Let $A=k[x_1, \ldots, x_n]$ be the polynomial ring generated by $x_1, \ldots, x_n$. Let $I$ be an ideal of $k[x_1, \ldots, x_n]$ (it is possible that $I$ is not homogeneous).

The algebra $A$ is a graded algebra: $A = \oplus_{i \ge 0} A_i$, where $A_i$ consists of degree $i$ homogeneous polynomials.

The algebra $A/I=k[x_1, \ldots, x_n]/I$ is a filtered algebra with the filtration $F_i(A/I) = (F_i(A)+I)/I$, where $F_i(A)=\oplus_{j \le i} A_j$.

Is associated graded algebra $$\mathrm{gr}(k[x_1, \ldots, x_n]/I)=\mathrm{gr}(A/I)=\oplus_{i \ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) \cap I)$$ isomorphic as a vector space to $k[x_1, \ldots, x_n]/I$?

Thank you very much.


1 Answer 1



Observe that $\mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(\mathrm{gr}(A/I)) = \oplus_{i=0}^r~\mathrm{gr}(A/I)_i$.

Further observe that $$\dim F_i(\mathrm{gr}(A/I) = \sum_{i=0}^r \dim \mathrm{gr}(A/I)_i = \sum_{i=0}^r (\dim F_i(A/I) - \dim F_{i-1}(A/I)) = \dim F_i(A/I).$$

Therefore $\mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.


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