Solvable Lie algebra with codimension 1 ideal There is an exercise in Humphreys's An Introduction to Lie Algebras and Representation Theory: "Any nilpotent Lie algebra contains a codimension 1 ideal".
The proof I am thinking of is the following.
Suppose the Lie algebra $L$ satisfies $L\neq[L,L]$. Then $\pi\colon L\to L/[L,L]$ is a projection onto an abelian algebra. Choose a codimension 1 subspace $I\subset L/[L,L]$. Then $I$ is an ideal and its preimage $\pi^{-1}(I)$ is necessarily an ideal in $L$, because any subspace containing the derived algebra is an ideal. Moreover, $\pi^{-1}(I)$ has codimension 1. $\square$
The only condition on $L$ is that $L\neq[L,L]$, which holds for solvable Lie algebras as well. Is this proof correct? How come Humphreys didn't set the exercise with "nilpotent" replaced by "solvable"?

The proof works indeed for all finite-dimensional solvable Lie algebras as above. 
We only need $[L,L]\neq L$. What does not work is, that every solvable Lie algebra $L$ admits a chain of ideals
$$
0=L_0\subseteq L_1\subset \cdots \subset L_n=L,
$$
i.e., is supersolvable. This is only true over algebraically closed fields
of characteristic zero - by Lie's theorem. So it can happen, that a solvable Lie algebra does not have an ideal of every possible codimension - see here for an example of a real solvable Lie algebra of dimension $4$ without an ideal of codimension $3$.
