Reductive Lie algebra

Remarks

The ideals of $\mathfrak g$ are exactly the subspaces that are invariant under the adjoint representation. Equivalent conditions, with $\mathfrak z$ the center of $\mathfrak g$:

1. $\mathfrak g = \mathfrak z \oplus [\mathfrak g, \mathfrak g]$ and $[\mathfrak g, \mathfrak g]$ is semisimple;
2. $\mathfrak g$ is the direct sum of an abelian ideal and a semisimple ideal;
3. the radical of $\mathfrak g$ equals $\mathfrak z$.

Example: the general linear Lie algebra $\mathfrak{gl}_n(k)$. It is the direct sum of the ideal $kI$ of scalar matrices, which is its center, and the semisimple ideal $\mathfrak{sl}_n(k)$, so condition 2 holds.

Non-example: the two-dimensional Lie algebra with basis $x, y$ and $[x, y] = y$. The line $ky$ is an ideal. A complementary line is spanned by some $x + cy$, and $[y, x + cy] = -y$ does not lie in it, so it is not an ideal. Hence the adjoint representation is not completely reducible.