[proofplan] We first verify that $(N \cap M_n)_{n \geq 0}$ is an $\mathfrak{a}$-filtration of $N$. We then form the Rees module $N^* = \bigoplus_{n \geq 0} (N \cap M_n)$ inside $M^* = \bigoplus_{n \geq 0} M_n$ and argue that $N^*$ is finitely generated over $R^*$. The key chain of reasoning is: $R$ noetherian implies $R^*$ noetherian; $(M_n)$ stable implies $M^*$ is finitely generated over $R^*$ by [Stability and Finite Generation of $M^*$](/theorems/2949); finitely generated over noetherian is noetherian; $N^*$ is a submodule of the noetherian module $M^*$, hence finitely generated; finite generation of $N^*$ implies stability of $(N \cap M_n)$ by the reverse direction of the same theorem. [/proofplan] [step:Verify that $(N \cap M_n)_{n \geq 0}$ is an $\mathfrak{a}$-filtration of $N$] We must check that $\mathfrak{a}(N \cap M_n) \subseteq N \cap M_{n+1}$ for all $n \geq 0$. Let $a \in \mathfrak{a}$ and $x \in N \cap M_n$. Then $ax \in \mathfrak{a} N \subseteq N$ (since $N$ is a submodule of $M$, it is closed under scalar multiplication) and $ax \in \mathfrak{a} M_n \subseteq M_{n+1}$ (since $(M_n)$ is an $\mathfrak{a}$-filtration). Therefore $ax \in N \cap M_{n+1}$, confirming \begin{align*} \mathfrak{a}(N \cap M_n) \subseteq N \cap M_{n+1} \quad \text{for all } n \geq 0. \end{align*} Moreover, $N \cap M_0 = N \cap M = N$ since $M_0 = M$, so $(N \cap M_n)_{n \geq 0}$ is a decreasing filtration of $N$ starting at $N$. [/step] [step:Form the Rees modules and show $N^*$ is a submodule of $M^*$] Define the Rees module of the filtration $(M_n)$: \begin{align*} M^* = \bigoplus_{n \geq 0} M_n, \end{align*} which is a graded module over the Rees algebra $R^* = \bigoplus_{n \geq 0} \mathfrak{a}^n$. Similarly, define \begin{align*} N^* = \bigoplus_{n \geq 0} (N \cap M_n). \end{align*} Since $N \cap M_n \subseteq M_n$ for all $n$, we have $N^* \subseteq M^*$ as graded abelian groups. The filtration property $\mathfrak{a}(N \cap M_n) \subseteq N \cap M_{n+1}$ established in the previous step ensures that $N^*$ is closed under the $R^*$-action, so $N^*$ is a graded $R^*$-submodule of $M^*$. [/step] [step:Show that $R^*$ is noetherian and $M^*$ is a noetherian $R^*$-module] Since $R$ is noetherian, the ideal $\mathfrak{a}$ is finitely generated, say $\mathfrak{a} = (a_1, \dots, a_s)$. The Rees algebra $R^*$ is generated as an $R$-algebra by $a_1, \dots, a_s$ in degree one, giving a surjective $R$-algebra homomorphism $R[T_1, \dots, T_s] \to R^*$ defined by $T_i \mapsto a_i$. By [Hilbert's Basis Theorem](/theorems/2904), $R[T_1, \dots, T_s]$ is noetherian, and a quotient of a noetherian ring is noetherian. Therefore $R^*$ is noetherian. Since the filtration $(M_n)$ is stable, [Stability and Finite Generation of $M^*$](/theorems/2949) (implication $(2) \Rightarrow (1)$) gives that $M^*$ is a finitely generated $R^*$-module. By [Modules over Noetherian Rings](/theorems/2903), $M^*$ is a noetherian $R^*$-module. [guided] We need to establish the noetherianity of $M^*$ as an $R^*$-module. This requires two ingredients: $R^*$ must be noetherian, and $M^*$ must be finitely generated over $R^*$. For the first: since $R$ is noetherian, $\mathfrak{a}$ is finitely generated, say $\mathfrak{a} = (a_1, \dots, a_s)$. The Rees algebra $R^* = \bigoplus_{n \geq 0} \mathfrak{a}^n$ is generated as an $R$-algebra by $a_1, \dots, a_s$ placed in degree one. This gives a surjection $R[T_1, \dots, T_s] \twoheadrightarrow R^*$ via $T_i \mapsto a_i$. Since $R$ is noetherian, [Hilbert's Basis Theorem](/theorems/2904) gives $R[T_1, \dots, T_s]$ noetherian, and any quotient of a noetherian ring is noetherian, so $R^*$ is noetherian. For the second: the filtration $(M_n)$ is stable by hypothesis. By [Stability and Finite Generation of $M^*$](/theorems/2949), specifically the implication "stable $\Rightarrow$ finitely generated," $M^*$ is a finitely generated $R^*$-module. Combining: $M^*$ is finitely generated over the noetherian ring $R^*$, so by [Modules over Noetherian Rings](/theorems/2903), $M^*$ is a noetherian $R^*$-module. This means every submodule of $M^*$ is finitely generated over $R^*$. [/guided] [/step] [step:Conclude that $N^*$ is finitely generated, hence $(N \cap M_n)$ is stable] Since $M^*$ is a noetherian $R^*$-module and $N^* \subseteq M^*$ is a submodule, $N^*$ is finitely generated over $R^*$. Now apply [Stability and Finite Generation of $M^*$](/theorems/2949) in the reverse direction (implication $(1) \Rightarrow (2)$), applied to the filtration $(N \cap M_n)_{n \geq 0}$ of $N$ with Rees module $N^*$. Since $N^*$ is a finitely generated $R^*$-module, the filtration $(N \cap M_n)_{n \geq 0}$ is stable. This completes the proof. [guided] We have established that $N^*$ is a submodule of the noetherian $R^*$-module $M^*$. Every submodule of a noetherian module is finitely generated, so $N^*$ is a finitely generated $R^*$-module. The final step is to convert the finite generation of $N^*$ back into stability of the filtration. We apply [Stability and Finite Generation of $M^*$](/theorems/2949) to the module $N$ with the $\mathfrak{a}$-filtration $(N \cap M_n)_{n \geq 0}$. The Rees module of this filtration is precisely $N^* = \bigoplus_{n \geq 0} (N \cap M_n)$. The theorem states: $(N \cap M_n)$ is stable if and only if $N^*$ is finitely generated over $R^*$. Since we have just shown $N^*$ is finitely generated, the filtration $(N \cap M_n)_{n \geq 0}$ is stable. Note the logical structure: we used the "stable $\Rightarrow$ finitely generated" direction for $M^*$ (to get noetherianity), and the "finitely generated $\Rightarrow$ stable" direction for $N^*$ (to get the conclusion). The hypothesis that $R$ is noetherian was essential for making $R^*$ noetherian, which was the bridge allowing us to pass from $M^*$ finitely generated to $M^*$ noetherian. [/guided] [/step]