 Solution

The derived series of a Lie algebra isThe algebra is a solvable Lie algebra if ${\mathfrak{b}}^{(m)}=0$ for some $m$. Here a bracket of subspaces means the linear span of their Lie brackets.

The Lie theorem states that a nonzero finite-dimensional representation of a complex solvable Lie algebra has a common eigenvector; equivalently, its operators can be simultaneously upper triangularized. Applied to an Irreducible Lie algebra representation, the common eigenvector gives a nonzero invariant line, which must be the entire module. A scalar action $\rho (x)=\lambda (x)$ respects the Lie bracket exactly when $\lambda ([x,y])=0$. Thus, in the finite-dimensional setting,Every functional displayed gives a one-dimensional irreducible, and two such modules are isomorphic exactly when their functionals agree.

The finite-dimensional qualification is essential here, although the printed wording does not repeat it. For example, let $[x,y]=y$ and act on $\mathbb{C}[t]$ by $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. These operators satisfy $[x,y]f=yf$. A nonzero submodule stable under $x$ is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ forces $p(t)\mid p(t-1)$, so the two polynomials, having the same degree and leading coefficient, are equal. In characteristic zero a polynomial invariant under a nonzero shift is constant. The module is therefore an infinite-dimensional simple module for the two-dimensional affine Lie algebra. It rules out extending the one-dimensional classification to unrestricted representations.