An Alternative Definition of Reductive Lie Algebra? I came across an alternative definition of reductive Lie algebra as follows:
$\mathfrak{g}$ is said to be reductive of all abelian ideals of it are contained in its center $Z(\mathfrak{g})$ and $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=0$.
My question is, does this definition coincide with the usual one with $Z(\mathfrak{g})=Rad(\mathfrak{g})$? In particular, how can one show that $\mathfrak{g}=Z(\mathfrak{g})\oplus \mathfrak{s}$ where the latter one is semisimple and in fact is indeed $[\mathfrak{g}, \mathfrak{g}]$? Thank you in advance!

Suppose that $\mathfrak{g}$ is reductive. Then its adjoint representation is semisimple, so that the subrepresentation $[\mathfrak{g}, \mathfrak{g}]$ has a complement $W$ by Weyl's theorem, i.e., we have
$$
\mathfrak{g}=[\mathfrak{g}, \mathfrak{g}]\oplus W.
$$ 
Because of $[\mathfrak{g},W]\subseteq [\mathfrak{g}, \mathfrak{g}] \cap W=0$ we have $W\subseteq Z(\mathfrak{g})$, so that $[\mathfrak{g}, \mathfrak{g}]+Z(\mathfrak{g})=\mathfrak{g}$. This sum is in fact direct, since the subrepresentation $[\mathfrak{g}, \mathfrak{g}]$ is semisimple, so that we have a complement $U$ to $[\mathfrak{g}, \mathfrak{g}]\cap Z(\mathfrak{g})$, and
$$
[\mathfrak{g}, \mathfrak{g}]=[\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}] +Z(\mathfrak{g})] 
=[\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}]]=[\mathfrak{g},U]\subseteq U,
$$
hence $[\mathfrak{g}, \mathfrak{g}]\cap Z(\mathfrak{g})=0$. It follows that 
$\mathfrak{g}=[\mathfrak{g}, \mathfrak{g}] \oplus Z(\mathfrak{g})$, with $[\mathfrak{g}, \mathfrak{g}]$ semisimple, and $Rad(\mathfrak{g})=Z(\mathfrak{g})$.
