Radical of a Lie algebra

In the mathematical field of Lie theory, the radical of a Lie algebra $$\mathfrak{g}$$ is the largest solvable ideal of $$\mathfrak{g}.$$

The radical, denoted by $${\rm rad}(\mathfrak{g})$$, fits into the exact sequence
 * $$0 \to {\rm rad}(\mathfrak{g}) \to \mathfrak g \to \mathfrak{g}/{\rm rad}(\mathfrak{g}) \to 0$$.

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

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

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

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

Related concepts

 * A Lie algebra is semisimple if and only if its radical is $$0$$.
 * A Lie algebra is reductive if and only if its radical equals its center.