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I have to find an example of non nilpotent Lie algebra $L$ and an ideal $I$ of $L$ such that $L/I$ is nilpotent. So we can take the algebra of $ 2 \times2$ matrix upper triangular and with null trace. So the matrix $(0,1),(0,0)$ is a nilpotent ideal of $L$ and $L/I$ is one-dimentional. But... is there another example?

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Your example is spot on. In general, you might take any soluble algebra which is not nilpotent. It will have a non-trivial quotient which is abelian, hence nilpotent.

In the category of trivial examples, the direct sum of a non-trivial nilpotent and a non-nilpotent algebra will do.

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But why in general is $L$ soluble? – ArthurStuart Feb 15 '13 at 21:55
What are you asking? In general $L$ is not soluble: take for example $\mathfrak{sl}_2\times(\text{some nilpotent Lie algebra)}$, which is not solvable and has a nilpotent quotient. – Mariano Suárez-Alvarez Feb 15 '13 at 21:59

There are as many other examples as you want in "the category of trivial examples". Let $L=I\oplus N$ the direct Lie algebra sum with any given nilpotent Lie algebra $N$ and an arbitrary non-nilpotent Lie algebra $I$. Then $I$ is an ideal with nilpotent quotient $L/I\cong N$, and $L$ is non-nilpotent.

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