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.
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).