Codimension-one ideal in a nonzero solvable Lie algebra

The derived algebra $[\mathfrak{g},\mathfrak{g}]$ is proper. Otherwise ${\mathfrak{g}}^{(1)}=\mathfrak{g}$, so every derived term would equal the nonzero algebra $\mathfrak{g}$, contradicting solvability in [L1]. Thus the abelianization $A=\mathfrak{g}\mathrm{/}[\mathfrak{g},\mathfrak{g}]$ in [L2] is nonzero and finite-dimensional.

Choose a hyperplane $H<A$: take one nonzero vector, extend it to a finite basis, and span all basis vectors except that one. The inverse image $\mathfrak{h}$ of $H$ in $\mathfrak{g}$ contains $[\mathfrak{g},\mathfrak{g}]$, so $[\mathfrak{g},\mathfrak{h}]\subseteq [\mathfrak{g},\mathfrak{g}]\subseteq \mathfrak{h}$ and $\mathfrak{h}$ is an ideal. By [L3], its codimension equals that of $H$, namely one. Only a finite basis extension is used, so algebraic closure and characteristic zero are unnecessary and no Choice principle is invoked.