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.
Best Answer
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.)