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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"?

Earthliŋ
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  • Your proof is correct and works for solvable Lie algebras. I don't know why Humphreys only set the exercise for nilpotent Lie algebras. – Tobias Kildetoft Nov 15 '13 at 09:18

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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 $3$.

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Dietrich Burde
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