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When trying to show that $I+J$ is nilpotent, whenever $I,J$ are nilpotent ideals of a Lie algebra $L$, I did it brute force: By induction we can show the following: An element of $(I+J)^{2N}$ is a sum of elements of the form $[x_1,[x_2,[\ldots,[x_{2N},x_{2N+1}]]\ldots]$, with $x_i \in I \cup J$. So at least $N$ of the $x_i$ lie in $I$ or in $J$. Using induction again, one can show that this product then lies in $I^N$ resp. $J^N$. Hence, if we choose $N$ large, it follows that $(I+J)^N = 0$.

I was wondering if there is some more elegant proof. Is it maybe easier to use Engel's theorem ?

  • Engel is massively overkill here: the result works over any commutative ring as a base ring, but I am pretty sure that Engel relies on the base ring being a field. – darij grinberg Jun 01 '15 at 19:14

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We have $(I+J)^{2m}\subseteq I^m+J^m$ for all $m\ge 1$, hence $(I+J)^{2m}=0$ for $m$ large enough. I find this proof very natural, but indeed there is another proof using Engel's theorem and another two lemmas from representation theory; so it is perhaps more elegant, but it is also more complicated. This depends on your taste, too. For the alternative proof see Lemma $5.3.3$, Lemma $5.3.4$ and Lemma $5.3.5$ here.

Dietrich Burde
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    The relation in the first sentence of your answer is a particular case of the inclusion $\left(I + J\right)^{a+b} \subseteq I^a + J^b$ for all $a \geq 1$ and $b \geq 1$. This latter inclusion, in turn, can be obtained from the stronger fact that $\left(I + J\right)^n \subseteq \sum_{k=0}^n \left(I^k \cap J^{n-k}\right)$ for all $n \geq 1$ (where both $I^0$ and $J^0$ mean the whole Lie algebra). And this stronger fact is easy to prove by induction over $n$. This is essentially ... – darij grinberg Jun 01 '15 at 19:30
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    ... a slick version of the proof the OP had in mind. No arguing about how terms look like, no "repeated expanding", no "shoving all brackets to the right", etc.. The moral of the story: Don't look for "more elegant proofs" before you have tried making your existing proof elegant. – darij grinberg Jun 01 '15 at 19:32
  • @darij I'm not sure if there is a typo here, but does that stronger fact actually hold? It does not seem true even for $n=1$. – Gregoire Rad May 04 '17 at 23:09
  • @GregoireRad: Sure? For $n = 1$, I am claiming that $\left(I + J\right)^1 \subseteq I^0 \cap J^1 + I^1 \cap J^0$, which is clearly true (seeing that $I^0$ and $J^0$ are the ambient Lie algebra). – darij grinberg May 05 '17 at 01:05
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    Hey Darij Grindberg, your claim is not true, even for n=1. Example: Let L be the vector space with basis e_1,...,e_4,x. Define the Lie bracket by [e_i,e_j]=0, [x,e_1]=[x,e_2]=0, [x,e_3]=e_1 and [x,e_4]=e_2 (for a more general statement see [Karin Erdmann, Mark J. Wildon - Introduction to Lie Algebras, page 33 exercise 3.1.]. Now take the two ideals I:=<e_1,e_2,e_4> and J:=<x,e_1,e_2,e_3>. Obviously I+J=L. But we have [L,L]=<e_1,e_2>, but [I,I]=0 and [J,J]=<e_1>. Which is a counterexample. –  May 17 '20 at 09:46
  • @ZPlaya: How is that a counterexample to anything Darij claimed? – Torsten Schoeneberg Dec 17 '20 at 20:37
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    @TorstenSchoeneberg Maybe I was confused, but the regular definitions are $J^1 := J'$ and $J^0 = J$... Not $J^1 = J$ and $J^0 =$ whole lie algebra... –  Dec 19 '20 at 11:10