A note on a theorem of Jacobson

Abstract

PROOF. In the first instance, let A have an identity or not. Let R. (L.) denote the right (left) multiplication in A, and D be a derivation of A. Then A has neither proper ideals nor proper D-ideals. In other words, the Lie algebra L (L') generated by Rz, L. (R., L. and the derivation D) is irreducible. Moreover, L'=L+ {aD } F (a vector space sum and not necessarily a direct sum), and L is an ideal of L'. Further, by a theorem of Jacobson [1, p. 47], we have L = C? [LL]; L' = C' E [L'L'] for centers C, C' of L, L'; [LL], [L'L'] are semisimple; [LL] is an ideal of [L'L']. Any transformation T in C commutes with the irreducible associative algebra generated by RX, L. and hence should be a multiple of the identity transformation I, by Schur's lemma. Now, if A contains an identity, C= FI, C'= FI; since the dimension of L' is at the most greater by unity than that of L, and since [L'L'] cannot have a one dimensional (abelian) ideal complementary to [LL], [LL] = [L'L']; i.e., L=L', or, DEL. Thus, every derivation of A is inner. REMARK 1. In case A is any simple nonassociative algebra, then CCC' and therefore [LL] = [L'L'] in this case also. If, in addition C = C' for every derivation D of A, then every derivation of A will be inner. Because F is algebraically closed, C=O or C=FI, C'=O or C'= FI. Since C = FI implies C'= FI = C and since every derivation is inner in this case as well as when C' = 0, the question raised at the outset boils down to the consideration of the only case C= 0, C' = FI. The plausibility of this case remains to be seen. Now, in the case of simple Lie algebra A over a field F of characteristic zero, L= {ad x}eA; L'=L+ { aD}JaEF. Since A is simple, the center of A = { x EA I ad x = 0 } = {0}. If ad y Ecenter C of L, then