# Relaxing a condition to prove that an associated graded ring is a domain implies the ring is a domain.

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?

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Let $k$ be a field, $V$ a non-zero vector space and let $R=k\oplus V$, with multiplication extending that of $k$ and such that $V\cdot V=0$. Consider the filtration such that $R^n=k$ for all $n\geq1$ and $R^0=0$. Then $\operatorname{gr}R=k$, a domain, and $R$ is not a domain.
Ah, so if we don't know the filtration is separated, we cannot conclude that $\text{gr} R$ is a domain implies $R$ is a domain. –  Adelaide Dokras Mar 4 '12 at 12:01
$R_0$ is the whole thing, and the $R_n$ with $n\geq1$ are equal to $V$. –  Mariano Suárez-Alvarez Mar 7 '12 at 20:09