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 1 L1 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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