Exercise 5, § 2

Let $A$ be a filtered commutative ring with filtration $(a)_m$ such that $a_0 = A$. (a) For the ring $\operatorname{gr}(A)$ to be generated by a family of elements whose degrees are bounded, it is necessary and sufficient that there exist an integer $q$ such that, for all $n$, $a_n = a_{n+1} + b_n$ where $b_n$ is the sum of the ideals $a_1^{\alpha_1} a_2^{\alpha_2} \ldots a_q^{\alpha_q}$ for all systems of integers $\alpha_i \geq 0$ ($1 \leq i \leq q$) such that $\sum_{i=1}^q i \alpha_i = n$. Then $a_n = a_{n+k} + b_n$ for all $k > 0$.

(b) For $\mathrm{gr}(A)$ to be Noetherian, it is necessary and sufficient that $A/a_1$ be Noetherian, that the condition of (a) hold and that for $i \leq q$ the $a_i/a_{i+1}$ be finitely generated $(A/a_1)$-modules (use Corollary 4 to Theorem 2 of no. 10).

(c) Let $K$ be a commutative field and $A$ the polynomial ring in two indeterminates $K[X, Y]$. A total ordering is defined on the set of monomials $X^m Y^n$ by setting $X^m Y^n \leq X^p Y^q$ if $m + n \leq p + q$ or if $m + n = p + q$ and $m \leq p$. Let $(M_k)$ be the sequence of monomials in $X, Y$ thus arranged in ascending order and let $a_n$ be the ideal of $A$ generated by the $M_k$ of index $k \geq n$. Show that (a) is a filtration compatible with the ring structure on $A$; with this filtration $\mathrm{gr}(A)$ admits divisors of 0 and is not Noetherian, although $A$ is a Noetherian integral domain (use the criterion in (b)); in particular, $\mathrm{gr}(a_1)$ (with the filtration induced on $a_1$ by that on $A$) is not a finitely generated $\mathrm{gr}(A)$-module, although $a_n$ is a finitely generated $A$-module (cf. § 1, no. 2, Corollary to Proposition 1).