12

M . L. RACINE

PROPOSITION 3. If $/K is a Jordan algebra such that the trace form

T(a,b) is non-degenerate then any order M of ^ is contained in a maximal

order.

PROOF. Since M is an o-module of integral elements, a, b € M

implies T(a, b) = T(a)T(b) - cr2(a, b)

€

o, Let M = {x

€

$ | T(x, M) C o } be the

dual module of M with respect to T. Since M is a finitely generated

o-module so is M. If M C M C M C . . . is a chain of orders, then

T(M.,M.) C o so M. C M. Therefore the chain must stop (o is Noetherian).

Hence M is contained in a maximal order.

q. e. d.

If the characteristic is not 2, $ is separable if and only if T(a,b) is

non-degenerate ([16] Theorem 5 p. 240). If 9 is simple exceptional then T

is non-degenerate irrespective of the characteristic of K ([29]). Assume that

s

$ is semisimple, «?=?, 0 . . . 0 ? with #. simple; 1 = V e., e, the

1 S 1 1L1 i i

unit element of #.. Let M be a maximal order of 3 and let M. be the

l ^ i

s

J

image of M under the projection of $ on p.. Since ) e, € M, e. e M.

and since M is a lattice of 5, M, is a lattice of $.. If x = x.,

J ] & l

y = V y. € M, yU = y y.U e M and y.U e M.. Therefore M. is an

i=i 1 x 1=1 1 x i ] x j ] J

order of $. and M 0 . . . 0 M is an order of p. But M C M 0 . . . 0 M .

So by maximality M = M, 0 . . . 0 M . Conversely given M. a maximal

I s

I

order of ?., 1 i s, M = M 0 . . . 0 M is a maximal order of #. For

if M C M' an order of $ then arguing as above the projection of M* on ^.

is an order containing M, and hence must be M.. In view of the discussion