I know a simple Lie algebra $L$ has only two ideals: $\lbrace0\rbrace$ and itself. If $\lbrace0\rbrace$ is the radical (i.e., maximal solvable ideal), we are done. If $L$ is the radical... where is the contradiction?

I'm sure this is really simple.

  • $\begingroup$ Note that the "maximal solvable ideal" does not exist in general; yet it's fine for finite-dimensional Lie algebras over fields. $\endgroup$ – YCor Apr 1 '17 at 18:00

The solvable radical $rad(L)$ is an ideal of $L$. So it must be $0$ or $L$, for $L$ being simple. If $rad(L)=L$, then $L$ is simple and solvable at the same time, hence $0$. If $L\neq 0$, then consequently $rad(L)=0$.

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  • $\begingroup$ A simple Lie algebra $L$ is perfect; that is, $L=L^2$. Hence, if it is also solvable, $L=L^{(n)}=0$. $\endgroup$ – David Towers Mar 29 '17 at 1:26

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