ARITHMETICS OF JORDAN ALGEBRAS

17

x

e

C. But C n L = U xO(L). Since C n L is compact

xcCnL

r

c n L = U X.CXD, x.

€

c n L.

i=l

l l

q. e. d.

Knebusch [24] has studied the orbits [x], x

e

C n L when the

characteristic of K is not 2.

§4. Maximal orders of Jordan division algebras over a complete discrete

valuation field.

Let #/K be a central Jordan division algebra, K a complete discret e

valuation field.

LEMMA 2. An element x

€

? is integral if and only if N(x)

e

o.

PROOF. Consider K(x), x ^ 0; this is a field. Since x is invertible

N(x) £ 0. Lemma 2 is then only a restatement of the same well known result

for fields (e.g . [6], p. 99; [40], p. 220).

q. e. d.

PROPOSITION 8. Let ^ be a Jordan division algebra over K a com -

plete discrete valuation field. The set of elements with integral norm is the

unique maximal order of ^.

PROOF. Let M = {x

6

p\ N(x) = o}. Since N(l) = 1, 1 e M; als o

2

N(x), N(y) € o imply N(xU ) = N(x)N(y) € o so xU eM. We wish to show

next that N(x), N(y) e o imply N(x + y)

€

o. Without los s of generality we

(x"

1

)

may assume | N(y)| | N(x)| where | | is the valuation on K. Let f '

be the x " isotope of p. Since N

( x

^(z) = N(x)

- 1

N(z), N

( x

\z) fi 0 for