Radical of a Lie algebra bracket itself

Question

Let $\mathfrak{g}$ be a Lie Algebra over $k$, $\mathfrak{n}$ its radical. Why is $[\mathfrak{n},\mathfrak{g}]$ the smallest of its ideals $\mathfrak{a}$ such that $\mathfrak{g}/\mathfrak{a}$ is reductive?

Answer 1

Let $\mathfrak{a}$ be an ideal such that $\mathfrak{g}/\mathfrak{a}$ is reductive. Then $\mathfrak{n}/\mathfrak{a} \subset \mathfrak{g}/\mathfrak{a}$ is a solvable ideal, and hence must be contained in the center of $\mathfrak{g}/\mathfrak{a}$ since it is reductive. So $[\mathfrak{n},\mathfrak{g}]/\mathfrak{a}=[\mathfrak{n}/\mathfrak{a},\mathfrak{g}/\mathfrak{a}]=0$ and we see that $[\mathfrak{n},\mathfrak{g}]\subset\mathfrak{a}$ as desired.