[proofplan] For part (1), we exhibit $G_{\mathfrak{a}}(R)$ as a finitely generated algebra over the noetherian ring $R / \mathfrak{a}$: the generators of $\mathfrak{a}$ project to degree-one elements whose monomials span every graded component, so [Hilbert's Basis Theorem](/theorems/2904) applies. For part (2), the stability condition provides an index $n_0$ after which $M_{n_0 + r} = \mathfrak{a}^r M_{n_0}$, and we show $G(M)$ is generated as a $G_{\mathfrak{a}}(R)$-module by the finite direct sum $\bigoplus_{n \leq n_0} M_n / M_{n+1}$, which is a finitely generated $R / \mathfrak{a}$-module because each $M_n$ is finitely generated over the noetherian ring $R$. [/proofplan] [step:Show $G_{\mathfrak{a}}(R)$ is a finitely generated $(R / \mathfrak{a})$-algebra] Since $R$ is noetherian, the ideal $\mathfrak{a}$ is finitely generated: write $\mathfrak{a} = (x_1, \dots, x_s)$ for elements $x_1, \dots, x_s \in \mathfrak{a}$. For each $i$, let $\bar{x}_i$ denote the image of $x_i$ in the degree-one component $\mathfrak{a} / \mathfrak{a}^2$ of the associated graded ring $G_{\mathfrak{a}}(R) = \bigoplus_{n=0}^{\infty} \mathfrak{a}^n / \mathfrak{a}^{n+1}$. We claim that $\bar{x}_1, \dots, \bar{x}_s$ generate $G_{\mathfrak{a}}(R)$ as an $(R / \mathfrak{a})$-algebra. It suffices to show that for each $n \geq 1$, the degree-$n$ component $\mathfrak{a}^n / \mathfrak{a}^{n+1}$ is spanned over $R / \mathfrak{a}$ by monomials $\bar{x}_{i_1} \cdots \bar{x}_{i_n}$ with $1 \leq i_1, \dots, i_n \leq s$. Since $\mathfrak{a} = (x_1, \dots, x_s)$, the power $\mathfrak{a}^n$ is generated as an $R$-module by the products $x_{i_1} \cdots x_{i_n}$. The image of $r \cdot x_{i_1} \cdots x_{i_n}$ (with $r \in R$) in $\mathfrak{a}^n / \mathfrak{a}^{n+1}$ equals $(r + \mathfrak{a}) \cdot \bar{x}_{i_1} \cdots \bar{x}_{i_n}$, since $\mathfrak{a}$ annihilates $\mathfrak{a}^n / \mathfrak{a}^{n+1}$ (any element of $\mathfrak{a} \cdot \mathfrak{a}^n$ lies in $\mathfrak{a}^{n+1}$ and hence vanishes in the quotient). Therefore $\mathfrak{a}^n / \mathfrak{a}^{n+1}$ is spanned over $R / \mathfrak{a}$ by these monomials. [guided] The associated graded ring $G_{\mathfrak{a}}(R) = \bigoplus_{n \geq 0} \mathfrak{a}^n / \mathfrak{a}^{n+1}$ has degree-zero component $R / \mathfrak{a}$, and the multiplication is induced by the product of representatives: for $a + \mathfrak{a}^{m+1} \in \mathfrak{a}^m / \mathfrak{a}^{m+1}$ and $b + \mathfrak{a}^{n+1} \in \mathfrak{a}^n / \mathfrak{a}^{n+1}$, the product is $ab + \mathfrak{a}^{m+n+1} \in \mathfrak{a}^{m+n} / \mathfrak{a}^{m+n+1}$. Why do the images $\bar{x}_1, \dots, \bar{x}_s$ of the generators of $\mathfrak{a}$ generate the entire graded ring? Because products of generators of $\mathfrak{a}$ give generators of $\mathfrak{a}^n$: since $\mathfrak{a} = (x_1, \dots, x_s)$, the $n$-th power $\mathfrak{a}^n$ is the $R$-module generated by all products $x_{i_1} \cdots x_{i_n}$. Passing to $\mathfrak{a}^n / \mathfrak{a}^{n+1}$, the $R$-module structure factors through $R / \mathfrak{a}$ since $\mathfrak{a} \cdot (\mathfrak{a}^n / \mathfrak{a}^{n+1}) = 0$. So $\mathfrak{a}^n / \mathfrak{a}^{n+1}$ is spanned as an $(R / \mathfrak{a})$-module by the monomials $\bar{x}_{i_1} \cdots \bar{x}_{i_n}$, which are exactly the degree-$n$ monomials in $\bar{x}_1, \dots, \bar{x}_s$. [/guided] [/step] [step:Conclude that $G_{\mathfrak{a}}(R)$ is noetherian via Hilbert's Basis Theorem] The ring $R / \mathfrak{a}$ is noetherian (it is a quotient of the noetherian ring $R$). Since $G_{\mathfrak{a}}(R)$ is a finitely generated $(R / \mathfrak{a})$-algebra (generated by $\bar{x}_1, \dots, \bar{x}_s$), [Hilbert's Basis Theorem](/theorems/2904) implies that $G_{\mathfrak{a}}(R)$ is noetherian. This completes the proof of part (1). [/step] [step:Identify the stability index $n_0$ and the candidate generating set for $G(M)$] For part (2), let $M$ be a finitely generated $R$-module and let $(M_n)_{n \geq 0}$ be a stable $\mathfrak{a}$-filtration of $M$. By stability, there exists $n_0 \geq 0$ such that $\mathfrak{a} M_n = M_{n+1}$ for all $n \geq n_0$. Iterating gives $M_{n_0 + r} = \mathfrak{a}^r M_{n_0}$ for all $r \geq 0$. The associated graded module is $G(M) = \bigoplus_{n=0}^{\infty} M_n / M_{n+1}$, a graded $G_{\mathfrak{a}}(R)$-module where the action of $a + \mathfrak{a}^{m+1} \in \mathfrak{a}^m / \mathfrak{a}^{m+1}$ on $x + M_{n+1} \in M_n / M_{n+1}$ is $ax + M_{m+n+1} \in M_{m+n} / M_{m+n+1}$. This is well-defined since $\mathfrak{a}^m M_n \subseteq M_{m+n}$ (by the filtration property, iterated $m$ times). Define the finite direct sum $Q = \bigoplus_{n=0}^{n_0} M_n / M_{n+1}$. We claim $Q$ generates $G(M)$ as a $G_{\mathfrak{a}}(R)$-module. [/step] [step:Show $Q$ generates $G(M)$ using the stability condition] It suffices to show that for all $n > n_0$, every element of $M_n / M_{n+1}$ lies in the $G_{\mathfrak{a}}(R)$-submodule generated by $Q$. Write $n = n_0 + r$ with $r \geq 1$. By stability, $M_n = \mathfrak{a}^r M_{n_0}$, so any element of $M_n$ is a finite sum of elements of the form $a_1 \cdots a_r \cdot m$ with $a_1, \dots, a_r \in \mathfrak{a}$ and $m \in M_{n_0}$. The image of such an element in $M_n / M_{n+1}$ equals the product $\bar{a}_1 \cdots \bar{a}_r \cdot \bar{m}$ in the graded module, where $\bar{a}_j \in \mathfrak{a} / \mathfrak{a}^2$ and $\bar{m} \in M_{n_0} / M_{n_0 + 1}$. Since $\bar{m} \in Q$ and $\bar{a}_j \in G_{\mathfrak{a}}(R)$, this product lies in the $G_{\mathfrak{a}}(R)$-submodule generated by $Q$. Since the components for $n \leq n_0$ are already contained in $Q$ and the components for $n > n_0$ are covered by the argument above, $Q$ generates $G(M)$ over $G_{\mathfrak{a}}(R)$. [guided] The stability condition $M_{n_0+r} = \mathfrak{a}^r M_{n_0}$ is the engine of this step. It asserts that beyond degree $n_0$, the filtration is determined by repeated multiplication by $\mathfrak{a}$. In terms of the associated graded module, this means the component $M_{n_0+r} / M_{n_0+r+1}$ is generated over $\mathfrak{a}^r / \mathfrak{a}^{r+1}$ (the degree-$r$ part of $G_{\mathfrak{a}}(R)$) by elements of $M_{n_0} / M_{n_0+1}$. To see why the product formula is correct: for $a_1 \cdots a_r \cdot m \in \mathfrak{a}^r M_{n_0} = M_{n_0+r}$, the image in $M_{n_0+r} / M_{n_0+r+1}$ is the product of $\bar{a}_1, \dots, \bar{a}_r$ and $\bar{m}$ in $G(M)$. This uses the fact that the graded module multiplication is compatible with the ring multiplication: the degree-$1$ elements $\bar{a}_j$ of $G_{\mathfrak{a}}(R)$ act on the degree-$n_0$ element $\bar{m}$ of $G(M)$ to produce a degree-$(n_0 + r)$ element of $G(M)$. Without stability, we would have no control over the graded components beyond $n_0$, and $G(M)$ need not be finitely generated. [/guided] [/step] [step:Conclude that $G(M)$ is finitely generated over $G_{\mathfrak{a}}(R)$] It remains to show that $Q = \bigoplus_{n=0}^{n_0} M_n / M_{n+1}$ is finitely generated as an $(R / \mathfrak{a})$-module. Since $M$ is a finitely generated module over the noetherian ring $R$, the module $M$ is noetherian by [Modules over Noetherian Rings](/theorems/2903). Each $M_n$ is a submodule of the noetherian module $M$, hence finitely generated as an $R$-module. Each quotient $M_n / M_{n+1}$ is therefore a finitely generated $R$-module. Since $\mathfrak{a} M_n \subseteq M_{n+1}$ (by the filtration property), the ideal $\mathfrak{a}$ annihilates $M_n / M_{n+1}$, so the $R$-module structure factors through $R / \mathfrak{a}$. Hence each $M_n / M_{n+1}$ is a finitely generated $(R / \mathfrak{a})$-module. The finite direct sum $Q = \bigoplus_{n=0}^{n_0} M_n / M_{n+1}$ is therefore a finitely generated $(R / \mathfrak{a})$-module. If $S$ is a finite generating set for $Q$ over $R / \mathfrak{a}$, then $S$ generates $G(M)$ as a $G_{\mathfrak{a}}(R)$-module (since $Q$ generates $G(M)$ over $G_{\mathfrak{a}}(R)$ and $R / \mathfrak{a} \subseteq G_{\mathfrak{a}}(R)$ as the degree-zero component). Hence $G(M)$ is a finitely generated graded $G_{\mathfrak{a}}(R)$-module. [/step]