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0 holds. It will follow that f (U xn1 , U xn2 , . ) = 0 holds in F(X) if we can show that the homomorphism of X onto Y is an isomorphism. We have seen that X is generated by U x and V x and Y is generated by Uy and Vy . Since U x → Uy and V x → Vy the isomorphism will follow by showing that Uy and Vy are algebraically independent over Φ. Now in Φ{Y}(q) we have Uy = yR yL , Vy = yR = yL where aR is b → ba and aL is a b → ab and yL and yR commute and are algebraically independent over Φ since if z ∈ Y, z y, then zkRk ylL = yl zyk and the elements yl zyk , l, k = 0, 1, 2, .

It is clear from the universal property of F(X) that of f (Un1 , Un2 , . . , . ) = 0 holds in F(X) then f (Uan1 , Uan2 , . ) = 0 holds in every quadratic Jordan algebra. Hence it suffices to prove f (Un1 , Un2 , . . , . ) = 0. Let Y be a set of the same cardinality as X and suppose x → y is a bijective mapping of X onto Y. Let Φ{Y} be the free associative algebra (with 1) gener- 39 ated by Y and let F s (Y) be the subalgebra of φ{y}(q) generated by Y. We have a homomorphism of F(X) onto F s (Y) such that x → y.

2) We use α[x, y, z] = [αx, y, z] = [xα, y, z] = [x, y, z]α for 62 x, y, z ∈ O, α ∈ N, and (xyx)z = x(y(xz)) (see the author’s book [2], pp. 18-19). We have to show that [xαx, y, z] = 0 if α ∈ N, x, y, z ∈ O. Since xx ∈ N this will follow by showing that [xαx, y, z] = [xx, y, z]α. For this we have the following calculation: [xαx, y, z] = [xαt(x), y, z] − [xαx, y, z] 1. Basic Concepts 44 = t(x)[x, y, z]α + (x(α(x(yz))) − (x(α(xy)))z = t(x)[x, y, z]α − (xα)[x, y, z] + (xα)((xy)z) − [xα, xy, z] = (xα)((xy)z) = g(x, y, z)α where g(x, y, z) = t(x)[x, y, z] − x[x, y, z] − [x, xy, z].