Isotopy and Identities in Alternative Algebras

In this paper we show how to construct an isomorphism between an alternative algebra A over a field of characteristic 6= 3 and its isotope A(1+c), where c is an element of Zhevlakov’s radical of A. This leads to the equivalence of any polynomial identity f = 0 in alternative algebras and the isotope identity f(s) = 0. Given an invertible element s of an alternative algebra A, we can form a new algebra by taking the same linear structure but a new multiplication x ∗s y = (xs)y. The resulting algebra, denoted A, is also alternative (see [1]) and is called an s-isotope. Associative and Cayley isotopes are always isomorphic [2], and an isotope of a finite-dimensional alternative algebra over an algebraically closed field of characteristic 6= 3 is isomorphic to the original algebra as well [1]. We consider the case of an arbitrary alternative algebra over a field of characteristic 6= 3 with nonzero Zhevlakov radical, and a particular choice of s, and construct an explicit isomorphism between A and A. As a consequence, we derive the equivalence of any polynomial identity and its isotope in an arbitrary alternative algebras over a field of characteristic 6= 3. The paper has benefited from many discussions with J. Ferrar, and I would like to thank him for his valuable help and encouragement. Theorem 1. There is a set of coefficients tj , i, j ≥ 0, such that t0 = 1 and for any integer m, any alternative algebra A over a field of characteristic 6= 3 and any elements a, b, c from A such that IdA(c) m = 0, the polynomial t(x) = ∑ i,j t i jc xc satisfies t(a) ∗1+c t(b) = t(ab). (1) Proof. We will show the way to calculate the coefficients tj such that (1) holds for any alternative algebra A. Clearly, t(a) ∗1+c t(b) = t(a)t(b) + (t(a)c)t(b) = ∑ tjt k l (c ac)(cbc) + ∑ tjt k l (c ac)(cbc) = ∑ tl (t i j + t i j−1)(c ac)(cbc), Received by the editors March 28, 1995. 1991 Mathematics Subject Classification. Primary 17D05. c ©1997 American Mathematical Society 1571