Is the semisimple part of $x \in L$ belong to $L$ when $L$ is a semisimple Lie algebra?

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 $$V$$)

• 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 2 at 16:29
• See this question. – Dietrich Burde Mar 2 at 17:45
• 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 2 at 17:53