The Arithmetic of Quaternion Algebra

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26 CHAPTER 2. QUATERNION ALGEBRA OVER A LOCAL FIELD Proof. We shall lead an absurdity. If Lnr was not contained in H, then for every x ∈ O, x /∈ R, the extension K(x)/K would be ramified. There exists then a ∈ R such that x − a ∈ P ∩ K(x). We could then write x = a + uy with y ∈ O. Iterating this procedure, the element x could be written as � n≥0 anun , an ∈ R. The field K(u) being complete would be closed. We thus had O ⊂ K(u). It is absurd. Corollary 2.1.7. The quaternion field H is isomorphic to {Lnr, π}.Its prime ideal P + Ou satisfies P 2 = Oπ. Its integer ring O is isomorphic to RL + RLu. The reduced discriminant d(O) of O equals n(P ) = Rπ. Proof. According to I, Corollary 2.2 and 2.4, we have H � {Lnr, x} where x ∈ K × but x /∈ n(L × nr). From (1), (2) of the beginning of this chapter it follows x = πy 2 where y ∈ K × . We can suppose x = π , then the first part of the corollary follows. Now suppose H = {Lnr, π}. The element u ∈ H satisfying I.(1) is the non zero minimal valuation, thus P = Ousatisfies P 2 = Oπ. The prime ideal Rπ is then ramified in O. according to Lemma 1.4, we have O = {h ∈ H, n(h) ∈ R}. similarly, RL = {m ∈ Lnr, n(m) ∈ R}. We can verify easily that, if h = m1 + m2u with mi ∈ Lnr, the property n(h) ∈ R is equivalent to n(mi) ∈ R, i = 1, 2. We can show too that O = RL + RLu. Using the formula involving the determinant in I, Lemma 4.7, we compute the reduced discriminant d(O). Since d(RL) = R, we see easily that d(O) = Rπ. Hence it follows d(O) = n(P ) or the different of O is O ⋆−1 = P . Definition 2.4. Let Y/X be a finite extension of field equipped with a valuation, and ring of it is AY , AX = X ∩ AY . Let PY , PX = PY ∩ AX be the prime ideals and kX, kY be the corresponding residue field. The residue degree f of Y/X is the degree [kY : kX] of the residue extension kY /kX. The ramification index of Y/X is the integer e such that AY PX = P e Y . We then deduce that the unramified quadratic extension Lnr/K has the ramification index 1, and the residue degree 2. The quaternion field H/K has the ramification index 2, and the residue degree 2. Let F/K be a finite extension of commutative field with ramification index e and residue degree f. We have ef = [F : K] because the cardinal of k is finite and RF /πRF � RF /πF e RF , if πF is a uniform parameter of F . Lemma 2.1.8. The following properties are equivalent: (1) f is even, (2) F ⊃ Lnr, (3) F ⊗ Lnr in not a field. Proof. For the equivalence (1) ←→ (2) see Serre [1],Ch. 1. For the equivalence (2) ←→ (3), it is convenient to write Lnr as the K[X]/(P (X)) where (P (X)) is a polynomial of degree 2. Thus F ⊗ Lnr equals F [X]/(P (X))F where (P (X))F is the ideal generated by (P (X)) in the polynomial ring F [X]. Since P (X) is a polynomial of degree 2, it is reducible if and only if it admits a root in F , i.e. if F ⊃ Lnr. Consider now HF � {F ⊗ Lnr, π}. If πF is a uniform parameter of F , it may as well suppose that π = πF e . According to I. Corollary 2.4, and Lemma 1, we find that, if e or f are even then we have HF � M(2, F ), hence F neutralizes H. Otherwise, that is to say if [F : K] is odd, HF � {F ⊗ Lnr, πF } ,where
2.1. CLASSIFICATION 27 F ⊗ Lnr is the unramified quadratic extension of F in Ks. Therefore HF is a quaternion field over F . The Theorem 1.2 is proved completely. For the use latter we make a remark here. Corollary 2.1.9. Every quadratic extension of K is isomorphic to a subfield of H. For an order of a maximal commutative subfield of H can be embedded maximally in H, if and only if it is maximal. The computation of Hilbert symbol Lemma 2.1.10. If the characteristic of k is different from 2, and if e is a unit of R which is not a square, then the set {1, e, π, πe} form a group of representations in K × of K × /K ×2 . Moreover Lnr is isomorphic to K( √ e). Proof. Consider the diagram 1 → R × 1 → R× → K × → 1 ↓ 2 ↓ 2 ↓ 2 R × 1 → R× → K × The vertical arrows represent the homomorphism h → h2 , and R × 1 1 + πa, a ∈ R}. We have [k × : k ×2 ] = 2, and R × 1 = R×2 1 because of (1 + πa) 1 2 = 1 + πa/2 + ... + C 1 2 n + ... = {h = converges in K. Thus [R × : R ×2 = 2, and [K × : K ×2 ] = 4. If e ∈ R × − R ×2 , R × ⊂ n(K( √ e)), and this characterize Lnr = K( √ e). Set ε = 1 if −1 is a square in K, and ε = −1 otherwise. Table of Hilbert symbol: a\b 1 e π πe 1 1 1 1 1 e 1 1 −1 −1 π 1 −1 ε −ε πe 1 −1 −ε ε Definition 2.5. Let p be an odd prime number, and a be an integer prime to p. The Legendre symbol ( a p ) is defined by ( a ) = p � 1, if a is a square mod p . −1, otherwise We see immediately that the Hilbert symbol (a, p)p of a, p in Qp equals Legendre symbol ( a p ). It is easy to compute Hilbert symbol (a, b)p of two integers a, b in Qp if p �= 2. We use the the computation rule of Hilbert symbol (Corollary 2.2) and � (a, b)p = 1, ( if p ∤ a, p ∤ b a p ), if p ∤ a, p�b
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