Suppose $\mathfrak{g}$ is a Lie algebra with radical $\text{Rad }\mathfrak{g}$ and let $\mathfrak{a}\subseteq\mathfrak{g}$ be a semisimple subalgebra. Is it necssarily the case that $\text{Rad }\mathfrak{g}\cap\mathfrak{a}=\{0\}$?
I am trying to prove that maximal semisimple subalgebras are Levi subalgebras, but I can't get past this step. If I know this to be true however, I can finish off the rest of the proof.
I would say yes. Let's take $x \in \text{Rad }\mathfrak{g}\cap\mathfrak{a}$ and then consider the ideal $\mathfrak{I}$ in $\mathfrak{a}$ generated by this element. Then the $\text{Rad }\mathfrak{a}$ contains $\mathfrak{I}$ and so is not void contraddicting the semi-semplicity of the subalgebra $\mathfrak{a}$
Yes. $\text{Rad }\mathfrak{g}\cap\mathfrak{a}$ is an ideal of $\mathfrak{a}$, because $\mathfrak{a}$ is a subalgebra and $\text{Rad }\mathfrak{g}$ is an ideal of $\mathfrak{g}$. Also, $\text{Rad }\mathfrak{g}\cap\mathfrak{a}$ is solvable, because it is a subalgebra of the solvable $\text{Rad }\mathfrak{g}$. But since $\mathfrak{a}$ is semisimple, its only solvable ideal is $\lbrace 0 \rbrace$.
