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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

2 Answers 2

up vote 5 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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Here is another argument. Note that $L$ cannot be solvable because of $[L,L]=L$. Now suppose that $I$ is a nonzero ideal, different from $L$. Hence $dim (I)=1$ or $dim(I)=2$. In the first case, $I$ is solvable and $L/I$ is of dimension $2$, hence solvable. It follows that $L$ is solvable, a contradiction. In the second case, $I$ is solvable and $L/I$ is $1$-dimensional, hence also solvable. Again it would follow that $L$ is a solvable extension of a solvable Lie algebra, hence solvable. A contradiction. It follows that $L$ has only the trivial ideals.

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