Basically title. Is there proof for the following statement, which doesn't rely on Lie's-Theorem?
For any solvable Lie-Algebra L, there exists a Flag of Ideals $0=I_0\subset \dots\subset I_n=L$ with $\dim I_i=i$.
Most textbooks I found always rely on Lie's-Theorem for it but I was wondering if there exists an independet proof for this fact.
By definition of solvable, $[L, L]$ is a proper ideal in $L$ (of dimension $n-k$, say) that is itself solvable. So if we can show that there exist $L = I_n \supset I_{n-1} \supset \ldots \supset I_{n-k} = [L, L]$ with $\dim I_i = i$ for $i = n, n-1, \ldots, n-k$, then the full result follows by induction.
The nice thing is: this is easy. Just take a flag of vector spaces $L = I_n \supset I_{n-1} \supset \ldots \supset I_{n-k} = [L, L]$ such that $\dim I_i = i$.
If we want to show that $I_i$ is an ideal we imagine that $X \in L$ and $Y \in I_i$. We want to show that $[X, Y] \in I_i$. Since also $Y \in L$ we have that $[X, Y] \in [L, L]$ and since $I_i \supset [L, L]$ we have that $[X, Y] \in I_i$.