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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.

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up vote 3 down vote accepted

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.

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Huh. I didn't see your answer show up, even though it looks like your finished before me! – Stephen Mar 4 '12 at 1:24
Weird, now the time stamps are reversed, and it claims I posted first. Am I going crazy? Not that it matter much... – Stephen Mar 4 '12 at 1:26
@Steve: Dear Steve, No, your answer came up just as I was writing the motivational discussion in mine, but I didn't look at it until after I posted mine. I was surprised but amused to see that we said the same thing (although you much more pithily), and added a comment to this effect (which doesn't show up in the time stamp because it was added within a minute of when I first posted my answer). Cheers, – Matt E Mar 4 '12 at 1:29
Dear Matt, thanks very much for the counter-example. Are $gr^1$ and $gr^2$ notation for the first and second components of the direct sum? (Sorry, I have not seen that until now.) – Jacqueline Pauwels Mar 4 '12 at 1:52
@Jakucha: Dear Jakucha, Yes, $gr^i$ is notation for the $i$th component of the direct sum. Regards, – Matt E Mar 4 '12 at 2:10

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.

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Thank you Steve. – Jacqueline Pauwels Mar 4 '12 at 1:52

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

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More generally, using Groebner bases one sees that quotients of polynomial rings admit filtrations such that the associated graded rings are quotients of the same polynomial ring by monomial ideals, and the latter are rarely domains. – Mariano Suárez-Alvarez Mar 4 '12 at 1:45
Thanks for this other idea. – Jacqueline Pauwels Mar 4 '12 at 1:53

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