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Let $A$ be a noetherian local ring with maximal ideal $m$. One says that an $A$-module $X$ is discrete if for every $x\in X$, there is a natural number $n$ such that $m^n.x=0$.

My question is: Given an $A$-module $M$, a submodule $N\subseteq M$ and a submodule $P\subseteq N$ such that $N/P$ is discrete, is there a submodule $Q\subseteq M$ such that $M/Q$ is discrete and $Q\cap N\subseteq P$?

In fact, the title of my question is the same as the title of §7.2 of Kato's paper "A generalization of local class field theory by using K-groups. I".

In that paper, the standard topology on an $A$-module $M$ is the group topology defined by taking as a base of open neighborhoods of $0\in M$ all $A$-submodules $N$ such that $M/N$ are discrete.

Kato then wrote :

"If $N$ is an $A$-submodule of $M$, the standard topology on $N$ over $A$ coincides with the restriction of the standard topology of $M$ over $A$."

I think this statement is equivalent to a positive answer to my question above. It may be a quite elementary argument but I'm just stuck. So, could anyone give some help?

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  • $\begingroup$ This looks related to the Artin-Rees lemma which should give both Kato's and your statement for finitely generated modules. Maybe one can somehow reduce to this case. $\endgroup$ Mar 10, 2014 at 12:52
  • $\begingroup$ @TorstenSchoeneberg: Thank you very much, Torsten! This is perhaps why A is assumed noetherian and really solves the finitely generated case. I'm still looking for a solution in the general case. $\endgroup$
    – Yong Hu
    Mar 10, 2014 at 13:57
  • $\begingroup$ But in general, isn't $A = \mathbb{Z}_p$, $M = \mathbb{Q}_p$, $N = \mathbb{Z}_p$ already a counterexample? Also, I cannot find the statement or a §7 in the paper you mention. $\endgroup$ Mar 11, 2014 at 9:12
  • $\begingroup$ Hi, Torsten. The paper of Kato I mentioned is the part I of his series of 3 articles on local class field theory. It is published in Journal of the Faculty of Science. University of Tokyo., Sec.IA, volume 26 (1979), no. 2, 303–376. The statement I cited can be found on page 360, as property (d). $\endgroup$
    – Yong Hu
    Mar 11, 2014 at 10:22
  • $\begingroup$ Hi again, Torsten. In your example, I think we can always take Q=P, for any submodule P of N with N/P discrete. $\endgroup$
    – Yong Hu
    Mar 11, 2014 at 10:32

1 Answer 1

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The assertion is indeed true in a more general context. My references are B. Stenström: Rings of Quotients, and Lam: Lectures on Modules and Rings.

Let a (right) linear topology on a ring $A$ be given by a family of (right) ideals $\mathfrak{F}$ (i.e. $\mathfrak{F}$ consists of all open right ideals of $A$). On any right $A$-module $M$ define the $\mathfrak{F}$-topology to be the one where a submodule $L \subseteq M$ is open iff for every $x \in M$ there is $I \in \mathfrak{F}$ such that $x I \subseteq L$.

In our setting, we have $\mathfrak{F} = \{ \text{ideals which contain some } \mathfrak{m}^n , n \in \mathbb{N} \}$.

Stenström Ch. VI Proposition 7.1 (due to Gabriel) gives equivalent criteria for a topology (given as $\mathfrak{F}$ as above) to satisfy what we want, namely:

$(*)$ Given a submodule $N \subseteq M$, the $\mathfrak{F}$-topology on $N$ is the subspace topology of the $\mathfrak{F}$-topology on $M$.

In Stenström, an $\mathfrak{F}$ with this property is called stable. One of the equivalent criteria is

$(**)$ If a module $M$ is $\mathfrak{F}$-discrete, then so is its injective hull $E(M)$.

To see that $(**) \Rightarrow (*)$, take like you did $P \subseteq N$ such that $N/P$ is $\mathfrak{F}$-discrete; we have to show that there is a submodule $Q \subseteq M$ such that $M/Q$ is $\mathfrak{F}$-discrete and $Q \cap N \subseteq P$. Indeed, with Zorn's lemma take a submodule $Q \subseteq M$ maximal with respect to the property $Q \cap N = P$. Now by maximality of $Q$ one can see that $M/Q$ is an essential extension of $(N + Q)/Q$, which in turn is $\mathfrak{F}$-discrete because it is isomorphic to $N/P$. So $(**)$ implies that $M/Q$ is $\mathfrak{F}$-discrete (because as essential extension, it "is" a submodule of the injective hull, and we obviously have that submodules of discrete modules are discrete).

Now Stenström Ch. VII Theorem 4.4 and Proposition 4.5 show that if our ground ring $A$ is commutative and Noetherian, indeed every Gabriel topology on $A$ is stable. (A Gabriel topology is a topology $\mathfrak{F}$ as above which additionally satisfies: Given any right ideal $\mathfrak{a}$, if there is $\mathfrak{b} \in \mathfrak{F}$ such that $(\mathfrak{a} : b) \in \mathfrak{F}$ for all $b \in \mathfrak{b}$, then $\mathfrak{a} \in \mathfrak{F}$. Using finite generation of the $\mathfrak{m}^n$, our $\mathfrak{F}$ is a Gabriel topology.)

To show that we indeed have $(**)$, one can use e.g. that for every indecompasable injective module $E$ with associated prime $\mathfrak{p}$, and every $x \in E$, there is a natural number $n$ such that $x \mathfrak{p}^n = 0 $. (A fact due to Noether herself, see Lam, Theorem 3.78). The associated primes of a module and any of its essential extensions (in particular the injective hull) coincide, and the associated primes of an $\mathfrak{F}$-discrete module are easily seen to be in $\mathfrak{F}$. We can conclude by writing our injective hull in question as direct sum of indecomposables.

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