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Let L be a finite-dim'l Lie algebra over an algebraically closed field of characteristic zero, and I be a solvable ideal of L. Prove that the ideal [L,I] is nilpotent.

My reasoning:

Consider the adjoint representation $ad: I \rightarrow gl(L)$ defined by $ x \mapsto ad_{x}: y \mapsto [x,y]$.

I solvable $ \implies ad(I)$ is also solvable

Lie's Theorem $\implies \exists z \in L $ such that $ad_{x}(z) = [x,z]=\lambda_{x}z$ $\forall x \in L$.

Now, to show that $[L,I]$ is nilpotent: Let $u_1,u_2 \in I$ and $v_1,v_2 \in L$.

$\left[[u_1,v_1],[u_2,v_2] \right]$

This is where I get stuck. I know that I need to show that $[I,L]^{n}={0}$ for some $n$ using the fact that there is a common simultaneous eigenvector but I don't know how.

nyj
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1 Answers1

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Let $rad(L)$ denote the sovable radical of $L$, i.e., the maximal solvable ideal of $L$. We have $I\subseteq rad(L)$, since $I$ is solvable. Now we can use the result $$[L,rad(L)]\subseteq nil(L),$$ where $nil(L)$ is the nilradical. The proof given here (which comes even before Lie's theorem) uses that all adjoint operators $ad_{[L,rad(L)]}(x)$ are nilpotent, so that $[L,rad(L)]$ is a nilpotent ideal by Engel's theorem. In particular, $[L,I]\subseteq nil(L)$ is a nilpotent ideal of $L$.

Dietrich Burde
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  • Thank you! We haven't learned about nilradicals yet. Do you have any hints of a way to show this using Lie's theorem? – nyj Apr 08 '16 at 17:27
  • You don't need the nilradical. It is enough to show that all $ad_{[L,rad(L)]}(x)$ are nilpotent, so that Engel's theorem gives you the result. Here you can use Lie's theorem. – Dietrich Burde Apr 08 '16 at 17:43
  • Here's my attempt:

    Consider the adjoint representation $ad:[L,rad(L)] \rightarrow gl(L)$ defined by $ad_{[x,y]}(z)=[[x,y],z]$

    Then $ {ad_{[x,y]} : x \in L, y \in rad(L) }$ is solvable and Lie's Thm $\implies $ there is some $ v \in L$ such that $[[x,y],v]=\lambda_{[x,y]}v$ for all $[x,y] \in [L,rad(L)]$

     – nyj Apr 09 '16 at 14:59