 : v p ( xp+θ py p y p ) = 2 and [L p : K p ] = 3 Q X (A Q ): inv(a(q)) = inv p1 (B p1 (Q)) 0. X (A Q ) Br = mod 1 16

64 Thanks! 17