Corollary

Let g be a Lie algebra and a an ideal of g. For the Lie algebra $g/a$ to be commutative, it is necessary and sufficient that $a \supset \mathscr{D}g$.

To say that $g/a$ is commutative amounts to saying that $\mathscr{D}(g/a) = \{0\}$. But $\mathscr{D}(g/a)$ is, by Proposition 4, the canonical image of $\mathscr{D}g$ in $g/a$.