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Let L be a 3-dimensional vector space over k with basis x,y,z. Given L an anti-commutative algebra structure by setting $[x,y]=z,[y,z]=x,[z,x]=y$

Prove that L is a simple Lie algebra.

So L is simple if 0 or L are the only ideals of L.

So assume, I is an ideal s.t. that $0 \not = v \in I$ and is not L. I suppose you would start with $v=ax+by+cz$ and then show what? that you can get everything?

I'm a bit unsure what to do, can someone give me a hint?

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You haven't proven that $L$ satisfies the Jacobi identity yet. –  Qiaochu Yuan Mar 1 '12 at 16:54
@QiaochuYuan That was an earlier part of the question. I've proven it satisfies jacobi identity and proven that central is 0. –  danielr900 Mar 1 '12 at 16:57
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up vote 4 down vote accepted

Let $0 \neq v = ax+by+cz \in I$; your aim is to prove that $x,y,z \in I$ because then $I$ is the whole of $L$. (in fact, it's enough to show that one of $x,y,z \in I$)

Here's my hint. Since $I$ is an ideal, $[x,v] \in I$. But you can work out $[x,v]$ using the definition of the Lie bracket. It will have a $y$ term and a $z$ term, but no $x$ term. So then apply $[y,-]$. You'll be down to a scalar multiple of $x$....

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