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Here $V $ is finite dimensional over $\mathbb{C}$.

I am trying to show that if a Lie algebra $L \subset \mathfrak{sl}(V)$ with irreducible natural representation has an abelian ideal $I$ then $I=\{ 0 \} $. I know that for any $\phi \in I $ that $V=\ker (\phi ^{\dim V }) $ is just the generalised eigenspace for $0$ and so every $\phi \in I $ is nilpotent.

Now letting $\phi \in I $, I need to show that $ \phi (v)=$ for all $v \in V $ but I am unsure how to start this to be honest.

Anonmath101
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    Also, this is the third related question to this topic you post in short time, plus you asked me to explain that in a comment to a fourth question. I am actually thinking about that and will reply, but that needs time. Maybe you should give yourself more time to think about these questions before you ask them here. I suggest that as a better learning strategy. – Torsten Schoeneberg Apr 19 '23 at 21:02
  • Yeah I should’ve said abelian in the question. I’ve spent quite a while on some of these questions. Sometimes just can’t seem to make any headway and need a start – Anonmath101 Apr 19 '23 at 21:04
  • I have now answered a more general version of your question (for solvable instead of abelian ideals) in an edit to https://math.stackexchange.com/a/3023372/96384. Also compare https://math.stackexchange.com/q/4132002/96384. The crucial and admittedly non-trivial step is the lemma mentioned in https://math.stackexchange.com/q/3784195/96384, which shows that certain weight spaces of the ideal action are invariant under the action of the entire Lie algebra. – Torsten Schoeneberg Apr 19 '23 at 21:26

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