# Trouble with “trivial” example of a semi-simple Lie algebra [duplicate]

If the radical of a Lie algebra is zero, we call it semi-simple. In the lecture notes that I'm following its stated that for any arbitrary Lie algebra (over a field with characteristic zero and finite dimensional) $$\mathfrak{g}$$ we have, $$\mathfrak{h}:=\mathfrak{g}/\mathrm{rad}(\mathfrak{g})$$ semi-simple. I have trouble seeing why.

I need to show that $$\mathrm{rad}(\mathfrak{h})=0$$. I think we can use the fact that if $$\mathfrak{I}$$ is a solvable ideal of $$\mathfrak{g}$$ such that $$\mathfrak{g}/\mathfrak{I}$$ is solvable, then $$\mathfrak{g}$$ is solvable, but I'm not really sure how...

• Are there any conditions on your Lie algebra? For example, over a field of characteristic zero? Finite dimensional? – Matt Feb 1 at 14:14
• @Matt Yes, I forgot to mention that its over a field with characteristic zero, finite dimensional! – Marius Jaeger Feb 1 at 14:16
• So that you know, in this (rather nice) case, you actually have that this quotient $\mathfrak{h}$ is isomorphic to an actual subalgebra of $\mathfrak{g}$. This follows from something called Levi's theorem. This is a useful theorem, because if you think about it, there is no reason to think that $\mathfrak{h}$, as you've defined it, should be a subalgebra of $\mathfrak{g}$ at all! – Matt Feb 1 at 14:21
• @Matt if semisimple is defined as "trivial radical", this works over arbitrary fields (and finite-dim Lie algebras). There's no need to use Levi factors. – YCor Feb 1 at 22:35

$$\mathfrak{g}$$ is semisimple if it has no non-zero abelian ideals.

Let $$\pi: \mathfrak{g} \rightarrow \mathfrak{h}$$ be the quotient map.

if $$\mathfrak{h}$$ is not semisimple, then $$(0)\subsetneq rad(\mathfrak{h})$$

therefore $$\pi^{-1}(0) = rad(\mathfrak{g}) \subsetneq \pi^{-1}(rad(\mathfrak{h}))$$

But $$\pi^{-1}(rad(\mathfrak{h}))$$ is an abelian ideal, this is a contradiction of the maximality of $$rad(\mathfrak{g}).$$

HINT:

Suppose for contradiction that $$\mathfrak{h}$$ is not semi-simple. Write down exactly what this means.

You can now utilize a particular fact, and show that $$\mathfrak{g}$$ would have to therefore have a non-zero solvable ideal. This gives a contradiction.