Given a Lie Algebra $L$ we say it is reductive if $\operatorname{Rad}L=Z(L)$.
How can we prove that $L$ is reductive iff it is an $\operatorname{ad}_L(L)$-module completely reducibile?
Suppose $L$ reductive and consider $\operatorname{ad}_L:L\to\mathfrak{gl}(L)$, then $\operatorname{ad}_L(L)\simeq L/Z(L)=L/\operatorname{Rad}L$ which is semisimple. Thus Weyl theorem says that $\operatorname{ad}_L(L)$ is completely reducible.
Now saying that $L$ is a $\operatorname{ad}_L(L)$-module, means that we have $\varphi:\operatorname{ad}_L(L)\times L\to L$, $\varphi(\operatorname{ad}_L(x),y)=\operatorname{ad}_L(x)y=[x,y]$.
It should be equivalent to say that $L$ is a representation for the Lie Algebra $\operatorname{ad}_L(L)$, i.e. we have $\varrho:\operatorname{ad}_L(L)\to\mathfrak{gl}(L)$, $\varrho(\operatorname{ad}_L(x))(y)=\varphi(\operatorname{ad}_L(x),y)=\operatorname{ad}_L(x)y=[x,y]$ for every $y\in L$.
But I don't know, from this, how can I write $L$ as a direct sum of irreducible representations of $\operatorname{ad}_L(L)$.
Can someone help me? Many thanks to all.
