Just a while ago a question was posted that for a filtration $R=R^0\supset R^1\supset R^2\supset\cdots$ on a commutative integral domain $R$, the associated graded ring $ \text{gr}(R)=\bigoplus_{n=0}^\infty R^n/R^{n+1} $ is not necessarily a domain.
I was playing with the converse, assuming the graded ring is a domain. I could only conclude that $R$ is a domain if $\bigcap R^n=0$. I did this by taking $x,y\in R$ nonzero. So $x\in R^n$ but $x\notin R^{n+1}$ for some $n$. Likewise $y\in R^m$ and $y\notin R^{m+1}$ for some $m$. Then the images are $\bar{x}\in R^n/R^{n+1}$ and $\bar{y}\in R^m/R^{m+1}$ are nonzero, so the product $ \bar{x}\bar{y}=\overline{xy}\in R^{n+m}/R^{n+m+1} $ is nonzero since $\text{gr}(R)$ is a domain. Then $xy\notin R^{n+m+1}$, so $xy\neq 0$, and $R$ is a domain.
Is there some way to still conclude this without making the extra assumption that $\bigcap R^n=0$? Or is it not true in general?