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This is 2.13 exercise in Erdmann and Wildon's book.

Define a center $$ Z(L) = \{ z\in L |\ [z,x]=0\ \forall \ x\in L \} $$

If $I$ is ideal of $L$ then let $$ B = C_L(I) = \{ z\in L|\ [z,x]=0\ \forall x\in I \} $$ (It is called by centralizer of $I$. So $B$ is ideal of $L$)

Consider the following conditions

(1) $Z(I)=0$

(2) If $D$ is a derivation on $I$ then $D = {\rm ad}_x$ for some $x\in I$

Then $$ L= I \oplus B$$

I have no skill or experience in Lie algebra. Thank you.

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I have an idea now : If $x\in L$ then ${\rm ad}_x$ is a derivation on $I$. By the condition it is a trivial derivation. –  Hee Kwon Lee Dec 12 '13 at 11:52

1 Answer 1

up vote 1 down vote accepted

By the definition of $B$, we have $[I,B]=0$.

If $x\in I\cap B$, then $x\in Z(I)=0$, so $I\cap B=0$.

For any $x\in L$, we can restrict $\mathrm{ad}_{x}$ on $I$. By (2), there exists $x_I\in I$ s.t. $\mathrm{ad}_{x}\big|_{I}=ad_{x_{I}}$. It's obivious that $x-x_{I}\in B$ since$$\mathrm{ad}_{x}\big|_{I}(y)=ad_{x_{I}}(y)\iff[x-x_I,y]=0$$for every $y\in I$.

To sum up, we get $L=I\oplus B$.

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