ARITHMETICS OF JORDAN ALGEBRAS
51
v (d c d ) v(c ) and v(c - d c d ) = v(c. ). Therefore
- - 1 -1 " v ( c i ) -1 - l - - 1
(c - d c d
2
) x $ . Let a = (c - d c d^) x, b = - c d^a.
-v(c
2
)
Then a , b
6
? and a
[ 2 3 ]
+ b
[ 1 3 ]
c M
4
. But ( e
n
+ e
2 2
) ( a
[ 2 3 ]
+ b
[ 1 3 ]
)
= ( C 1 S + d l b ) 1 3 + (V + ° l b , 2 3 = (C 1 ~ V l l d 2 ) a i 3 + ( d 2 " d 2 ) a 2 3 ~~ X l 3 e M 4 '
/ 0 © \
© ©
Similar arguments show that C M' Multiplying this out,
r r
r r
using r 0 and M' M' C © we obtain the desired results .
q. e. d.
We are now in a position to complete the proof of Theorem 3. Assume
that conditions (1) - (3) are satisfied and that we do not have the already
ruled out exception. So L = L J_ L , L i-modular, L (i+1 )-modular
(L could be empty). Assume first that rank L. ^ 1, j = 1, 2, and let
rank L = s. We claim that
© . . . © *P .
(5)
M' © . . . © T3 . . . ?
? . . . ? ©
. . .
©
? . . . ? © . . . ©
s
First any plane satisfies the hypothesis of Lemma 15 unles s & ha s no
symmetric prime and the plane is i-modular with i odd. In that case , since
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