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We say $\frak{h}$ is a characteristic ideal of a lie algebra $\frak{g}$ if $[\frak{h},\frak{g}]\subset\frak{h}$, and $D(\frak{h})\subset\frak{h}$ for every derivation $D\in Der(\frak{g})$.

The theorem states that if $\frak{h}$ is a characteristic ideal of a lie algebra $\frak{g}$, then every ideal of $\frak{h}$ is also an ideal in $\frak{g}$. Essentially, if we let $\frak{a}$ be an ideal in $\frak{h}$, we want to show that $[\frak{a},\frak{g}]\subset \frak{a}$, but I don't know what derivation $D\in Der(\frak{g})$ should be used?

Sid Caroline
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If I am right, then let $x \in \mathfrak{g}$, $a\in\mathfrak{a}$, $\operatorname{ad}_x \colon \mathfrak{h}\rightarrow\mathfrak{h}$ defined by $\operatorname{ad}_x(h)=[x,h]$ is a derivation (Jacobi). This implies that $\operatorname{ad}_x(a)=[x,a]\in \mathfrak{a}$ since $\mathfrak{a}$ is a characteristic ideal of $\mathfrak{h}$. This implies that $\mathfrak{a}$ is an ideal of $\mathfrak{g}$.

Jendrik Stelzner
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Tsemo Aristide
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    I wonder if there exists a counter example to my original statement. – Sid Caroline Nov 18 '18 at 08:03
  • @SidCaroline yes you have a counterexample of the form $K^2\rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2\times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra. – YCor Nov 18 '18 at 11:24