Let L be a finite-dim'l Lie algebra over an algebraically closed field of characteristic zero, and I be a solvable ideal of L. Prove that the ideal [L,I] is nilpotent.
My reasoning:
Consider the adjoint representation $ad: I \rightarrow gl(L)$ defined by $ x \mapsto ad_{x}: y \mapsto [x,y]$.
I solvable $ \implies ad(I)$ is also solvable
Lie's Theorem $\implies \exists z \in L $ such that $ad_{x}(z) = [x,z]=\lambda_{x}z$ $\forall x \in L$.
Now, to show that $[L,I]$ is nilpotent: Let $u_1,u_2 \in I$ and $v_1,v_2 \in L$.
$\left[[u_1,v_1],[u_2,v_2] \right]$
This is where I get stuck. I know that I need to show that $[I,L]^{n}={0}$ for some $n$ using the fact that there is a common simultaneous eigenvector but I don't know how.
Let $rad(L)$ denote the sovable radical of $L$, i.e., the maximal solvable ideal of $L$. We have $I\subseteq rad(L)$, since $I$ is solvable. Now we can use the result $$[L,rad(L)]\subseteq nil(L),$$ where $nil(L)$ is the nilradical. The proof given here (which comes even before Lie's theorem) uses that all adjoint operators $ad_{[L,rad(L)]}(x)$ are nilpotent, so that $[L,rad(L)]$ is a nilpotent ideal by Engel's theorem. In particular, $[L,I]\subseteq nil(L)$ is a nilpotent ideal of $L$.