If a Lie algebra L decomposes as a direct sum of its derived subalgebra and its center, is L reductive? A Lie algebra $L$ is said to be reductive if for any ideal $\mathfrak{a}$ of $L$, there is an ideal $\mathfrak{b}$ of $L$ such that $L=\mathfrak{a}\oplus\mathfrak{b}$. 
It is known that a reductive Lie algebra decomposes as $L = L'\oplus Z(L)$, where $L'$ is the derived subalgebra of $L$ and $Z(L)$ is the center of $L$. Is the converse true? 
The answer is "yes" if $L'$ is semisimple; can we deduce that $L'$ is semisimple just from the decomposition $L = L'\oplus Z(L)$?
 A: We cannot deduce that. Let $L$ be a finite dimensional semisimple Lie algebra and $V$ be an irreducible representation of $L$ where $1<\dim V<\infty$. Define $L \times V$ a Lie algebra structure by $([X, u], [Y, v])=([X,Y], Xv-Yu)$. Then you can check the followings:

*

*$[L \times V, L \times V]=L \times V$, i.e. $L$ is perfect.

*$Z(L \times V)=0$

*$V=\{ 0 \} \times V$ is an ideal of $L \times V$. $\nexists$ ideal $I$ of $L \times V$ such that $L \times V = V \oplus I$.

This shows that $L$ is  not semisimple (as it is not reductive). We can also prove that $L$ is not semisimple by calculating its radical.
Claim. $Rad(L \times V) = V$.
proof. It suffices to show $Rad(L \times V) \subset V$. If $(X, u) \in Rad(L \times V)$ with $X \neq 0$, we can prove inductively that there exists $0 \neq X_n \in L$ such that $(Y, 0) \subset (Rad(L \times V))^{(n)}$ which contradicts to the solvability, by the following lemma:
Lemma. Let $I$ be a nonzero ideal of a semisimple Lie algebra $L$. Then $[I, I] \neq0$.
proof. By Weyl's theorem, there exists an ideal $J$ of $L$ such that $L=I \oplus J$. If $[I, I]=0$, we have $L=[L,L]=[I,I]+[J,J] \subset [J,J] \subset J$ so contradiction occurs.
In fact, you can deduce $L=[L, L] \oplus Z(L)$ if $[L, L]$ is semisimple. It follows that $L$ is reductive if $[L, L]$ is semisimple. To prove this, view $L$ as a $[L, L]$-module. By Weyl's theorem, $L=[L, L]\oplus M$ for some $[L, L]$-submodule $M$ of $L$. Since $[L, L] \cap Z(L)=0$, it suffices to show the following claim:
Claim. $M \subset Z(L)$.
proof. Let $x \in M$ be given, and suppose there exists $y \in L$ such that $[x, y] \neq 0$. Since $[L, L]$ is semisimple and $0 \neq [x, y] \in [L, L]$, there exists $z \in [L, L]$ such that $[z, [x, y]] \neq 0$. However, Jacobi identity yields $[z, [x, y]] = [x, [z, y]] + [y, [x, z]]$. Here $[x, [z,y]]=0$ since $[z,y] \in [L, L]$ and $L=[L, L] \oplus M$. Similarly, $[y,[x,z]]=[y,0]=0$. Hence $[z, [x,y]]=0$; contradiction.
