I found the following statement in Humphreys' book, 'Introduction to Lie Algebras and Representation Theory', in the first chapter, section 3.1:

"Notice that for arbitrary (Lie algebra) L, $L$ / Rad$L$ is semisimple."

And as substantiation of the statement he cites the proposition "If $I$ is a solvable ideal of a Lie algebra $L$ such that $L/I$ is solvable, then $L$ itself is solvable" which was proven earlier. I could prove the proposition that he's referring to but I could not, after hours of trying, see how the statement follows from the proposition. I'd much appreciate some help on this.

  • $\begingroup$ It is certainly implicit that the Lie algebra is finite-dimensional. However it seems to work over any field. $\endgroup$ – YCor Feb 1 at 22:34

Let $J$ be a solvable ideal of $L/rad(L)$ and $p:L\rightarrow L/rad(L)$ the quotient map. Denote $U=p^{-1}(J)$, it is an ideal (sub-Lie algebra) of $L$ which contains $rad(L)$, you have $U/rad(L)\simeq J$ and $rad(L)$ is solvable. The proposition that you quote implies that $U$ is solvable, since $rad(L)$ is the maximal ideal of $L$, $U=rad(L)$ and $J=0$. We deduce that $L/rad(L)$ is semi-simple.

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  • $\begingroup$ Did you mean "U/rad(L)≃J"? $\endgroup$ – Tuneer Chakraborty Jul 10 '17 at 19:21
  • $\begingroup$ yes, I means that $\endgroup$ – Tsemo Aristide Jul 10 '17 at 19:22

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