Let $k$ be a field with $\operatorname{char} k = 0$. Let $L$ be a Lie $k$-algebra. Then the universal envelope $U(L)$ is a PI-algebra iff $L$ is abelian.


Remark:
PI-algebra means polynomial identity algebra, an algebra that satisfies a nonzero polynomial.


Obviously, if $L$ is abelian, then $U(L)$ is commutative. But the reverse direction is difficult for me. I also know that this is a result in an article "Two remarks on PI-algebras, V.N.Latysev". But I can not find the article.



The universal envelope U(L) is a PI-algebra iff L is abelian
The article is available [here](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=smj&paperid=4949&option_lang=eng), it seems that it hasn't been translated. The result you're interested in is Теорема 2; the proof is to notice that a non-abelian Lie algebra in characteristic zero contains as a subalgebra either a two-dimensional non-abelian Lie algebra or a three-dimensional Heisenberg Lie algebra, and their universal envelopings are not PI.


**UPD:** This only works for finite-dimensional Lie algebras.


