Let $\mathfrak{g}$ be a Lie algebra over a field $F$, not necessary of characteristic $0$. We say that an ideal $\mathfrak{a} \unlhd \mathfrak{g}$ is characteristic if it is stable under any derivations, i.e. $\delta(\mathfrak{a}) \subseteq \mathfrak{a}$ for any $\delta \in \mathrm{Der}(\mathfrak{g})$. We also denote the radical $\mathfrak{r}(\mathfrak{g})$ to be the (unique) maximal solvable Lie subalgebra of $\mathfrak{g}$.
Is it true that $\mathfrak{r}(\mathfrak{g})$ is a characteristic ideal of $\mathfrak{g}$?
I have a few ideas on tackling this problem but were mostly to no avail. The most convincing approach appears to be showing that $\delta(\mathfrak{r}(\mathfrak{g}))$ is solvable, and hence must lie inside the radical. But I can't seem to proceed much further from here. Even if I were to assume $\mathfrak{r}(\mathfrak{g})$ is abelian, the furthest I can go is that given $x,y \in \mathfrak{r}(\mathfrak{g})$: $$ [\delta(x),\delta(y)] = \delta([\delta(x),y]) - [\delta^2(x),y] $$ which does not seem to help much.
Any help is appreciated.
