Let $L$ be a lie algebra, $I$ be an ideal of $L$ and $B$ be the $B$ be the centralizer of $I$ in $L$:
Let $B = C_L(I) = \{x \in L | [x,a]=0$ for all $a \in I\}$
I have shown that $B$ is an ideal of $L$. Now, make two further assumptions:
- $Z(I)=0$.
- If $D: I \rightarrow I$ is a derivation of $I$, then $D = ad(a)$ for some $a \in I$
Prove that under these conditions we have that $L = I \oplus B$
So it was simple to show that $I \cap B = \{0\}$. If $z \in I \cap B$ then we have both that $[a,z]=0$ for all $a \in A$ and that $[a,z] \neq 0$ for all $a \in A$ and thus $z = 0 $
Now for the hard part. I was trying to get condition 2 involved, so I considered the derivation $ad([x,b]): I \rightarrow I$ where $x \in I$, $b \in B$.
We can see that:
$ad([x,b])(a) = [[x,b],a] = -[[b,a]x]-[[a,x],b]= -[0,x]-[a',b]=0$
Furthermore, by condition (2) we have that $ad([x,b])=ad(a)$ for some $a \in A4.
Anyway, these are just some ideas. Insights / walthroughs appreciated
Let $x\in L$, the restriction of $ad_x$ to $I$ is a derivation, there exists $i\in I$ with $[x,y]=[i,y], y\in I$ it implies that $[x-i,y]=0$ and $x-i\in B$