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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?

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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.

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    $\begingroup$ You also want to say that $\mathfrak{g}/[\mathfrak{g},\mathfrak{n}]$ is reductive (which probably requires characteristic zero). $\endgroup$ – YCor Dec 17 '15 at 23:51

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