The Arithmetic of Quaternion Algebra

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58 CHAPTER 3. QUATERNION ALGEBRA OVER A GLOBAL FIELD everywhere, hence ZL(s) = ZK(s) 2 has a double pole at s = 1, this implies that L is not a field and then a ∈ K ×2 . n = 3, q(x, y, z) = ax2 + by2 + z2 , up to equivalence on K. Choosing H to be the quaternion algebra associating to {a, b}, the principle is equivalent to the characterization of matrix algebra. n ≥ 4, It turns back by induction to the precedent cases, cf. Lam [1],p.170. Since JH and ZK satisfy the functional equations: JH(s) = (−1) |Ram(H)| � p∈Ramf (H) ZK(s) = D s−1/2 K ZK(s). We obtain a functional equation for ZH: Np 1−s · ZH(2 − s), ZH(s) = (D 4 HN(dH) 2 ) 1 2 −s (−1) |Ram(H)| ZH(1 − s) which, if comparing it with the functional equation (thm.2.2)ZH = D 1 2 −s H ZH(1− s) obtained directly when H is a field, shows that DH = D 4 K N(dH) 2 , but immediately: Property II. The number of places ramified in a quaternion algebra is even. In the case of characteristic different from 2, this statement is equivalent to the reciprocal formula of Hilbert symbol. Corollary 3.3.3. (Reciprocal formula of Hilbert symbol). Let K be a global field of characteristic different from 2. For two elements a, b of K × , let (a, b)v be their Hilbert symbol on Kv. We have the product formula � (a, b)v = 1 where the product takes on every place v of K. v Application: 1) Choosing K = Q and for a, b two odd prime numbers, one can verifies the process for obtaining the quadratic reciprocal formula 2)Computation of symbol (a, b)2. The Hilbert symbol of two rational numbers a, b on Qp can be computed easily with the rule described in II,§1. We shall calculate (a, b)2 by using the product formula:(a, b)2 = � v�=2 (a, b)v. Before proving the property of existence of a quaternion algebra of the given local Hasse invariants, we extract some consequences of property I and property II above. The extension L/K are always assumed to be separable. Corollary 3.3.4. (Norm theorem in quadratic extension). Let L/K be a separable quadratic extension, and θ ∈ K × . For θ to be a norm of an element in L if and only if θ is a norm of an element in Lv = Kv ⊗ L for every place v excluding possibly one. Proof. The quaternion algebra H = {L, θ} is isomorphic to M(2, K) if and only if θ ∈ n(L) by I,2.4., and for that if and only if Hv � M(2, K) for every place v excluding possibly one byu property I and II. Since Hv � {Lv, θ}, the corollary is proved.
3.3. CASSIFICATION 59 Corollary 3.3.5. (characterization of neutralized field). an extension of finite degree L/K neutralize a quaternion algebra H over K if and only if Lw neutralize Hv for every place W |v of L. Proof. For L neutralizing H if and only if L ⊗ H � M(2, K), According to Property I, if and only if that for every place w of L we have (L ⊗ H) w�M(2,Lw . Using then the equality (L⊗H)w=Lw ⊗Hv if v = w|K; the second tensor product is taken on Kv. Lemma 3.3.6. Let K be a local field, and L = K(x) be a separable quadratic extension of K. Let f(x) be the minimal polynomial of x over K: f(x) = (X − x)(X − ¯x) = X 2 − t(x)X + n(x). If a, b ∈ K are enough near to t(x), n(x) respectively , then the polynomial g(X) = X 2 − aX + b is irreducible over K and has a root in L. Proof. If K = R, the discriminant t(x) 2 − 4n(x) is strictly negative, hence a 2 − 4b is so if a and b are enough near to t(x) and n(x). If K �= R, let y ∈ Ks such that y 2 = ay + b. If ||a|| < A and ||b|| < A, where A is a strictly positive constant, the inequalities finally prove that ||y|| < A. We have (y − x)(y − ¯x) = (t(x) − a)y = (n(x) − b), and hence can take ||(y − x)(y − ¯x)|| also small as we like if choosing a and b sufficiently near to t(x) and n(x). But x �= ¯x since the extension is separable, and it is possible to choose a and b such that ||y − x|| < ε, ||y − ¯x|| > varepsilon. There does not exist a K-automorphism f such that f(x) = ¯x, f(y) = y ! Thus K(y) ⊃ K(x), and because of [K(y) : K] ≤ 2, so K(x) = K(y). This lemma and the theorem of approximation (thm. 2.2) allow us to obtain Lemma 3.3.7. It exists a separable quadratic extension L/K such that Lv/Kv is equal to a given separable extension for v belonging to a finite set of places. Theorem 3.3.8. Let L/K be a quadratic extension and n the norm of L/K which is extended to ideals. We have [K × )] = 2. A : K× n(L × A Proof. Let χ be a character of K × A being trivial on K× n(L × and Zv = K × A ∩ {K× n(L × v ) � w�=v K× v } is closed in K × A , because of χ = � xv, and K v∈V × n(L × � A = Zv. v∈V We shall also prove the inequality [K × A : K× n(L × A element iv of K × A not belongs to K× n(L × A ): iv = (xw), withxw = A ). Locally χ2 v = 1, ] ≤ 2. We construct an � 1, if w �= v uv, where uv /∈ n(L × v ) if w = v , for every place v of K such that Lv is a field. This element does not belong to K × n(L × A ). If it belongs to K× n(L × a ), it would exist an element x ∈ K × such that x /∈ n(L × v ) but x ∈ n(L × w), ∀w �= v. it contradicts with 3.4.
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