By Richard D. Schafer

Concise examine provides in a quick house a few of the very important principles and ends up in the speculation of nonassociative algebras, with specific emphasis on substitute and (commutative) Jordan algebras. Written as an advent for graduate scholars and different mathematicians assembly the topic for the 1st time. "An vital addition to the mathematical literature"—Bulletin of the yankee Mathematical Society.

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Prove: A flexible algebra J is a noncommutative Jordan algebra if and only if any one of the following is satisfied: (38) (39) (40) (41) (x2 y)x = x2 (yx) for all x, y in J; x2 (xy) = x(x2 y) for all x, y in J; 2 2 (yx)x = (yx )x for all x, y in J; + J is a (commutative) Jordan algebra. We see from (41) that any semisimple algebra (of characteristic = 5) satisfying the hypotheses of Theorem 11 is a noncommutative Jordan algebra. Since (35 ) and (36) are multilinear, any scalar extension AK of a noncommutative Jordan algebra is a noncommutative Jordan algebra.

Then F has characteristic p, J+ is the pn -dimensional (commutative) associative algebra J+ = F [1, x1 , . . , xn ], xpi = 0, n ≥ 2, and multiplication in J is given by n (50) fg = f · g + ∂f ∂g · · cij , i,j=1 ∂xi ∂xj cij = −cji , where at least one of the cij (= −cji ) has an inverse. Proof: Since J = F 1 + N, every element a in J is of the form (51) a = α1 + x, α ∈ F , x ∈ N. By (51) every associator relative to the multiplication in J+ is an associator (52) [x1 , x2 , x3 ] = (x1 · x2 ) · x3 − x1 · (x2 · x3 ), xi ∈ N.

Proof: It is sufficient to prove that H(C3 ) is not special. For, if J were special, then J ∼ = J ⊆ A+ with A associative implies JK = K ⊗J ∼ = K ⊗ J ⊆ K ⊗ A+ = (K ⊗ A)+ = AK + so that H((CK )3 ) ∼ = JK is special, a contradiction. Suppose that H(C3 ) is special. There is an associative algebra A (of possibly infinite dimension over F ) such that U is an isomorphism of H(C3 ) into A+ . For i = 1, 2, 3 define elements ei in A and 8-dimensional subspaces Si = {di | d ∈ C} of A by (26) xU = ξ1 e1 + ξ2 e2 + ξ3 e3 + a1 + b2 + c3 JORDAN ALGEBRAS 39 for x in (24); that is, for ξi in F and a, b, c in C.