The Arithmetic of Quaternion Algebra

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.