The Arithmetic of Quaternion Algebra

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64 CHAPTER 3. QUATERNION ALGEBRA OVER A GLOBAL FIELD Proof. According to the definition of lattice (I.4), for a given lattice Y , there exists a, b ∈ K × such that aY ⊂ X ⊂ bY . For almost every v /∈ ∞, av, bv are units. Thus Xp = Yp, p.p. We shall prove V ↦→ (Vp) p/∈S is surjective. If (Zp) p/∈S is a set of local lattices which are almost everywhere equal to Xp, we then set Y = � p/∈S (H ∩ Zp). We want to prove Y is a lattice, and Yp = Zp. There exists a ∈ R such that aXp ⊃ Zp ⊃ a −1 p for every p /∈ S. It follows aX ⊃ Y ⊃ a −1 X, hence Y is a lattice. Since S �= ∅, according to 1.4, H is dense in � p/∈S Hp. From this we have H ∩ (πZp) = Y is dense in πZp. in particular, Y is dense in Zp, thus Yp = Zp if p /∈ S. We now prove Y ↦→ (Yp) p/∈S is injective. let Z � p/∈S (Yp ∩ H). We claim that Y = Z. It is true that Y ⊂ Z, and there exists a ∈ R such that aZ ⊃ Y ⊃ Z. Let z ∈ Z. There exists y ∈ Y very near to z by p-adic for every place p /∈ S, such that a is not a unit in Rp. In fact, we have Yp = Zp if p /∈ S, and we utilize the approximation theorem 1.4. There exists then y ∈ Y such that y − z ∈ aZ. We conclude that z ∈ Y . The proposition is proved. Definition 3.16. A property ⋆ of lattice is called a local property when a lattice Y has the property ⋆ if and only if Yp has the property ⋆ for every p /∈ S. Examples of local property: The properties for a lattice to be 1. an order, 2. a maximal order, 3. an Eichler order, i.e. the intersection of two maximal orders, 4. an ideal, 5. an integral ideal, 6. a two-sided ideal, are the local properties. This can be deduced easily by the proposition 5.1. We utilize that if I is an ideal, then its left order Ol(I) (cf. I.4, above Prop.4.2) satisfies Ol(I)p = Ol(Ip) for all p /∈ S. Definition 3.17. The level of Eichler order O is an integral ideal of R, denoted by N such that Np is the level of Op ∀p /∈ S. Corollary 3.5.2. Let I be an ideal of H,and O be an order of H. n(I) denotes the reduced norm of I, and d(O) the reduced discriminant of O. Then we have n(Ip) = n(I)p and d(Op) = d(O)p. Proof. If (f) is a finite system of generators of I/R, by the definition (I. above the lemma 4.7) n(I) is the R-ideal generated by n(f). Moreover (f) is also a finite system of generators of Ip/Rp. It follows that n(Ip) = n(I)p. By the definition in I,§4, above the lemma 4.9, I ∗ = {x ∈ H|t(xf) ∈ R, ∀f}. By the proposition 5.1 we obtain (Ip) ∗ = (I ∗ )p. Replacing I by O, and taking the reduced norm, we see that d(O)p = n(O ∗−1 )p = [n(O ∗ ) −1 ]p = n(O ∗ ) −1 p = n(O ∗−1 p = d(Op).
3.5. ORDERS AND IDEALS 65 From II.1.7 and II.2.3 we obtain a characterization of maximal orders by their reduced discriminant. This permit us to use it for the construction of a maximal order, or to distinguish whether a given order is maximal. Corollary 3.5.3. For an order O to be a maximal order if and only if its reduced discriminant to be equal to d(O) = � p. p∈Ram(H),p/∈S Set d(O) = D; the reduced discriminant of an Eichler order of level N is equal to DN. However, the Eichler orders are not characterized by their reduced discriminant unless it is square-free. Since (D, N) = 1, it is equivalent to say that N is square-free. See exercise 5.3. Example: let H be the quaternion field over Q of reduced discriminant 26, i.e. the field generated over Q by i, j satisfying i 2 = 2, j 2 = 13, ij = −ji. In fact, The Hilbert symbol (2, 13)v for the valuations v of Q are (2, 13)∞ = 1, (2, 13)13 = ( 2 13 ) = −1, (2, 13)p = 1, if p �= 2, 13 and the product formula � (2, 13)v = 1 gives (2, 13)2 = 1. We can show that O = Z[1, i, (1+j)/2, (i+ij)/2] is a maximal order if and only if it makes sure that 1. O is a ring, 2. the elements of O are integers; the reduced trace and the reduced norm are integers, 3. O is a Z-lattice, Q(O) = H (the last property is obvious for it), 4. The reduced discriminant of O equals 26. Addition table: The trace of the sum of two integers is an integer , we verify that the norm remains integer in the following table. i (1 + j)/2 (i + ij)/2 i 2i i + (1 + j)/2 i + (i + ij)/2 (n = −8) (n = −5) (n = 4) (1 + j)/2 ∗ 1 + j (1 + i + j + ij)/2 (n = −12) (n = 3) (1 + ij)/2 ∗ ∗ 1 + ij (n = 24) Multiplication table : The norm of the product of two integers is integer, we verify in the following table that the reduced trace remains integer too, and the product is stable in O. left\right i (1 + j)/2 (i + ij)/2 i 2 (i + ij)/ 1 + j (1 + j)/2 (i − ij)/2 = i − (i + ij)/2 (7 + j)/2 = 3 + (1 + j)/2 −3i (i + ij)/2 1 − j = 2 − 2(1 + j)/2 (7i + ij)/2 = 3i + (i + ij)/2 7
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