# Associated graded ring of a Fermat cubic

Let $R$ be a graded Fermat cubic, i.e. $R$ is a graded ring given by $$R=\mathbb{C}[x,y,z]/(x^3+y^3+z^3),$$ with a standard grading $\operatorname{deg}(x)= \operatorname{deg}(y)=\operatorname{deg}(z)=1$. Let me denote irrelevant ideal by $m$ $$m=\bigoplus_{i \geq 1} R_i.$$ Is it possible to give an explicit description of associated graded ring $$gr_m(R)=\bigoplus_{i\geq 0}m^i/m^{i+1}$$ of $R$?

Affine cone over Fermat cubic in $\mathbb{P}^2$ is singular, but singularity is isolated an correspond to the maximal ideal $m$. So, geometrically it would give a description of special fiber in the blow-up of the affine cone at singular point i.e. it is natural to expect that singularity of $gr_m(R)$ is milder than original singularity.

Upd: After some thinking I come to conclusion that it should be true $R \cong gr_m (R)$, if $R$ is generated in degree one. With the same prove as was pointed out in the answer. (Am I right?) But this is not true in general as very simple example $R=\mathbb{C}[x, y]/(x^3+y^2)$, where $\operatorname{deg}(x)=2$, $\operatorname{deg}(y)=3$ shows.

• Isn't it just isomorphic to the original ring? Nov 12, 2013 at 14:58

In your case $\mathfrak m^n=\bigoplus_{i\ge n}R_i$ (in general $\mathfrak m^n\subseteq \bigoplus_{i\ge n}R_i$, but in this particular case they are equal) and therefore $\mathfrak m^n/\mathfrak m^{n+1}\simeq R_n$. As neilme pointed out this means that the associated graded ring is isomorphic to $R$.