$L / \operatorname{Rad} L$ is semisimple for an arbitrary Lie algebra $L$

Question

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) $L$, $L / \operatorname{Rad} L$ 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.

Answer 1

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