# Doubt in the proof of Ado's theorem

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?

## 1 Answer

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 $$nil({\cal a})\subset nil({\cal g})$$.