The adjoint representation splits into simple ideals

A subspace $U\subseteq \mathfrak{g}$ is stable under the adjoint action exactly when $[\mathfrak{g},U]\subseteq U$, which is exactly the ideal condition. Therefore [L1] says every ideal has an ideal complement.

An irreducible adjoint summand is a nonzero ideal with no nonzero proper ideal of $\mathfrak{g}$. It cannot be abelian, since that would be a solvable ideal of a semisimple algebra. Its ideal complement commutes with it, so an ideal inside the summand is also an ideal of $\mathfrak{g}$; hence the summand is simple. Conversely, each simple factor from [L2] has no proper adjoint submodule and is irreducible. The zero algebra has the empty decomposition.