Let $\mathfrak{g}$ by a Lie algebra, and $I$ an ideal of $\mathfrak{g}$.
Say that $\mathfrak{g} / I$ is semisimple. Is it then true that the radical of $\mathfrak{g}$ is contained in $I$?
This seems to be like it should be true, because otherwise consider the image of $\mathrm{rad}(\mathfrak{g})$ in $\mathfrak{g} / I$. If $\mathrm{rad}(\mathfrak{g}) - I$ is non-trivial, then surely it must be solvable ideal of $\mathfrak{g} / I$, which would lead to a contradiction? Is this logic sound?
It is the good idea, let $p:g\rightarrow g/I$ the quotient map $p(rad(g))$ is a solvable ideal, since $g/I$ is semi-simple, $p(rad(g))=0$ and $rad(g)\subset I$.