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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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I've seen a theorem in Commutative Algebra from Atiyah and Mac Donald in chapter 10 which proves this. I'm not completely sure what the proof is, but you should try to prove something that of $(M_n)$ and $(N_n)$ are stable $\mathfrak{a}$-filtrations of $M$ then they have bounded differences, i.e. there exists $n_0 > 0$ such that $M_{n_0 + n} \subseteq N_m$ and $N_{n_0 + n} \subseteq M_n$. Then apply this to the sequences $(\mathfrak{a}^nN)$ and $(\mathfrak{a}^nM \cap N)$. Hope this helps. –  Noud Dec 16 '11 at 13:53
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. –  Giovanni De Gaetano Dec 16 '11 at 16:06

1 Answer 1

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

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