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Let $U(L)$ be the enveloping algebra of a Lie algebra $L$. How can I prove that $U(L)$ hasn't zero divisiors (e.g. if $xy=0$, $x,y \in U(L)$ then $x=0$ or $y=0$)?

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Do you have any assumption on $\rm L$ ? –  Damien L Feb 15 '13 at 21:54
    
Not... it is only a Lie algera. –  ArthurStuart Feb 15 '13 at 21:56

1 Answer 1

The standard argument is to show that the associated graded algebra with respect to the usual filtration is a polynomial ring.

Since being an integral domain is something that goes from the associated graded algebra to the algebra, this works.

This should be done in pretty much every textbook dealing with the subject.

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Using Poincare-Birkho-Witt theorem I do it... but is there an easier proof? –  ArthurStuart Feb 15 '13 at 22:01
    
I doubt it. ${}$ –  Mariano Suárez-Alvarez Feb 15 '13 at 22:03

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