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
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