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If $L$ is a semi-simple Lie algebra, then $L=L'$.

Since $L$ is semi-simple we can write it as a direct sum of simple ideals $L_i$, i.e. $L=\oplus_{i=1}^r L_i$. Then $L'=\oplus_{i=1}^r L_i'$ and since every $L_i$ is simple we have $L_i'=L_i$ and consequently $$L'=\oplus_{i=1}^r L_i'=\oplus_{i=1}^r L_i=L$$

This proof seems surprisingly short. So my question is, did I make any mistakes or did I forget any crucial assumptions? Is there perhaps a more elementary proof?

Phil-ZXX
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  • It is correct. The reason it can be done in such a short way is that you invoke some fairly heavy machinery when you decompose it as a direct sum of simple Lie algebras. I have a feeling this can also be done more directly using the definition of semisimple, but I will need to think a bit about it. – Tobias Kildetoft May 20 '13 at 14:32
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    We are assuming characteristic $0$ here to use the machinery Tobias mentions. For a counterexample in positive characteristic, see https://math.jhu.edu/~sakellar/automorphic-files/liestructure.pdf, remark 2.2. – Torsten Schoeneberg Oct 17 '22 at 17:50

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