ARITHMETICS OF JORDAN ALGEBRAS 29
L gives rise to maximal orders of «? write L a s in Lemma 7, then
E(L) =
-1
ct.
1
0
-1
*,
1
-1
G, CU
1 2
-1
o,
1
Q Q
a
1
Q,
-1 -1 -1
2 2 1 2
-1 -1 -1 -1
Q
a , a
Q
ct
\n n 1 n n 2
A typical element of M = # (K0 , L ) n E ( L ) is
h °
o e,
6 - ( 3
a (3
e . o
-7 a 0 £
2 * "
with ^ o, a o n o~ 1 o
1
= o (a1 D Q
2
) , P
6
Q
2
n a
i
= Q
2
,
7
a" 1 n a" 1
= Qj , 6 e o n o
1
Q
2
C o. Fix a then M maximal and a C a imply
Q2 = a so that Q Q = o and p

a , 6

o. Repeat the process to get
n
Q = o 1 i n. That M = $ n E(L), L = 0 Y (ox. 0 ay.), {x.,y. } a
1 1
i
L j
-
l l l l
1=1
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