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).
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