The Arithmetic of Quaternion Algebra

34 CHAPTER 2. QUATERNION ALGEBRA OVER A LOCAL FIELD
Lemma 2.3.4. (Hijikata [1]). Let f, f ′ be two maximal inclusions of B in On.
Let nf = ˜ hnf, where ˜ �
0 1
hn is the inner automorphism induced by
πn �
.
0
(1) f is equivalent to f ′ modulo N(On) if and only if f is equivalent to f ′ or
n ′ × f modulo O n . If n = 0, the equivalence modulo N(On) coincides with the
equivalence modulo O × 0 .
(2) Let x, x ′ ∈ E and fx, fx ′ defined as that in the above lemma. Then f x is
equivalent to fx ′ modulo O× n if and only if x ≡ x ′ mod(πr+n ).
(3) If π−2n (t2 − 4n) is a unit in R (rep. not a unit in R), Then fx is equivalent
to nfx ′ if and only if x = t − x′ mod(πr+n ) ( rep. x ≡ t − x ′ mod(πr+n and
p(x ′ ) ≇ 0mod(πn+2r+1 ))
Proof. (1) is obvious. (2): if x ≡ x ′ mod(π r+n , we put a = π −r (x − x ′ ) and
�
1
u =
a
�
0
. Hence u ∈ O
1
× n and ufx(g)u−1 �
′ x
=
⋆
�
r π
. Inversely, suppose
⋆
that fx is equivalent to fx ′ modulo O× n . Since every element of O × n is upper
triangular modulo πn , if u ∈ O × n , π−r (ufx(g)u−1 − x) has the same diagonal
modulo πn with π − r(fx(g) − x), then x ≡ x ′ mod(πn+r ). (3): If π−n−2rf(x ′ )
is a unit, and nfx ′(g) satisfies the condition (3) of the above lemma, then it
�
′ t − x π
is equivalent to
r
�
. Besides, from (2) fx is equivalent to nfx ′
−π−rf(x ′ ) x ′
modulo O × n if and only if x = t − x ′ mod(πr+n ). If π−n−2rf(x ′ ) is not a unit,
we set u =
� �
1 b
0 1
for b ∈ R, and u n fx(g)u −1 = (xij). Modulo π n+r , then
xij = t−x ′ , x12 = b(2x ′ −t)−π −n+rf(x ′ ). Therefore if π−r (2x ′ −t) is a unit, or
in an equivalent way if π−2r (t2 − 4n) is a unit, we can choose b so that π−rx12 �
′ t − x π
is a unit, and the new (xij) is equivalent to
r
�
modulo O × n .
−π−rf(x ′ ) x ′
Finally, suppose that π−n−2rf(x ′ ) and π−2r (t − 4n) are not units, then if we
note that O × n is generated modulo πn by the diagonal matrices and the matrices
� �
1 b
of form , we see that for all the u ∈ O
0 b
× n , if unfx ′(g)u−1 = (xij), x12π−r is never a unit hence nfx ′ can not be equivalent to fx modulo O × n .
From these two lemmas we deduce the following proposition which allow us
to compute the number of maximal inclusion of Bs in On modulo the group
of inner automorphism induced by G = N(On) or O × n . The theorem 3.2 is a
consequence of it.
Proposition 2.3.5. (1) B can be embedded maximally in On if and only if E
is non-empty.
(2) The number of maximal inclusion of B in On modulo the inner automorphisms
induced by O × n equals the cardinal of the image of E in R/(π n+2r R) if
On = O0 is maximal, or if π −r (t 2 − 4m) is a unit. otherwise, the number is the
sum of the last cardinal and the cardinal of the image of F = {x ∈ E|p(x) ≡
0mod(Rπ n+2r+1 )} in R/(π n+2r R).
Proof. The proof of theorem 3.2. Suppose O = O0 is a maximal order. Since
N(O) = K × O × , the number of the maximal inclusion modulo the inner automorphism
induced by a group G with O × ⊂ G ⊂ N(O) depends not on G.