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Let $\mathfrak{g}$ be a reductive Lie algebra, i.e. every solvable ideal $\mathfrak{a}$ is contained in the center $\mathfrak{z}(\mathfrak{g})$. Then $[\mathfrak{g},\mathfrak{g}]$ is semisimple, i.e. there is no nonzero solvable ideal.

My proof would go as follows:

Let $\mathfrak{a}$ be a solvable ideal in $\mathfrak{g}$. The kernel of $ad(\mathfrak{g})$ is exactly $\mathfrak{z}(\mathfrak{g})$ which induces an isomorphism $ad(\mathfrak{g}) \cong \mathfrak{g}/\mathfrak{z}(\mathfrak{g}) $. Since $\mathfrak{a}$ is solvable it lies in $\mathfrak{z}(\mathfrak{g})$ and is therefore $0$ in $\mathfrak{g}/\mathfrak{z}(\mathfrak{g})$ which means that it is $0$ in $ad(\mathfrak{g})$. Therefore $ad(\mathfrak{g})$ is semisimple and since $ad(\mathfrak{g}) \cong [\mathfrak{g},\mathfrak{g}]$, the claim follows.

Is this proof correct?

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1 Answer 1

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I believe what you call $ad(g)$ is the image of the adjoint map $ad:g\rightarrow gl(g)$ defined by $ad(x)(y)=[x,y]$. The assumption that $ad(g)$ is isomorphic to $[g,g]$ is not always true, consider for example a solvable Lie algebra whose center is trivial.( An example of this is the Lie algebra of the group of affine transformation of the real line). For such an algebra, $ad(g)$ is isomorphic to $g$ and is not nilpotent, and $[g,g]$ is nilpotent.

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