Borel subalgebras contain solvable radical

Question

Let $L$ be a Lie algebra and let $B$ be a Borel subalgebra (a maximal solvable subalgebra) of $L$. I want to understand why $\operatorname{Rad} L \subseteq B$.

In his proof, Humphreys (Introduction to Lie Algebras and Representation Theory - pag 83) says that, since $B$ is a solvable subalgebra and $\operatorname{Rad}L$ is a solvable ideal of $L$, then $B+\operatorname{Rad}L$ is a solvable subalgebra.

Why should it be true? I know in general that the sum of two solvable ideal is a solvable ideal, but I don't know how to apply this in my situation. Thanks.

Answer 1

Take the commutator of this stuff with itself again and its an induction argument on the length of the derived series (the longer one of the two).

I just want to make this step in Ravi's answer clearer, since I don't think it is obvious (well, at least to me...).

Using the fact that $\text{Rad }L$ is an ideal, we have $$[B+\text{Rad }L,B+\text{Rad }L]\subset [B,B]+\text{Rad }L.\hspace{1cm}(*)$$ If we use the notation $L^{(0)}=L$, $L^{(n)}=[L^{(n-1)},L^{(n-1)}]$, then $(*)$ implies that for each $n$, $$[B^{(n-1)}+\text{Rad }L,B^{(n-1)}+\text{Rad }L]\subset B^{(n)}+\text{Rad }L.$$ So we can prove by induction that $$\begin{aligned}(B+\text{Rad }L)^{(n)}&\subset B^{(n)}+\text{Rad }L.\end{aligned}$$ Now, since $B$ is solvable, $B^{(n)}=0$ for some $n$, so $(B+\text{Rad }L)^{(n)}\subset \text{Rad }L$. Since $\text{Rad }L$ is solvable, we conclude that $(B+\text{Rad }L)^{(n)}$ is solvable. Thus $B+\text{Rad }L$ is solvable.

Answer 2

We have $$[B+\text {Rad} L, B+\text {Rad} L] = [B,B]+[B,\text{Rad} L]+[\text{Rad} L,B]+[\text{Rad} L,\text{Rad}L].$$

The first term in in the commutator of B, the second two terms are in Rad L and the last one is in the commutator of Rad $L$. Take the commutator of this stuff with itself again and its an induction argument on the length of the derived series (the longer one of the two).