Consider two quadratic algebras based on the vector space $\mathbb{R}^3$ with basis $x,y,z$. The antisymmetric tensor algebra $\Lambda \mathbb{R}^3$ obviously has zero divisors, e.g. $(x)(xy)=0$, but the symmetric tensor algebra $S \mathbb{R}^3$ has not, as it is isomorphic to the polynomial functions on $x,y,z$. For $\Lambda \mathbb{R}^3$ the existence of zero divisors is obvious at degree 1, as $x^2=0$.

We could ask this more generally for a possibly noncommutative quadratic algebra with relations $R\subset V\otimes V$ - when is the quadratic algebra $Q(V,R)$ an integral domain? To give a more specific question, if $Q(V,R)$ an not integral domain, then are there always examples of zero divisors in degree 1, i.e. $x,y\in V$ so that $xy\in R$?

Apologies if this is well known - could you give me a reference for a proof or counterexample if it is! The question arose from trying to generalise the behaviour of the symbols of differential operators, where classically $SV$ is an integral domain, to operators with other relations.