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.

Solution 1:

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.

Solution 2:

A slightly simpler example: find a filtration on $k[x,y,z]/(xy-z^2)$ such that the associated graded ring is $k[x,y,z]/(xy)$.