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I need an example about ideal from lie algebra $\mathfrak{sl}(2,\mathbb{C})$ except trivial ideal and $\mathfrak{sl}(2,\mathbb{C})$ itself, can someone help me?

I try to make ideal except trivial ideal and $\mathfrak{sl}(2,\mathbb{C})$ itself, but I got stuck.

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    $\begingroup$ There aren't any... sl(2,$\mathbb{C}$) is a simple Lie algebra. This Lie algebra is small enough that it isn't too hard to prove directly. $\endgroup$ – Ted Jan 6 '13 at 7:22
  • $\begingroup$ @Ted can you give me an example not simple lie algebra and its ideal? $\endgroup$ – user46309 Jan 6 '13 at 10:03
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    $\begingroup$ Take the 2-dimensional Lie algebra with bracket defined by $[x,y]=y$. The the scalar multiples of $y$ is an ideal. $\endgroup$ – PAD Jan 6 '13 at 10:06
  • $\begingroup$ @Ted Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Jun 12 '13 at 11:12
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We claim that $\mathfrak{sl}(2,\mathbb{C})$ is a simple Lie algebra.

To see this, note that $\mathfrak{sl}(2,\mathbb{C})$ is 3-dimensional with basis $\{H, X, Y\}$ with the Lie brackets $$[H,X] = 2X$$ $$[H,Y] = 2Y$$ $$[X,Y]=H.$$

From the above relations, if an ideal contains one of $H$, $X$, or $Y$, then it is all of $\mathfrak{sl}(2,\mathbb{C})$, so it is sufficient to show that a nonzero ideal contains a nonzero multiple of one of $H$, $X$, or $Y$.

So suppose an ideal contains a nonzero element $U = aH + bX + cY$. Then the ideal also contains $$[U,X] = -cH + 2aX $$ $$[U,Y] = bH + 2aY$$ and thus also $$[[U,X],X] = -2cX$$ $$[[U,Y],Y] = 2bY$$ Hence, if either $b \ne 0 $ or $c \ne 0$ we are done, and if $b=c=0$ then we must have $a \ne 0$ (since $U \ne 0$) and hence $U$ is a nonzero multiple of $H$ and we are also done.

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