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• W2024600114 abstract "Let A be a Malcev algebra, B be an ideal of A and J2(B) = J(B, A, A) where J(B, A, A) is the linear subspace of A spanned by all elements of the form J(x, y, z) =(xy)z +(yz)x + (zxry, x € B,y, z e A. For k > 1, define J^*l(B) = /(/*(B), A, A). Then B is called /~-potent if there exists an integer N > 1 such that J~(B) = 0. Now let A be a Malcev algebra over a field of characteristic 0 such that the radical R of A is /,-potent. Then R is complemented by a semisimple subalgebra and all such complements are strictly conjugate in A. The proofs follow those in the Lie algebra case. In recent years the theory of Malcev algebras has greatly advanced. However, the status of the Wedderburn principal theorem (Levi theorem) and accompanying Malcev-Harish-Chandra theorem does not appear to have been settled. The following special case, when the radical is /2-potent, would seem to be of interest. In this situation the treatment is much like the Lie algebra case. All Malcev algebras and all modules are assumed finite dimensional over a field of characteristic 0. We recall the following terminology. Let A be a Malcev algebra and define R to be right multiplication by x. For x, y, z £ A, let ]ix, y, z) = ixy)z + iyz)x + izx)y = zi~ R + [R , R ]). ' J ' ' ' xy x y For x, y £ A, let A(x, y) = [Rx, R 1 Rx and NIA) = {z £ A; zMx, y) = 0 V x, y £ A i. NIA) is called the /-nucleus of A and is an ideal of A. Also let Dix, y) = [R , R 1 + R A x y xy and Dix, y) is a derivation of A. If B l. Then B is called /.,-potent if there exists an integer N > I such that Received by the editors March 12, 1974 and, in revised form, April 29, 1974. AMS (MOS) subject classifications (1970). Primary 17A30, 17B05. Copyright © 1975, American Mathematical Society 1 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 2 E. L. STITZINGER }2IB) = 0. Note that we have slightly altered the definition of /2-potent (see [6, p. 444]). We prove the following. Theorem 1. Let A be a Malcev algebra over a field of characteristic 0. Suppose that the radical R of A is J2-potent. Then there exists a semisimple subalgebra S of A such that A = RG)S. The usual way of showing the Wedderburn principal theorem is to prove the case when the radical R is a minimal ideal of A such that R =0, and then the general case follows by a standard argument [l, p. 87] provided that one can obtain in the general case an ideal B of A such that B is properly contained in R and IR/B) = 0. For Malcev algebras this approach is possible since R + JIR, R, A) is an ideal of A properly contained in R (provided R 4 0) by [5, Theorem 1, p. 228]. Hence it suffices to prove our result holds in the case that R is a minimal ideal of A. Furthermore, if R is /,potent in A and is a minimal ideal in A, then JIR, A, A) is an ideal of A [6, Theorem 3.5I and is properly contained in R, hence JIR, A, A) = 0 and R C NIA). Then the natural representation of A on R is a homomorphism, i.e., ib) [Rx, R 1 = ib)Rx for all b e R, x, y £ A. Since R2 = 0, we consider R as an A = A/R-module and the associated representation is still a homomorphism. Since (A) = A, A is a supplement of R in A and if A is properly contained in A, then A is a complementary subalgebra of R and the result holds in this case. Hence we may assume that A — A. Summarizing this paragraph we may consider the case when (1) R is a minimal ideal of A, (2) ib)[R , R 1 = bR fot all b £ R, x, y e A, x' y xy J (3) A2=A. We first consider the critical case when A is Lie. Then AA(x, y) C R and RAix, y) = 0. Consequently (4) A(x, y)Mu, v) = 0= MxMu, v), y) holds in A. Following the ideas of [6], one obtains the following identity for any Malcev algebra [A(x, y), Mu, v)1 = MxMu, v), y) + A(x, yMu, 12)) (5) + 6RKx,y,uv)-6A{uv*y) as follows : License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use MALCEV ALGEBRAS WITH /2-POTENT RADICAL 3 [A(x, y), Mu, v)1 = [A(x, y), Diu, v)1 2[A(x, y), R 1 y J uv = MxDIu, v), y) + A(x, yO(22, v)) 3Muv, xy) + A(x, yiuv)) + My, iuv)x) by the proof of [6, Proposition 8.14, p. 454] and [6, 2.35, p. 432]. Then A(xD(z2, v), y) + A(x, yDiu, v)) = MxMu, v), y) + 2Mxiuv), y) + Mx, yMu, v)) + 2A(x, yiuv)). Substitution then gives [A(x, y), Mu, v)1 = MxMu, v), y) + Mx, yMu, v)) lAiuv, xy) + 3A(x, y(z2i;)) + 3A(y, iuv)x). Using [6, 2.32, p. 432] on the last two terms gives (5). Now for the algebra under consideration, from (4) and (5) we obtain the identity (6) Rrl . = Muv, xy). ix,y,uv) ' Since A = A, (6) yields the identity (7) tjix, y, z) = tMz, xy) = /(/, z, xy) which holds in A. We have the usual criterion for R to have a complementary subalgebra in A [l, pp. 86—89]. That is, let o be a linear map from A into A suchthat aa = a fot all a € A and define (8) gib, c) = TAA -ibcA £ R fot all b, c e A~. Since R = 0, R is an A-module under the product ra = ra and because of (2), the associated representation is a homomorphism. Then R has a complementary subalgebra if and only if there exists a linear mapping p of A into R such that (9) gib,c) = bpc-Ab-ibAp. We collect some properties of g. Since A is antisymmetric, (10) gib,b) = 0 which yields gib, c~) = —g(c, b ). Next write ~brb~2 = ib^b2)a + gdj, bA and compute iblb^)b'i = ib~b2)alci + gibvb2)b^ = ((e1I2)fc3)<T + gib~b2, ?3) + gibv b2)bZ. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use" @default.
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• W2024600114 title "Malcev algebras with $Jsb{2}$-potent radical" @default.
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