# showing that a lie algebra is the direct sum of two ideals.

I am trying to do an exercise(2.13) in Wildon and Erdmann's Intro to Lie Algebras and I'm stuck. The meat of the question is to show that under the following conditions

i. the center, $Z(I)=0$
ii. if $D:I\to I$ is a derivation, then $D= \mathrm{ad}~x$ for some $x\in I$

a lie algebra, $L$ is the direct sum of an ideal $I$ of L and the centraliser $B$ of $I$ in $L$. That is, I'm required to show that $L=I\oplus B$.

With the help of the Jacobi identity, I've been able to show that $B$ is an ideal of $L$. My difficulty is in showing the direct sum. I know I have to show that $B\cap I = \{0\}$ and every element in $L$ can be written as a sum of elements in $B$ and $L$.

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$Z(I) = \{0\}$ is equivalent to the intersection of the centralizer of $I$ and $I$ being trivial. So $B \cap I = \{0\}$. Now let $y \in L$. By (ii) there exists $x \in I$ such that $\operatorname{ad} y\vert_I = \operatorname{ad} x\vert_I$. Then $y - x \in B$ since $[y-x, z] = [y,z] - [x,z] = 0$ for all $z \in I$. Thus $y = x + (y-x)$ with $x \in I$ and $y-x \in B$.

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can u explain your first sentence a bit. –  May Oct 23 '12 at 17:56
Suppose $x \in B \cap I$. Then since $x \in B$, $[x,y] = 0$ for all $y \in I$. Since $x \in I$, this implies $x\in Z(I) = \{0\}$. So $B \cap I = \{0\}$. –  Eric O. Korman Oct 23 '12 at 17:58