Using Engel's Theorem and Lie's Theorem, one can easily establish the following result:


Let $ \frak{g} $ be a finite-dimensional Lie algebra over an algebraically closed field $ \mathbb{F} $ of characteristic $ 0 $. If $ \frak{g} $ is solvable, then $ [{\frak{g}},{\frak{g}}] $ is nilpotent.


In order to apply the two theorems stated at the beginning, one must assume that (i) $ \mathbb{F} $ is algebraically closed, (ii) $ \mathbb{F} $ has characteristic $ 0 $, and (iii) $ \frak{g} $ is finite-dimensional. If we relax each of these three conditions in turn, are there certain well-known counterexamples?



On nilpotency of the derived subalgebra of a solvable Lie algebra
* Condition (i) can be removed, as already observed by Daniel.
* In positive characteristic you can find a counterexample in the book "J. Humphreys: Introduction to Lie algebras and representation theory" (Chapter 2, Section 4, page 20, Exercise 4), so condition (ii) cannot be relaxed.
* Finally, let $H={\mathbb F}x+{\mathbb F}y+{\mathbb F}z$ be the Heisenberg algebra with basis $x,y,z$, where $z$ is central in $H$ and $[x,y]=z$. Consider the ring of polynomials ${\mathbb F}[t]$ as a left $H$-module with $x$ acting as $d/dt$, $y$ acting by multiplication by $t$, and $z$ acting as the identity. Now consider the split extension $L=H\ltimes {\mathbb F}[t]$. Then $L$ is solvable of derived length 3, but the derived subalgebra $[L,L]={\mathbb F}z+{\mathbb F}[t]$ is not nilpotent. Thus condition (iii) cannot be removed.


