Doubt in the proof of Ado's theorem

Question

I am currently going through a proof of Ado's theorem. I am stuck in one step. Suppose $\mathfrak{g}$ is a solvable Lie algebra which is not nilpotent. Then one can show that there is an ideal $\mathfrak{a}$ of codimension 1 which contain $nil\mathfrak{g}$. Then we have a complementary vector subspace $\mathfrak{h}$ so that $\mathfrak{g}=\mathfrak{a} \oplus \mathfrak{h}$ as vector space. Then the nilradical $nil\mathfrak{g} = nil\mathfrak{a}$. Why is this true? Clearly, since $nil\mathfrak{g} \subset \mathfrak{a}$ we have $nil\mathfrak{g} \subset nil\mathfrak{a}$. How can I get the other inclusion?

Answer 1

The quotient ${\cal g}/nil(g)$ is a reductive algebra. Since ${\cal g}$ is solvable, this reductive algebra is commutative. This implies that every vector space which contains $nil({\cal g})$ is an ideal of ${\cal g}$. We deduce that $nil({\cal a})$ is a nilpotent ideal of ${\cal g}$ since $nil({\cal g})$ is the maximal nilpotent ideal of ${\cal g}$, we deduce that $nil({\cal a})\subset nil({\cal g})$.