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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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You cannot relax that hypothesis (which is usually described by saying that the filtration is separated) The graded ring does not see what happens in that intersection, so it provides no information about it.

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.

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The example is of an increasing filtration... you can change it into one with a decreasing filtration as in your question without problems. – Mariano Suárez-Alvarez Mar 4 '12 at 10:33
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
What is the decreasing filtration which I suppose has nonzero intersection to show that the condition is necessary? – Adelaide Dokras Mar 7 '12 at 19:43
$R_0$ is the whole thing, and the $R_n$ with $n\geq1$ are equal to $V$. – Mariano Suárez-Alvarez Mar 7 '12 at 20:09

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