Corollary 1

Let g be a solvable Lie algebra. The nilpotent radical of g is $\mathscr{D}g$. If $\rho$ is a finite-dimensional simple representation of g, $\rho(g)$ is commutative and the associative algebra L generated by 1 and $\rho(g)$ is a field of finite degree over K.

Here $r = g$, whence $s = \mathscr{D}g$. Hence $\varphi(\mathscr{D}g) = \{0\}$, which shows that $g' = \rho(g)$ is commutative. Every element $\neq 0$ of $L$ is invertible by Schur’s Lemma; hence $L$ is a field.