The Arithmetic of Quaternion Algebra

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68 CHAPTER 3. QUATERNION ALGEBRA OVER A GLOBAL FIELD 1. The class number h(D, N) of the ideals to the left of an Eichler order of level N in a quaternion algebra H/K of reduced discriminant D is equal to h. 2. The type number of the Eichler order of level N in H is equal to t(D, N) = h/h ′ (D, N), where h ′ (D, N) is the class number of the two-sided ideals of an Eichler order of level N. 3. h ′ (D, N) is equal to the class number in the restrict sense of the ideals belonging to the group generated by the square of the ideals of R, The prime ideal dividing D and the prime ideals I such that I m ||N with a odd power. Proof. The reduced norm induces a mapping: O × A \H× A /HK �→ n R A ×\K × A /KH, which is surjective , since n(H × v ) = K × v if /∈ Ram∞(H) , and injective if v ∈ R∞(H), R × A ⊃ K× v , since H × A ⊂ O× AH× K by the approximation theorem 4.3 for H1 and n(Op×) = R × p if p /∈ S. It follows the theorem and the part (1) of the corollary. We have n(N(Op)) = � K × p , if p|D or if pm K ||N with m odd ×2 p R × p , . otherwise It follows that the group of reduced norms of two-sided ideals of an Eichler order of level N is generated by the squares of ideals of R and the prime ideals I which divide D, or such that I m ||N with an odd power m. The class number of two-sided ideals of O is equal to the class number in the restrict sense of the norms of two-sided orders.It is then independent of the choice of O (among the orders of the same level). The type number of orders of a given level is then equal to the quotient of the class number of ideals ( this number is independent of the level) by the class number of two-sided ideals of an order of the level. Exercise 5.5-5.8. We consider an Eichler order O, an element x ∈ O, a two-sided ideal I of O , such that x is prime to I, that is to say, n(x) is prime to n(I). We shall give a generalization of the theorem of Eichler on the arithmetic progressions: Proposition 3.5.8. The reduced norm of the set x + I equals to the set KH ∩ {n(x) = J} where J = I ∩ R if S satisfies C.E. Proof. We verify easily that � it is true�locally. If x = 1, we utilize: n 1 + π x 0 a) the trivial relation n( ) = 1 + π 0 1 nx, b) if Hp/Kp is a field, then Hp � {Lnr, u} by II,1.7, and we have a well-known result (Serre [1]) that the units of Lnr being congruent to 1 modulo pn is sent surjectively to the units of Kp being congruent to 1 modulo pn .
3.5. ORDERS AND IDEALS 69 If x �= 1 and x is is a unit in Op for every place p such that Ip �= Op, and then we turn back to the precedent case. If Ip = Op we use n(Op) = Ip. It follows the global result from the local result by means of 4.1 and 4.3. Choose y ∈ KH ∩ {n(x) + J} such that – z ∈ H,n(z) = y, is integer except for possibly z ∈ S, – hv ∈ Ov, n(hv) = y, ∀v ∈ V and hp ∈ x + Ip if p /∈ S. There exists u ∈ H 1 K very near to z−1 hv ∈ H 1 v excluding possibly a place z ∈ S. The element zu of reduced norm y can be chosen such that zu ∈ O and zu ∈ x + I. Corollary 3.5.9. For every Eichler order O we have n(O) = KH ∩ R. The proposition allows to decide whether a maximal order is euclidean. The non-commutativity is obliged to distinguish the notion of euclidean order by right and left. Definition 3.19. An order O is right euclidean if for every a, b ∈ O there exist c, d ∈ O with a = bc = d, d = 0 or Nn(d) < Nn(b) where N is the norm defined by N(x) = Card(R/Rx) if x ∈ R. We define the left euclidean in a natural way. Definition 3.20. We say that R is euclidean modulo W , where W is a set of real places of K, if for every a, b ∈ R there exist c, d ∈ R with a = bc + d, d = 0 or Nd < Nb and d is positive for the places w ∈ W . Theorem 3.5.10. If R is euclidean modulo Ram∞(H), every maximal Rorderof H is left and right euclidean when S satisfies C.E. Proof. Let a, b belong to an order O of H. There exist x, y ∈ R such that n(a) = n(b)x + y with y = 0 or N(y) < Nn(b) and y ∈ KH. If n(a), n(b) are prime to each other, y �= 0, and according to 5.9, it exists d ∈ O such that a ∈ I + dwith n(d) = y, I ∩ R = Rn(b) where I is a two-sided ideal of O. We can verify easily that I ⊃ bO from this it follows that there exist c, d ∈ O with a = bc + d, Nn(d) < Nn(b). Coming along n(a), n(b) being prime to each other, we suppose that O is a maximal order. We begin by observing that we can assume a, b have no common left divisors, if one is interested in the right euclidean. The maximal R-orders are principal if R is euclidean modulo Ram∞(H). We can suppose also that the irreducible divisor P = Ox of the left ideal Oa are distinct from that of the ideal Ob. We shall show that it exists an element x ∈ O such that n(b) and n((a − bx) are prime to each other, then the theorem will be proved. Let P be an irreducible divisor of O n(b) in O. If b ∈ P then a /∈ P and for every x ∈ O , a − bx /∈ P . If b /∈ P , then a − bx ∈ P and a − bx ′ ∈ O implies b(x − x ′ ) ∈ P , Hence (x − x ′ ) ∈ P . Therefore there exists an infinity of x ∈ O such that a − bx /∈ P . The number of irreducible divisors of O n(b) is finite, thus we can find x with the property a − bx /∈ P, ∀P |O n(b). Then n(b) and n(a − bx) are prime to each other.
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The Arithmetic of Quaternion Algebra

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