Absolute-Valued Algebraic Algebras Are Finite-Dimensional

Composition algebras of degree two

“…Let A be a finite dimensional composition algebra, and x, y two commuting elements of A. Then n(x)y 2 +n(y)x 2 =2(x,y)xy (12) This result also extends to three elements. If we set y = x 2 in (12) we obtain…”

“…This solution will exist if the determinant of the matrix of the coefficients is nonzero, that is, n(x)(x, x 3 ) -(x, x 2 ) 2 ^ 0; but this is precisely what is verified by any element of T, so for any x e T we obtain that x 4 , x 2 x 3 e span(x, x 2 , x 3 ). Moreover, by (12) we can ensure that x 3 x 3 and x 2 x 2 also lie in this subspace, so it is closed under products; in other words, alg(x) -span(x, x 2 , x 3 ) Vx e T. By density this holds for any x in A. Now the proposition follows from Lemma 7.…”

“…On the other hand, equation (12) with y >-> x 3 , x i-> x and y >-+ x 3 , x i-* x 2 gives n(x)x 3 x 3 + n(x) 3 x 2 = 2(x, x 3 )x 4 n(x) 2 x 3 x 3 + n(x) 3 x 2 x 2 = 2(x 2 , x 3 )x 3 x 2 .…”

“…More recently, it has been proved, by means of topological arguments, that the dimension of any algebraic absolute valued algebra is finite [12].…”

Two-graded absolute valued algebras

“…The work [30], by Angel Rodríguez Palacios, is an excellent survey of the actual state of the art. The following references are also fundamental for the reader: [4,5,15,19,[21][22][23]28,29]. In some cases, the results arising in the literature give conditions on an absolute valued algebra assuring that such an algebra is finite-dimensional.…”