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Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{C}$ and let $\operatorname{Rad}(\mathfrak{g})$ denote the radical of $\mathfrak{g}$; the unique maximal solvable ideal of $\mathfrak{g}$.

If $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, I want to know about the relation between $\operatorname{Rad}(\mathfrak{g/h})$ and $\operatorname{Rad}(\mathfrak{g})/\operatorname{Rad}(\mathfrak{h})$; what are the necessary and sufficient conditions for $\operatorname{Rad}(\mathfrak{g/h})$ to be isomorphic to $\operatorname{Rad}(\mathfrak{g})/\operatorname{Rad}(\mathfrak{h})$? Is $\operatorname{Rad}(\mathfrak{g/h})$ always isomorphic to $\operatorname{Rad}(\mathfrak{g})/\operatorname{Rad}(\mathfrak{h})$?

Juan Lafontaine
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1 Answers1

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To simplify the notation write $R$ for ${\rm Rad}$ and let the Lie algebra be $L$ and the ideal $I$. Put $X/I=R(L/I)$. Then $R(L)\cap I=R(I)$, so we have $(R(L)+I)/I\cong R(L)/R(I)$. Clearly $R(L)+I\subseteq X$, so, if $L=R(L)\dot{+} S$ is the Levi decomposition of $L$, $X=R(L)+ X\cap S$. But $X\cap S$ is a semisimple ideal of $S$, and so $X\cap S= (X\cap S)^{(n)} \subseteq I$ for some $n \in \mathbb{N}$. Hence $X \subseteq R(L)+I$ and $R(L)/R(I)\cong (R(L)+I)/I= X/I=R(L/I)$.

David Towers
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