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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