I am trying to prove the following.
Proposition Let $g$ be a Lie algebra such that the first derived algebra $Dg$ is a proper ideal of $g$. Consider the quotient $g/D^1 g$, and show that the preimage of any codimension 1 subspace is
- an ideal, and
- has codimension 1.
I could show that the preimage is indeed an ideal, but I need some help with the part about the dimensions:
So suppose that $\phi: x \mapsto [x]$ is the quotient projection. And write $[h] + Dg$ for a codimension 1 subspace of $g$ (here $h\subset g$). Then the preimage of this subspace is just the linear span $h+Dg\leq g$. Let $y+z \in h+Dg$ with $y\in h$ and $z\in Dg$. Let $x\in g$. We want to show that $[x, y+z]\in h$, ie that $h + Dg \lhd g$. We can write $x$ as $x_0+x_1$ for $x_0 \in Dg$, $x_1\not\in Dg$. Then $$[x,y+z] = [x_0, y+z] + [x_1, y] + [x_1,z] \in h + Dg$$ The inclusion follows since $Dg$ is an ideal so the first and third terms belong $Dg$; and $[x_1, y] \in D_g$, as $g/Dg$ is abelian.
The dimension of the preimage of a codimension $1$ vector subspace by a surjective linear map $f:V\rightarrow W$ has codimension $1$. Let $L$ be a subspace of $W$ of codimension $1$, there exists $y$ such that $y+L=W$, write $f(x)=y$. For every $z\in V$, $f(z)=u+ay, u\in L$ write $u=f(u')$, this implies that $f(z-u'-ax)=0$ and $z-ax-u'\subset Ker f\subset f^{-1}(L)$. We deduce that $V=f^{-1}(L)+x$.