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Let $L$ be a finite dimensional semisimple Lie algebra over a field of charcteristic $0$ and algebraically closed. I have learned that the map $$ ad: L \rightarrow \ Der(L) $$ is an isompsphism, where $adx(y) = [x,y]$ and $Der(L)$ denotes all the derivations on $L$. Furthermore, I have learned that if $x \in L$ is semisimple then $adx:L \rightarrow L$ is semisimple. I was wondering from these information does it actually follow that $x_s \in L$? (where $x_s$ is the unique semisimple part of $x$)

Clarification: Given $x \in L$ we can always write it as $x = x_s + x_n$ where $x_s$ is the semisimple part and $x_n$ is the nilpotent part and I know that they both lie in $gl(V)$ but do they lie in $L$? (and I am viewing $L$ as a subalgebra of $gl(V)$ for some finite dimensional vector space $V$)

Xander Henderson
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Johnny T.
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    I think this is true; this seems to follow from the fact that $L$ is the Lie algebra of an algebraic group. Probably there's then a more direct proof. This is maybe done in Bourbaki. – YCor Mar 02 '19 at 16:29
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    See this question. – Dietrich Burde Mar 02 '19 at 17:45
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    This is definitely done in Bourbaki, for any field of characteristic $0$, and probably in most other sources at least for $\Bbb C$. See Dietrich Burde's link and comments for an exact reference. Note that if you view $L$ as subalgebra of "some" $gl(V)$ you should worry about that being independent of the choice of $V$ (which it is, but that's part of the proof). – Torsten Schoeneberg Mar 02 '19 at 17:53

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