Notes on Lie Ideals of Simple Artinian Rings

Abstract

Let R be a ring. If we replace the original associative product of R with their canonic Lie product, or [a, b] = ab − ba for every a, b in R, then R would be a Lie ring. With this new product the additive commutator subgroup of R or [R, R] is a Lie subring of R. Herstein has shown that in a simple ring R with characteristic unequal to 2, any Lie ideal of R either is contained in Z(R), the center of R or contains [R, R]. He also showed that in this situation the Lie ring [R, R]/Z[R, R] is simple. We give an alternative matrix proof of these results for the special case of simple artinian rings and show that in this case the characteristic condition can be more restricted.Proof. If L ⊆ T i for some index i, then by the preceding lemma Lj)th position of this latter matrix, hence u i = u j , for all i = j. In other words, L ⊂ {uI ; u ∈ [D, D]}. Note also that for all d ∈ D, the matrices dE ij ∈ E(D) ⊂ [R, R], where i = j, so for all uI ∈ L the matrices [uI , dE ij ] are in L. This also implies that ud − du = 0 for all d ∈ D, and so u ∈ F . In other words, L ⊆ FI or L ⊆ Z[R, R]. This contradiction completes the proof. Lemma 6. Let R = M n (D), where either n = 2 and char D = 2, or n ≥ 3 and let T i , for a fixed i (1 ≤ i ≤ n), be as defined in Lemma 4, then S i , the Lie ring generated by the set {dE ri ; d ∈ D, r = i}, is a Lie ideal of T i , such that [R, R] = {[S i , A]; A ∈ [R, R]} . In other words, [R, R] is the Lie ideal closure of S i (the least Lie ideal containing S i in [R, R]).Proof. The fact that S i is a Lie ideal of T i can be checked easily. For the last claim, by Lemma 1, it suffices to check that the Lie ideal closure of S i produces E(D). In case n ≥ 3 note that [E ri , E ij ] = E rj , for all r = i, j = i and r = j. Also 1350038-4 J. Algebra Appl. 2013.12. Downloaded from www.worldscientific.com by MONASH UNIVERSITY on 04/11/15. For personal use only.