If we take $L$ a finite dimensional Lie algebra on $\mathbb{R}$, $A$ a Lie sub-algebra and $I$ an ideal of $L$ such that $L=A \oplus I$ as vector spaces.
We have that $rad(L)=rad(A) \oplus rad(I)$ as vector spaces. How can we prove this result?
PS: The hypothesis of this statement may be too strong, but I prefer to mention them to be sure.
If you apply the Levi decomposition to $A$ and $I$, you obtain $A=rad(A)\oplus s(A)$ and $I=rad(I)\oplus s(I)$ where $s(A)$ and $s(I)$ are semi-simple.
Since $I$ is an ideal, for every $x\in L$, $[x,rad(I)]$ is solvable ideal of $I$, thus $rad(I)$ is an ideal contained in $rad(L)$, since $rad(A)$ is resoluble, there exists $n$ such that $rad(A)^n=0$ where $rad(A)^2=[rad(A),rad(A)]$ and $rad(A)^{n+1}=[rad(A)^n,rad(A)]$ this implies that $[rad(A)\oplus rad(I),rad(A)\oplus rad(I)]^n\subset rad(I)$, we deduce that $rad(A)\oplus rad(I)\subset rad(L)$, since $L/(rad(A)+rad(I))$ is semi-simple, we deduce that $rad(L)=rad(A)\oplus rad(I)$.