I am recently learning Lie Algebras. I was just curious if there is any relation between the solvable radical of any Lie algebra L with that of Der(L) which has also Lie algebra structure. The main reason of my curiosity comes from the question that is there any relation between semisimplicity of L and Der(L)? I have seen the result that given L is semisimple, we must have Der(L) is isomorphic to ad(L) and L and so this implies that rad(Der(L))= rad(L)=0. So in general, how rad(Der(L)) and rad(L) are related? or any chance that semisimplicity of Der(L) implies that of L?
Thanks!
If $L$ is semisimple over a field of characteristic zero, then ${\rm Der}(L)={\rm ad}(L)$, and $\ker(ad)$ is trivial. Hence ${\rm Der}(L)$ and $L$ are isomorphic, so that their solvable radicals are isomorphic, too.
In general, the relation is not so clear, because we speak of different Lie algebras. For example if $L$ is abelian, then ${\rm rad}(L)=L$, but ${\rm Der}(L)={\rm End(V)}$ is reductive with solvable radical being the center of $\mathfrak{gl}_n(K)$, i.e., the scalar multiples of the identity.
If ${\rm Der}(L)$ is semisimple, then it has no nonzero solvable ideals. Hence the ideal ${\rm ad}(L)$ is not solvable, because for ${\rm ad}(L)=0$ we know already that $L$ is abelian, so that ${\rm Der}(L)$ is not semisimple. Hence also $L$ is not solvable, otherwise its homomorphic image would be solvable too. Furthermore we have
Proposition: Let $L$ be a finite-dimensional Lie algebra of characteristic zero such that ${\rm Der}(L)$ is simple. Then $L$ is simple.
Corollary: If ${\rm Der}(L)$ is semisimple, so is $L$.
Proof: See the article Derivations of Lie Algebras by J. de Ruiter, page $303$. The proof uses Theorem $3$ and the Levi decomposition.