“…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.…”