Let $L$ be a nonperfect Lie algebra (i.e. $L \ne [L, L]$) which is not nilpotent. Is there a ideal of $L$ such as $M$ such that intersection drived subalgebra of $L$ and $Z(M)$ is nonzero?


Yes, there is such an ideal. Take $L$ to be the $2$-dimensional nonabelian Lie algebra with basis $e_1,e_2$ and Lie bracket $[e_1,e_2]=e_1$ and the ideal $M=\langle e_1\rangle$. Then $L$ is not nilpotent, $L\neq [L,L]$ and $M$ is an ideal with $Z(M)=[L,L]\neq 0$.

In general, however, this need not be true. Take $L=\mathfrak{gl}_2(\Bbb{C})=\mathfrak{sl}_2(\Bbb{C})\oplus \Bbb{C}$. Then $L$ is not perfect and not nilpotent, but $Z(M)$ is either zero or $\Bbb{C}$ for all ideals $M$, and the intersection with $[L,L]$ is zero.

  • $\begingroup$ Thank you for your help. $\endgroup$ – Afsaneh Dec 18 '18 at 11:35
  • $\begingroup$ Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L? $\endgroup$ – Afsaneh Dec 18 '18 at 11:38
  • $\begingroup$ In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above. $\endgroup$ – Dietrich Burde Dec 18 '18 at 12:18

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