ARITHMETICS OF JORDAN ALGEBRAS 11
Also M' is an o-lattice of $ which contains 1. We wish to show that M'
is an order. Let a. L, b. c M, i = 1, 2; (*a, + b . )U ,, =
i ' i ' 1 1 ota +b
« a U + *a,U
L
+ « a U
L
+ b , U + b U
u
+ b U . Since
1 « a
2
1 « a
2
, b
2
1 b
2
1 ora2 1 ara2, b
2
1 b
2
a a
0
9{r'1\ U = 0 . Also b.U . c M and aa.U , «(M n R^" 1 *) = «L.
2 a a
2
1 b
2
1 b
2
Let a' = a U , . Then a' ^ ( r " 1 ' , so a' o a = 1U , e &(r*. There -
1 1 l , b
2
1 ' 1 2 a i * a 2
2 2 2
fore, by (2) and (3), a a ^
fa
= a { a ^ b ^ = a { b ^ a ^ =
« Z [(b
2
o a
x
) o a
2
- { a
l b 2
a
2
} ] = " V ^ ^ ) . a
2
- b ^ ^ ^ ] =
" 2 K ° a 2 " b 2 U a
r
a
2
1 = ° S i n c e a i ° a 2 a n d b 2 U a
r
a
2
* W = °-
Finally by (3) a b ^
b
= " { a ^ * ^ } = a[(a
2
0 ^ ) 0 ^ - { b ^ b ^ ]

«L
(r-1)
sinc e a
0
o h = a0LL , c M n ft and (a
n
o b_ ) © b =
Z 1 Z 1 , D .
Z 1 Z
(a0 o bjCJ, , c M n / "
1
' = L.
2 1 l , b
2
q . e . d .
We consider next th e converse, namely: if $ is semisimple does it
contain maximal orders? Suppose # is simple then its centroid T is a finite
field extension of K. In general £ th e integral closure of 0 in r need
not be a finitely generated o-module (see for example [42] p . 443). The same
problem is encountered in classica l associativ e arithmetic where one assume s
either that © is a finitely generated o-module (this holds if r/ K is separable)
or that the algebra is central .
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