The enveloping algebra of a finite dimensional Lie algebra has no zero divisor

Asked Mar 04 '13 at 15:34

Active Mar 04 '13 at 15:55

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Let $L$ be a complex, finite dimensional Lie algebra.

It is well-known that the graded associative algebra of the enveloping algebra $U(L)$ is isomorphic to the symmetric algebra $S(L)$. Therefore $U(L)$ has no non-zero zero-divisors.

But I really want to know is: can we prove this without the Poincare-Birkhoff-Witt theorem?

Thanks !

edited Mar 04 '13 at 15:55

rschwieb

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asked Mar 04 '13 at 15:34

user64963

Hi: can you suggest a reference where I can read about the well-known fact? – rschwieb Mar 04 '13 at 15:56
HI: I read the book: <> 2ed by Anthony W. Knapp; page.223, proposition 3.16. – user64963 Mar 04 '13 at 16:00

The enveloping algebra of a finite dimensional Lie algebra has no zero divisor

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