TITLE: Solvable equivalent to nilpotency of first derived Lie algebra?
QUESTION [5 upvotes]: The Wikipedia "Solvable Lie Algebra" page lists the following property as a notion equivalent to solvability: $\mathfrak{g}$ is solvable iff the first derived algebra $[\mathfrak{g},\,\mathfrak{g}]$ is nilpotent.
I can't see this at all and it does sound dodgy: solvability and nilpotency different by "only one rung" in the derived series. Is this correct (I suspect I'm missing something trivial- I can't seem to find anything relevant in Knapp "Lie Groups: Beyond an Introduction" (mine is reprint of the first edition))? If not universally correct, is it correct with further assumptions about the field which $\mathfrak{g}$ is a vector space over? (For example, a field of characteristic nought, algebraically closed, ...). Can someone either give a proof, a counterexample or a reference to one? 
I'd also like to put a reference on the Wikipedia page quotation, or clearly have it stricken from the page if incorrect.

REPLY [3 votes]: There is a counterexample given in Jacobsons book, i.e., an example of a finite-dimensional solvable Lie algebra $L$ in characteristic $p>0$ such that $[L,L]$ is not nilpotent.
More concretely, this Lie algebra $L$ has a basis $(x,y,e_1,\ldots ,e_p)$ with Lie brackets
\begin{align*}
[x,y]   & = x, \\
[x,e_1] & = e_p, \\
[x,e_i] & = e_{i-1}, \; i\ge 2,\\
[y,e_i] & = (i-1)e_i, \; 1\le i\le p.
\end{align*}
Since $V=\operatorname{span}(e_1,\ldots ,e_p)$ is solvable (even abelian) and the quotient $L/V$ is solvable, so is $L$. But $[L,L]=kx\oplus V$ is not nilpotent.
We have $[L,L]^1=[L,L]^2=\cdots =V$.