= Solution
The <derived series of a Lie algebra> is
$$
\mathfrak b^{(0)}=\mathfrak b,\qquad \mathfrak b^{(j+1)}=[\mathfrak b^{(j)},\mathfrak b^{(j)}].
$$
The 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,
$$
\boxed{\text{irreducibles are }\mathbb C_\lambda,\qquad \lambda\in(\mathfrak b/[\mathfrak b,\mathfrak b])^*.}
$$
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.
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