Does $R$ a domain imply $\operatorname{gr}(R)$ is a domain? 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.
 A: No. Let $R=\mathbb{C}[x,y]/(y^2-x^3-x^2)$ be the coordinate ring of a singular cubic. Filter $R$ by total degree in $x$ and $y$. Check that $R$ is a domain (i.e., the cubic is irreducible) but that $(x-y)(x+y)=x^3=0$ in the associated graded ring.
What is going on: analytically locally, the curve is reducible near $0$: there are branches approximated by $y-x$ and $y+x$. So the completion wrt the filtration above is not a domain, and hence its associated graded, which is the same as the associated graded of the original ring, cannot be a domain.
A: A slightly simpler example: find a filtration on $k[x,y,z]/(xy-z^2)$ such that the associated graded ring is $k[x,y,z]/(xy)$.
A: The answer is no.  Here is a counter-example:
Let $R = \mathbb C[x,y]/(y^2 - x^3 -x^2),$ with the filtration being the one induced by the filtration by degree on $\mathbb C[x,y]$ (i.e. take the quotient
filtration on $R$).   This is the same as the $\mathfrak m$-adic filtration on $R$, where $\mathfrak m$ is the ideal generated by $x$ and $y$.
The $R$ is a domain, but $gr(R)$ is not: $y^2 - x^2 = (y - x)(y+x) = x^3$,
and so $y-x$ and $y + x$ are non-zero elements of $gr^1$ whose product vanishes
(because the element $x^3$ has vanishing image in $gr^2$).
Here is where this example came from:
If $R$ is any ring and $\mathfrak m$ a maximal ideal in $R$, then the associated graded ring $gr(R)$ with respect to the $\mathfrak m$-adic filtration on $R$ is the same as the associated graded $gr(\widehat{R})$, where $\widehat{R}$ is the $\mathfrak m$-adic completion of $R$, again endowed with the $\mathfrak m$-adic filtration.  
Now it can happen that $R$ is a domain but $\widehat{R}$ is not, e.g. because $R$ is the affine ring of an irreducible variety, which has more than one analytic branch passing through the point corresponding to $\mathfrak m$.  The example I chose was perhaps the simplest one of this kind: $R$ is the affine ring of a nodal curve, and $\mathfrak m$ is the maximal ideal corresponding to the node, which has two analytic branches passing through it.
Added: This is identical to the example Steve just gave, and we were both led to it for the same reason.
