DETERMINATION OF A+ FOR THE SIMPLE FLEXIBLE ALGEBRAS* BY RICHARD E. BLOCK UNIVERSITY OF CALIFORNIA (RIVERSIDE) Communicated by A. Adrian Albert, July 26, 1968 1. Introduction.-We consider nonassociative algebras which are finite- dimensional over a field of characteristic $ 2. For any algebra A (with product written as xy), the symmetrized algebra A+ is defined as the algebra with the same underlying vector space as A and with product xmy = '/2(xy + yx). The algebra A is called flexible if (xy)x = x(yx) for all x,y in A. One of the principal problems in the theory of noncommutative power-associative algebras has been the determination of A+ when A is a simple flexible power-associative algebra. Using idempotent decomposition methods, a number of authors have determined A + in various cases. However, as in other situations, the degree 2 case has been especially difficult and this one case, at characteristic p, remains open. In this note we shall give a uniform determination of the structure of A +, in particular solving the above problem for degree 2 algebras. The result leads in principle to a determination of A. The proof is chiefly based on the author's very general new theorem on differentiably simple algebras' and does not assume power- associativity. The flexible algebras (introduced by Albert2) provide a natural setting for the application to noncommutative algebras of idempotent decompositions in commutative algebras. Included among the flexible algebras are all commuta- tive and anticommutative algebras (in particular, Jordan and Lie algebras) and alternative algebras, as well as the noncommutative Jordan algebras (defined by Schafer3 as flexible algebras with (x2y)x = x2(yx) for all x,y in A). For the con- cepts of degree, nodal, etc., mentioned below, see for example the book by Schafer.4 If A is a finite-dimensional flexible strictly power-associative non-nil simple algebra over a field F of characteristic # 2, 3, then A has a unity element and hence a degree t. If t > 2, work of Schafer3 and Oehmke' shows that A+ is a simple Jordan algebra. At characteristic 0, Kokoris6 has shown that A + is simple if t = 2, and Kleinfeld and Kokoris7 have obtained the same result if t = 1 (Anderson8 has given another proof of this last case). At characteristic p when t = 2, Oehmke' has shown that A+ is simple if A is noncommutative Jordan, and Tsai9 has obtained the same conclusion if A is partially stable; when t = 1, Kokoris'0 has shown that A + need not be simple but (at least when F is algebraically closed, so that A is nodal) A + _B,(F) for some n _ 0 where B (F) is the (commutative associative) truncated polynomial algebra F [Xi,..., Xn]/(Xlp... Xnp) (and Bo(F) = F). The case in which A is a nil simple algebra seems not to have been discussed in the literature. (Here, by saying that A is simple we merely mean that A has no proper ideals and A22 0. We do not exclude consideration of nil simple algebras-flexible power-associative ones may all be anticommutative, and the present result shows that this would be true if there are no commutative nil simple algebras.) 394 Downloaded by guest on October 1, 2021 VOL. 61, 1968 MATHEMATICS: R. E. BLOCK 395 2. The Main Result.-In any algebra A we write (x,y,z) for the associator (xy)z - x(yz) and [x,y] for the commutator xy -yx. Thus the flexible law is equivalent to the property that (x,y,z) + (z,y,x) = 0 for all x,y,z in A. If A is flexible, the mapping d. defined by dxy = [x,y] (y E A) is a derivation of A+, i.e., dx(y * z) = (day) *z + y dz (y,z £ A). In fact, the latter is equivalent to (x,yz) + (z,y,x) + (x,z,y) + (y,z,x) = (y,x,z) + (z,x,y). (This in turn, with x = z, gives flexibility.) Let D = {dIx E A }. If I is a D-ideal of A + (i.e., an ideal invariant under all d_ in D), then I (considered as a subset of A) is an ideal of A, and conversely. Therefore A is simple and not anticommutative if and only if A + is D-simple (A + is called D-simple if it has no proper D-ideals and A+2 # 0; this implies that it is differentiably simple, i.e., that it is a nontrivial algebra with no proper ideals invariant under all deriva- tions). The following is the main result of this paper. THEOREM. Let A be a finite-dimensional algebra over a field F of characteristic $ 2. If A is flexible, simple, and not anticommutative, then either A + is simple or the characteristic is prime and A + -- B.(E) for some n > 0 and some field E over F (in the latter case A has a 1, and if A is regarded as an algebra over its center, then some scalar extension AK is nodal, with AK+- Bm(K) for some m > 0). We now give the proof. The author's main theorem on differentiably simple algebras' states that any finite-dimensional differentiably simple algebra is either simple or has characteristic p and is isomorphic to the tensor product S 0 Bn(F) over F for some simple S and some n > 0. The remarks above show that this applies to A +. Thus we may assume that the characteristic is p and A + = S 0 Bn(F), n > 0, where S = S 0 1 is a (commutative) simple sub- algebra of A+. The centroid of an algebra is by definition the associative algebra of all linear transformations of the algebra which commute with all left and right multiplications by elements of the algebra. For the present we assume that S is central, i.e., that its centroid consists of just the scalar mappings. Then we may identify the centroid C of S 0 Bn(F) with Bn(F) in the following way. For each b in B = Bn(F), define a mapping Tb of S 0 B into itself by rb(5 0 b') = s 0 (b b ). Then T is a one-one homomorphism of B into C. If c E C and {bi I is a basis of B, writing c(s 0 1) = 2 ic(s) 0 b, we see that each ci is in the centroid of S and hence there is a b, in B such that c(s 0 1) = s 0 b, for all s. But then c[(s s') 0 (1 b)] = (s ps') 0 (bi.b) and hence r is onto. We also use the known facts' that for each d in D the mapping d' of C into C defined by d'(c) = [d,c](c C C) is a derivation of C and that C is D'-simple where D' = {d' C D}. We claim that if a E A, then tada is a derivation of A+, where ta denotes multiplication by a in A +: tax = am x. Indeed, the flexible law implies that a(ax) - a(xa) + (ax)a - (xa)a = a(ax) - (xa)a = -(a,a,x) + a2x- xa2 - (x,a,a) = [a2,x], so that tda = (1/2) da2. We write C = Bn(F) = F[xi,. .,x,.] where xiP = 0, i = 1,. .,n. Then the Downloaded by guest on October 1, 2021 396 MATHEMATICS: R. E. BLOCK PRoc. N. A. S. radical R of C is (x,. .. ,x.) and there is some a in A of the form u 0 b (u E 8, b £ B.(F)) such that da'R is not contained in R. Hence there is some c in Csuchthatda'c = 1 +r,r£R. But (tada)'C = tadaC - Ctada = tadaC - tacda = ta(1 + r) E C. Applying this to s 0 1, we get (u * s) X (b + b * r) = s X bi for some bi in B (F) and all s in S. For an si in S such that u5si 5 0 this gives u-s, = as,(aCE F), bi = a (b + b-r), and u-s = as for all s. Hence S has a 1 and u = al. For any s in S either d'8,,b R or d'(,,+ u) ®9 b R is not contained in R. In the first case s C Fl, and in the second s + u E F1, so again s E Fl. Therefore S = F1 and A+ = Bn(F). Now suppose that S is not necessarily central. Regarding A as an algebra over its centroid, take the scalar extension to the algebraic closure K. Then AK is flexible and simple, AK+ _ Bm(K) for some m by the previous case, A + is asso- ciative, and hence S, being simple and commutative, is a field. The unit element 1 of A + is also a unit element of A since 2tDd, = di implies that di = 0. Then since AK is power-associative,7 it is nodal. This completes the proof. COROLLARY. A finite-dimensional simple nil noncommutative Jordan algebra of characteristic $2 is anticommutative. This corollary follows immediately from the theorem above and the theorem of Albert that nil Jordan algebras are nilpotent. PROPOSITION. Let B be a commutative algebra of characteristic $ 2, with products written as x y. Then there is a flexible algebra A with A + = B if and only if there is a linear mapping 5: x - d, of B into the derivation algebra of B such that d,(x) = 0 for all x in B; multiplication in A is given by xy = x.