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Suppose $A$ is a Noetherian ring, $M$ a finite $A$-module and $N \subset M$ a submodule. Furthermore, $\mathfrak{a} \subset A$ is an ideal.

Consider the $\mathfrak{a}$-adic topology on M, i.e. the filtration $... \mathfrak{a}^3M \subset \mathfrak{a}^2M \subset \mathfrak{a}M \subset M$ and the associated linear topology on $M$. Analogously, $N$ is topologized via the filtration $... \mathfrak{a}^3N \subset \mathfrak{a}^2N \subset \mathfrak{a}N \subset N$.

How can I show that the $\mathfrak{a}$-adic topology on $N$ is equivalent to the subspace topology of $N \subset M$?

I may use Artin-Rees, i.e.

$\exists n_o \in \mathbb{N}:\forall n > n_0: \mathfrak{a}^nM \cap N = \mathfrak{a}^{n-n_0}(\mathfrak{a}^{n_0}M \cap N)$

Thanks for any suggestions.

Edit: $A$ is commutative.

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    Indeed. It is Theorem 10.11 of "Introduction to Commutative Algebra" of Atiyah & Mac Donald. You have to find your way a little about stable filtrations of Modules. But it doesn´t seem to be too hard.2011-12-16

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This follows almost immediately from Artin--Rees. It may help to state Artin--Rees in the weaker form $\mathfrak a^n N \subset \mathfrak a^n M \cap N \subset \mathfrak a^{n-n_0} N.$