Brauer Groups and the Hasse Principle Mckenzie West RANT 3/18/2013 For the discussing here, k is a eld, and all k-algebras are nite-dimensional algebras over k.A majority of these notes are gathered from [2] and [4]. 1 Central Simple Algebras Denition 1.1 A k-algebra is simple if it has no non-trivial two-sided ideals. Example 1.2 For all n > 0, Mn(k) is simple. For all i; j, the matrix Eij of all zeros except a 1 in the i; jth position is in Mn(k). If A = faijg 2 A, a two sided ideal of Mn(k) then EijAEij = aijEij is the matrix of zeros except the i; jth entry of A remains as aij. Use elementary matrix operations to conclude that the entries of matrices in A form an ideal in k. That is, A = Mn(I) for some I ⊆ k a two sided ideal. Since k is a eld, I = (0) or I = k. Theorem 1.3 (Wedderburn's Theorem) Every simple k-algebra, A, has ∼ A = Mn(D); ∼ for an integer n and division k-algebra D, unique up to isomorphism. Moreover, A ⊗k Mn(k) = Mn(A). Proof: Idea: Let M be a simple A-module, get D, the centralizer of A in EndA(M). Then show ∼ opp A = EndD(M) = Mn(D ). For more, see [3] IV.1.15 and IV.2.1. Denition 1.4 A k-algebra is central if its center is exactly k. Example 1.5 The matrix rings Mn(k) are central k-algebras. Each A 2 C(Mn(k)), the center of Mn(k), has AEii = EiiA. That is, the ith column of A with zeros everywhere else equals the i row of A with zeros everywhere else. This implies that the only non-zero entries are on the diagonal. Further AE1i = E1iA, so each of the diagonal entries are equal. Example 1.6 The Hamiltonian quaternions H = C + iC + jC + ijC are a central simple C-algebra. We have a homomorphism H −! M2(C); given by p −1 0 0 −1 i 7! p and j 7! : 0 − −1 1 0 1 Note that p p w + −1x −y − −1z w + xi + yj + zij 7! p p : y − −1z w − −1x p p p This is an isomorphism since the equations w + −1x = a, w − −1x = d, −y − −1z = b, and p a b y − −1z = c have unique solutions for every 2 M ( ). c d 2 C Theorem 1.7 For a k-algebra, A, the following are equivalent, (i) A is a central simple algebra. ∼ (ii) There is a nite separable extension L of k and n ≥ 1 such that A⊗k L = Mn(L) as L-algebras. ∼ (iii) For some n ≥ 1, A ⊗k ks = Mn(ks) as ks algebras, where ks is the separable closure of k. ∼ (iv) There is a eld extension L of k and n ≥ 1 such that A ⊗k L = Mn(L) as L-algebras. ∼ (v) There is a nite dimensional central division k-algebra D such that A = Mr(D) for some r ≥ 1. The D and r are uniquely determined by A. Proof: See Theorem 1.3, and drop some isomorphisms like a boss. Denition 1.8 A eld L as in (iv) is called a splitting eld for A We now have a category, CSAk, of central simple algebras over k with the morphisms being k-algebra homomorphisms. opp For A 2 CSAk, dene A , the opposite algebra of A, to be the k-algebra with the same underlying set and algebra as A but with multiplication a · b = ba for a; b 2 Aopp. Certainly, opp A 2 CSAk. Additionally, if A; B 2 CSAk then A ⊗k B 2 CSAk, too. 2 The Brauer Group For A; B 2 CSAk, say A ∼ B if one of the following (equivalent conditions) holds: ∼ ∼ (a) There exist m; n ≥ 1 and division algebra D 2 CSAk such that A = Mm(D) and B = Mn(D) as k-algebras. ∼ (b) There exist m; n ≥ 1 such that Mn(A) = Mm(B). Dene Br k to be the set CSA= ∼, of similarity classes. The operations A; B 7! A ⊗k B and opp A 7! A on CSAk induce a group structure on Br k. Moreover, Br k is abelian. We call the abelian group Br k the Brauer group of k. If A 2 CSAk, and L is a nite extension of k, we have that A ⊗k L 2 CSAL. Any two-sided ideals of A⊗k L must come from A and a eld, and thus must be trivial. Then the map A 7! A⊗k L induces a homomorphism Br k ! Br L. Note that as a result, Br is a covariant functor from elds to abelian groups (i.e. it preserves arrows). 3 The Hasse Principle Suppose X is a curve, or more generally, a variety over k. Then we say X satises the Hasse Principle if X(kv) 6= ; for all places v implies X(k) is non-empty. Theorem 3.1 (Hasse, Minkowski, see [1]) Projective curves dened by one quadratic form over a number eld k satisfy the Hasse Principle. Example 3.2 The curve 2y2 = x4 − 17z4; does not satisfy the Hasse Principle over Q. Certainly, the curve has real solutions. Recall the Hasse bound: p Smooth curves of genus 1 over Fp, p ≥ 3, have at least p + 1 − 2 p > 0 points in Fp. Thus by Hensel's Lemma, we have points in Qp for all primes p 6= 2; 17. Over F2, (1; 1; 1) is a smooth point, and over F17, (1; 3; 0) is a smooth point. Since these are smooth points, we can use Hensel's Lemma here to extend them to points in Q2 and Q17, respectively. Now suppose there is a solution (X; Y; Z) in Q. We can assume that (X; Z) = 1 and Y > 0. If q j Y is prime, then X4 = 17Z4 (mod q). Hence 17 is a square modulo q and by quadratic reciprocity, q is a square modulo 17. Additionally, 2 and -1 are squares modulo 17, so Y is a square modulo 17. Write 2 . Then 4 4 . We conclude that is a fourth Y = Y0 (mod 17) 2Y0 = X (mod 17) 2 power modulo 17. But it is not! Thus there is no solution in Q. 4 The Brauer-Manin Obstruction In cohomology, there is a map inv 2 for any Galois extension , called the L=k : H (L=k) ! Q=Z L=k invariant map. See [3] or [2] for more information on this. This will motivate the following Theorem. Theorem 4.1 If ks is the separable closure of k, then there is an isomorphism of abelian groups Br 2 × (k) ' H (k; ks ): Proof: (Partial) Suppose A 2 CSAk. By denition, we have an isomorphism φ : Mr(ks) ! A⊗k ks. σ In fact, Gk = Gal(ks=k) acts on φ, where φ is the map such that the diagram φ Mr(ks) −−−−! A ⊗k ks ? ? σ? σ? y y σφ Mr(ks) −−−−! A ⊗k ks commutes. Set −1 σ Aut ξσ := φ ( φ) 2 ks−alg(Mr(ks)): Note that every matrix automorphism is inner, so Aut . For , ks−alg(Mr(ks)) ' P GLr(ks) σ; τ 2 Gk we have that −1 στ −1 σ σ −1 στ σ −1 τ σ ξστ = φ ( φ) = φ ( φ) φ ( φ) = ξσ (φ φ) = ξσ ξτ : Thus ξ is a 1-cocyle. When we change the isomorphism, φ, we are simply composing with auto- morphisms of Mr(ks). This will change ξ to a cohomologous cocycle. Thus we have an element in 1 H (Gk; P GLr(ks)) the depends on A. It follows from descent that 2 ∼ 1 fA 2 CSAk : dim A = r g=k − iso: = H (Gk; P GLr(ks)): The short exact sequence × 1 −−−−! ks −−−−! GLr(ks) −−−−! P GLr(k2) −−−−! 1; yields another, 1 1 2 × 0 = H (Gk; GLr(ks)) −−−−! H (Gk; P GLr(ks)) −−−−! H (Gk; ks ): We will show that the composition map Br 2 × is a surjection. (It is an injection as Φ: (k) ! H (Gk; ks ) a result of cohomology, see [5].) That is, if , a 2-cocycle with values in ×, then −1 as;t ks as;t = pss(pt)pst where ps 2 GLq(ks) is a 1-cocycle. Suppose V is a vector space over ks with basis es for s 2 G. Set to be . Then ps 2 homks (V; V ) ps : et ! as;test pss(pt)eu = as;tus(at;u)estu; and as;tpst = as;tast;uestu: The desired equality follows since as;t is a cocycle. Remark: This same proof works to classify all sorts of objects, such as elliptic curves with a xed j-invariant. From global class eld theory, we have an exact sequence Br L Br 0 −−−−! (k) −−−−! v (kv) −−−−! Q=Z −−−−! 0 The last non-zero map is given by the invariant maps as discussed previously. Extend the idea of a central simple k-algebra to an algebra over X(k) for a curve X that allows us to pass from X(k) to Br(k) and from X(kv) to Br(kv). Now we have a bigger and better version of the commutative diagram above, Q X(k) −−−−! v X(kv) ? ? ? ? y y Br L Br 0 −−−−! (k) −−−−! v (kv) −−−−! Q=Z −−−−! 0 If X(k) is non-empty, then mapping down to Br(k) and over to Q=Z gives 0 by the exactness of the sequence.