On complex irreducible representations of Lie algebras

Question

In my lecture notes I have stumbled upon the following:

"Indeed, the statement [Lie's theorem] immediately implies that any such representation [complex representation of a solvable Lie algebra] contains an invariant subspace of dimension one, which in particular implies that irreducible complex representations are automatically one-dimensional and trivial."

Two questions here:

1. Why is the prerequisite "solvable" no longer needed in the last statement? Or is it just true for any representation of a solvable Lie algebra?

2. I get that it must be one-dimensional (cf. for example Complex irreducible representation of solvable lie algebra). But why does it follow that such representation must be trivial?

Answer 1

1. Solvability is still needed, because it is false otherwise. Take $L=\mathfrak{sl}(2)$. It has an irreducible representation in every dimension. Moreover, the field has to be algebraically closed of characteristic zero; otherwise there are counterexamples.

2. Sometimes "trivial" just means $1$-dimensional, even though the action is not by zero. Compare with the group case, which was answered here: Why does the trivial representation have degree 1?