The Levi's decomposition in Lie theory and the problem of classification Lie algebras

Question

It is well-know that: "any finite-dimensional Lie algebra over a field of characteristic zero can be expressed as a semidirect sum of a semi-simple subalgebra and its radical (its maximal solvable ideal). It reduces the task of classifying all Lie algebras to obtaining the classification of semi-simple and of solvable Lie algebras". I am wondering about that "reduction". Let's consider the following simple example: let $L=M\otimes_\phi N$ is a semidirect product of Lie algebras where $M \subseteq L$ and $N\subseteq L$ are commutative Lie subalgebras of $L$ and the Lie structure on $L$ depends on $\phi$. If so, followed by the above claim, $\phi$ does not play any role in the problem of classifying such Lie algebras?

Answer 1

The Levi decomposition of $L$ is given by $$ L=\mathfrak{s}\ltimes_{\phi} {\rm rad}(L), $$ where $\mathfrak{s}$ is a semisimple subalgebra of $L$, and ${\rm rad}(L)$ the maximal solvable ideal of $L$. Suppose now we have classified all semisimple Lie algebras $\mathfrak{s}$, and we would know all solvable ones (which we don't). Then we know all $L$.

So, given a Lie algebra $L$, we need to find its Levi decomposition. There is an algorithm, how to compute it. The map $\phi$ must be a Lie algebra homomorphism $\phi\colon \mathfrak{s}\rightarrow {\rm Der}({\rm rad}(L))$. Then we know $L$ from this decomposition, because we know $\mathfrak{s}$ and the solvable ideal.