Radical of a Lie algebra

The radical, denoted by ${\displaystyle \mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{g})}$, fits into the exact sequence

${\displaystyle 0\to \mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{g})\to \mathfrak{g}\to \mathfrak{g}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{g})\to 0}$.

where ${\displaystyle \mathfrak{g}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{g})}$ is semisimple. When the ground field has characteristic zero and ${\displaystyle \mathfrak{g}}$ has finite dimension, Levi's theorem states that this exact sequence splits; i.e., there exists a (necessarily semisimple) subalgebra of ${\displaystyle \mathfrak{g}}$ that is isomorphic to the semisimple quotient ${\displaystyle \mathfrak{g}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{g})}$ via the restriction of the quotient map ${\displaystyle \mathfrak{g}\to \mathfrak{g}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{g}).}$

A similar notion is a Borel subalgebra, which is a (not necessarily unique) maximal solvable subalgebra.

Definition

Let ${\displaystyle k}$ be a field and let ${\displaystyle \mathfrak{g}}$ be a finite-dimensional Lie algebra over ${\displaystyle k}$. There exists a unique maximal solvable ideal, called the radical, for the following reason.

Firstly let ${\displaystyle \mathfrak{a}}$ and ${\displaystyle \mathfrak{b}}$ be two solvable ideals of ${\displaystyle \mathfrak{g}}$. Then ${\displaystyle \mathfrak{a}+\mathfrak{b}}$ is again an ideal of ${\displaystyle \mathfrak{g}}$, and it is solvable because it is an extension of ${\displaystyle (\mathfrak{a}+\mathfrak{b})/\mathfrak{a}\simeq \mathfrak{b}/(\mathfrak{a}\cap \mathfrak{b})}$ by ${\displaystyle \mathfrak{a}}$. Now consider the sum of all the solvable ideals of ${\displaystyle \mathfrak{g}}$. It is nonempty since ${\displaystyle \{0\}}$ is a solvable ideal, and it is a solvable ideal by the sum property just derived. Clearly it is the unique maximal solvable ideal.