Suppose you want to construct a perfect Lie algebra with a nonabelian solvable radical $\mathfrak{r}$, say with a commutator series of length 2. What are the conditions that guarantee the Lie algebra will be perfect? Is it sufficient to have an irreducible representation of a semisimple Lie algebra $\mathfrak{s}$ on $\mathfrak{r}/[\mathfrak{r},\mathfrak{r}]$ and $[\mathfrak{r},\mathfrak{r}]$?



perfect Lie algebra with a nonabelian solvable radical
Let $K$ be a field of char. zero, $\mathfrak{g}$ a f.dim. Lie algebra over $K$. Let $\mathfrak{r}$ be its solvable radical, $\mathfrak{s}=\mathfrak{g}/\mathfrak{r}$ (which is semisimple) and $V=\mathfrak{r}/[\mathfrak{r},\mathfrak{r}]$ (which is naturally an $\mathfrak{s}$-module). Then $\mathfrak{g}$ is perfect iff $V^\mathfrak{s}=\{0\}$, by a simple verification using the existence of a Levi factor. (In particular, no assumption is needed on the action of $\mathfrak{s}$ on $[\mathfrak{r},\mathfrak{r}]$, and irreducibility is not the relevant condition.) 


