Suppose you have a filtration $R=R^0\supset R^1\supset R^2\supset\cdots$ on a commutative ring $R$. This gives the associated graded ring $ \text{gr}(R)=\bigoplus_{n=0}^\infty R^n/R^{n+1}. $ From my reading, I know multiplication is defined on homogeneous elements in the following way. If $a\in R^m$ and $b\in R^n$, then $a+R^{m+1}\in R^m/R^{m+1}$ and $b+R^{n+1}\in R^n/R^{n+1}$, then $ (a+R^{m+1})(b+R^{n+1})=(ab+R^{m+n+1}). $
Something I've been wondering though, is if $R$ is an integral domain, is it true that $\text{gr}(R)$ is an integral domain? I've heard that the converse is true, but I'm curious about this direction.
What I find confusing, is suppose $x,y\in\text{gr}(R)$, and $xy=0$. Now $x=\sum(a+R^m)$ and $y=\sum (b+R^n)$, and so we can multiply these element by using distributivity and the definition of multiplication on homogeneous components. But in multiplying out such an arbitrary sum, it seems very difficult to conclude where $x=0$ or $y=0$ or not to conclude whether or not $\text{gr}(R)$ is a domain. Thanks.