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The question is in the title: how can I show $\mathfrak{sl}_n(\mathbb{C})$ is simple?

In every book I scoured, they say $\mathfrak{sl}_n(\mathbb{C})$ is simple but they do not provide a proof! Is there a proof which does not resort to Lie groups?

Erdmann and Wildon book on Lie algebras, this is an exercise and it appears quite early on, before proving theorems like Cartan's Cariterion or before developing Killing forms. So, is there a way to show the result in a fairly basic way?

Thanks.

Shaun
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    I don't know any elementary proof which is not simply calculations. But doing those calculations is a really good exercise that will give you a good feeling for what is going on, so I advise you to do it anyway. So you should check that any non-zero ideal is in fact the entire algebra by playing a bit around with a suitable basis. – Tobias Kildetoft Nov 03 '13 at 12:51
  • @TobiasKildetoft by suitable basis, do you mean the standard basis $E_{ij}$? Can I do it by induction? –  Nov 03 '13 at 13:01
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    Indeed that basis (well, those with $i\neq j$ as well as the $E_{ii} - E_{jj}$). I am not aware of a nice way to do induction, but try to write up what happens when you take the Lie bracket between two of those basis elements, and play around a bit with them (note that at some point it will indeed be important that you are working over a field of characteristic not dividing $n$). – Tobias Kildetoft Nov 03 '13 at 13:04
  • @TobiasKildetoft Thank you. I'll try it! –  Nov 03 '13 at 13:08

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