Codimension-one ideals in nilpotent Lie algebras

If $\dim \mathfrak{g}=1$, the only proper subalgebra is $0$, which is itself an ideal of codimension one. The case $\mathfrak{g}=0$ has no proper subalgebra and is vacuous.

Assume $\dim \mathfrak{g}>1$ and the assertion for all nilpotent Lie algebras of smaller dimension.

By [L1], choose $0\mathrm{\ne }z\in Z(\mathfrak{g})$ and put $\mathfrak{a}=kz$. Then $\mathfrak{a}$ is a one-dimensional central ideal.

The quotient $\bar{\mathfrak{g}}=\mathfrak{g}\mathrm{/}\mathfrak{a}$ is defined by [L3], has smaller dimension by [L4], and is nilpotent by [L2].

If $z\in \mathfrak{h}$, then $\bar{\mathfrak{h}}=\mathfrak{h}\mathrm{/}\mathfrak{a}$ is proper in $\bar{\mathfrak{g}}$. By step 1.2 there is a codimension-one ideal $\bar{\mathfrak{m}}$ containing it. Its inverse image $\mathfrak{m}$ under the quotient map is an ideal containing $\mathfrak{h}$, and [L4] gives codimension one.

If $z\mathrm{\notin }\mathfrak{h}$, set $\mathfrak{s}=\mathfrak{h}+\mathfrak{a}$; centrality makes this a subalgebra. If $\mathfrak{s}=\mathfrak{g}$, then $\mathfrak{h}$ has codimension one and $[\mathfrak{g},\mathfrak{h}]=[\mathfrak{h}+\mathfrak{a},\mathfrak{h}]\subseteq \mathfrak{h}$, so it is the required ideal. If $\mathfrak{s}$ is proper, then $\mathfrak{s}\mathrm{/}\mathfrak{a}$ is proper in $\bar{\mathfrak{g}}$; apply step 1.2 there and take the inverse image as in step 3.1.

The alternatives $z\in \mathfrak{h}$ and $z\mathrm{\notin }\mathfrak{h}$, including both subcases of the latter, exhaust all possibilities and each yields a codimension-one ideal containing $\mathfrak{h}$. The only selections were one central witness and finitely many induction witnesses, so no Choice principle is used.