Radical and nilradical of the affine Lie algebra

By [L3], $\mathfrak{a}$ is solvable. It is an ideal of itself, so the universal property [L1] gives $\mathrm{rad}(\mathfrak{a})=\mathfrak{a}$.

The line $ky$ is an ideal because $[x,y]=y$ and $[y,y]=0$. It is abelian and therefore nilpotent, so [L2] gives $ky\subseteq \mathrm{nilrad}(\mathfrak{a})$.

Let $I$ be an ideal not contained in $ky$. It contains a vector $v=ax+by$ with $a\mathrm{\ne }0$. Since $I$ is an ideal, $[v,y]=ay$ lies in $I$, whence $y\in I$; then $x=a^{-1}(v-by)\in I$. Thus $I=\mathfrak{a}$, which is not nilpotent by [L3]. Consequently every nilpotent ideal is contained in $ky$.

Step 1.2 shows that $ky$ is a nilpotent ideal, and step 1.3 shows that it contains every nilpotent ideal. The universal property [L2] therefore gives $\mathrm{nilrad}(\mathfrak{a})=ky$. Together with step 1.1 this proves both displayed identities. The only division is by the explicitly nonzero scalar $a$; no choice principle is used.