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Let $\mathfrak{g}$ be a finite-dimensional Lie algebra, and denote by $U(\mathfrak{g})$ its universal enveloping algebra. It appears to be a consequence of the Poincaré-Birkhoff-Witt Theorem that $U(\mathfrak{g})$ has no zero-divisors. All sources I look at consider this to be either obvious or an easy exercise. But to be honest I'm baffled by this problem.

Attempt 1: Take a basis $e_1,\ldots,e_n$ of $\mathfrak{g}$, so that $U(\mathfrak{g})$ is generated by all terms of the form $e_1^{k_1} \cdots e_n^{k_n}$. Take two elements $x$ and $y$ in $U(\mathfrak{g})$, and suppose $x y = 0$. Our goal is to show that $x$ or $y$ is trivial. Thanks to Poincaré-Birkhoff-Witt, $x$ and $y$ are a finitely sum of the form $$x = \sum_{k_1,\ldots,k_n} a_{k_1\cdots k_n} e_1^{k_1} \cdots e_n^{k_n}$$ and $$y = \sum_{k_1,\ldots,k_n} b_{k_1 \cdots k_n} e_1^{k_1} \cdots e_n^{k_n}$$ We can then expand the product $xy$, and for each product of the form $$e_1^{k_1} \cdots e_n^{k_n} e_1^{k'_1} \cdots e_n^{k'_n}$$ we can use the rule $e_i e_j + e_j e_i = [e_i,e_j]$ finitely many times to find an expression of it with respect to the basis that PBW gives us. Working all this out, the expression for $xy$ entirely gets out of hand, and it is not at all clear that it won't vanish for non-trivial $x$ and $y$.

Attempt 2: It is often hinted that you should use the associated graded ring of $U(\mathfrak{g})$ in some way. Many times it is stated that, as a consequence of PBW, this graded ring is a polynomial ring, and that therefore there aren't zero divisors. Both logical steps elude me, and I have no idea how to proceed in this direction.

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    $\begingroup$ Do you read German? If so, you can find a proof in II. Kapitel, Folgerung 2.9 of Hans-Jürgen Schneider, Hopfalgebren, version 0.1.99 of my scribe notes. This is on page 312 currently. The main engine of the argument (besides the PBW theorem) is the lemma (part of Lemma 2.8) saying that if the associated graded of a filtered algebra is a (non-commutative) integral domain, then so is the algebra itself. This is also in Corollary 2.3.9 (ii) of Dixmier's Enveloping algebras. $\endgroup$ – darij grinberg Jun 7 at 15:46
  • $\begingroup$ Those are some remarkably detailed notes! Thank you. Aside --- how is your first name pronounced if I may ask? $\endgroup$ – user542740 Jun 25 at 13:13
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    $\begingroup$ Thank Hans-Jürgen Schneider, who essentially wrote a finished book on the blackboard, in a good old German tradition. I just fixed the occasional typo and added the occasional detail as well as a few sections which, in hindsight, are diluting the quality of the notes. My first name is pronounced in the German fashion; the IPA translation would probably be [dɑrɪj] (but I don't know IPA well enough to make guarantees). $\endgroup$ – darij grinberg Jun 25 at 13:34
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The trick is that you can look at just the highest-degree terms. Note that $$e_1^{k_1}\dots e_n^{k_n}\cdot e_1^{j_1}\dots e_n^{j_n}=e^{k_1+j_1}\dots e_n^{k_n+j_n}+\text{lower degree terms}$$ since all the other terms come from replacing two of the factors by their bracket and thus have lower degree. So the highest degree part of $x\cdot y$ is just obtained by multiplying the highest degree parts of $x$ and $y$ as if the $e_i$ all commuted (i.e., treating them as variables in an ordinary polynomial ring). Since a polynomial ring in $n$ variables has no zero divisors, the highest degree part of $x\cdot y$ is nonzero (assuming $x$ and $y$ are nonzero) and thus $x\cdot y$ is nonzero.

Or, without picking a basis, if $x$ has degree $d$ and $y$ has degree $e$, we look at the image $x'$ of $x$ in $A_d$ and the image $y'$ of $y$ in $A_e$ where $A$ is the associated graded algebra of $U(\mathfrak{g})$. By definition of the ring structure on $A$, $x'\cdot y'$ is the image of $xy$ in $A_{d+e}$. If $A$ has no zero divisors and $x$ and $y$ are nonzero, then $x'$ and $y'$ are nonzero and thus $x'y'$ is nonzero, so $xy$ must be nonzero. (This argument shows more generally that if the associated graded ring of a filtered ring has no zero divisors, then neither does the original ring.)

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  • $\begingroup$ Damn, that single first sentence was all I needed to make it click. Thanks. $\endgroup$ – user542740 Jun 25 at 13:11

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