ARITHMETICS OF JORDAN ALGEBRAS 39

PROOF. Induct on the rank of L. If rank L = 1 there is nothing to

prove. Consider {| h(x, y)| | x, y c L}. Since L is finitely generated

sup{|h(x , y)| | x , y

e

L } is attained. If |h(x, x)| | h ( u , v ) | for all u , v

e

L ,

L = Ox is a component. Indeed if y € L, y = ax + v, v e (&L ) ,

h(x, ax + v) = h(x, x) a + h(x, v) = h(x, x) a and |h(x, x ) a | |h(x, x)| t h e r e -

J_

fore a e 0 and v e (&L ) nL. If | h ( x , x ) | is not maximal for any x

G

L

then | h(x , x )| is maximal for some x t x x x

e

L and

L = Ox + Ox is modular. A primitive vector y = a x + a x of L must

have at leas t one a. a unit of 0, so | h(x., y)| = | h(x , x )| , 1 i i j 2

sinc e max(| h(x , x )| , |h( x , x ) | ) |h(x , x ) | . Hence L is

v(h(x x ))-modular. Moreover L- is a component. If y eL, y = y, + v,

y € $L , v

e

(&L ) . We wish to show that y e L . Therefore we may assum e

y t 0. Pick d e $ such that dy is a primitive vector of L . So

v(h(x x )) v(h(x x ))

h(dy 1^) = ? . But h(y

f

L ) = h(y,L

1

) C h(L,L) = ? by

v ( h ( x

r

x

2

) )

the assumptions on x , x Therefore r =dh( y ,L )

v(h(x , x

2

) )

±

C d*P and v(d) 0. Hence y e L and v = y - y e L n (JSL ) .

q. e. d.

Planes L ~ (

H

) w i t n l b l = I o| and | a | , | d | | b | are called

subnormal; (the reason behind the terminology will become clear in

Chapter III).