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This is inspired by that question by Andreas Thom.

Let $L$ be a finitely generated Lie ring (or Lie algebra over a field) such that $L=[L,L]$, that is the Abelianization of $L$ is 0. Is it true that $L$ is generated as an ideal by one element?

If the answer is "no", can we consider the universal enveloping algebra of the counterexample to get a counterexample to the question about associative rings?

Same question can be asked for Jordan rings and other classical non-associative rings.

Update Consider the same question for arbitrary (non-associative) rings. That case should be easier.

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  • $\begingroup$ We thought about Lie algebras too, but could not see any chance to control the number of generators as a 2-sided ideal. $\endgroup$
    – Andreas Thom
    Commented Dec 6, 2011 at 20:37
  • $\begingroup$ How about general non-associative algebras? $\endgroup$
    – user6976
    Commented Dec 6, 2011 at 20:39
  • $\begingroup$ I do not know. I have no experience with general non-associative algebras. It is already hard to construct idempotent examples; the free non-associative ring on $n$ idempotents does not look idempotent. $\endgroup$
    – Andreas Thom
    Commented Dec 7, 2011 at 12:47
  • $\begingroup$ @Andreas: An "idempotent" ring is a ring $R$ with the property that $R*R$ generates $R$ as an Abelian group (i.e. every element is a sum of products). Certainly the free idempotent generated ring is "idempotent". I do not see how it can be generated as an ideal by one element. $\endgroup$
    – user6976
    Commented Dec 7, 2011 at 13:59

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