Vtyjkyuk

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

In his proof, Humphreys (Introduction to Lie Algebras and Representation Theory - pag 83) says that, since $B$ is a solvable subalgebra and $operatornameRadL$ is a solvable ideal of $L$, then $B+operatornameRadL$ 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 $$[B+text Rad L, B+text Rad L] = [B,B]+[B,textRad L]+[textRad L,B]+[textRad L,textRadL].$$

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).

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 $textRad L$ is an ideal, we have $$[B+textRad L,B+textRad L]subset [B,B]+textRad L.hspace1cm(*)$$ If we use the notation $L^(0)=L$, $L^(n)=[L^(n-1),L^(n-1)]$, then $(*)$ implies that for each $n$, $$[B^(n-1)+textRad L,B^(n-1)+textRad L]subset B^(n)+textRad L.$$ So we can prove by induction that $$beginaligned(B+textRad L)^(n)&subset B^(n)+textRad L.endaligned$$ Now, since $B$ is solvable, $B^(n)=0$ for some $n$, so $(B+textRad L)^(n)subset textRad L$. Since $textRad L$ is solvable, we conclude that $(B+textRad L)^(n)$ is solvable. Thus $B+textRad L$ is solvable.