The classical Artin-Rees lemma tells the following. Let $R$ be a Noetherian commutative ring and $I\subset R$ be an ideal. Let $M$ be a finitely generated $R$-module and $N\subset M$ be a submodule. Then there exists an integer $m\ge0$ such that for all $n\ge0$ the following equality of two submodules in $N$ holds:
$$
 I^{n+m}M\cap N=I^n(I^mM\cap N).
$$
The usual proof is based on the Hilbert Basis Theorem: essentially, one uses the fact that the Rees ring $\bigoplus\_{n=0}^\infty I^n$ is graded Noetherian (since it is a quotient ring of the ring of polynomials over $R$ spanned by some finite set of generators of the ideal $I$).


I would like to specialize the Artin-Rees lemma to principal ideals $I=(s)\subset R$, and then extend it from multiplicative subsets of the form $\{1,s,s^2,s^3,\dotsc\}\subset R$ to other multiplicative subsets $S\subset R$.


So let $R$ be a Noetherian commutative ring and $S$ be a multiplicative subset in $R$. Assume for simplicity that $S$ is (at most) countable and all the elements of $S$ are regular (nonzero-divisors) in $R$.


Let $M$ be a finitely generated $R$-module and $N\subset M$ be a submodule. Does there exist an element $t\in S$ such that, for every $s\in S$, the equality of two submodules in $N$
$$
 stM\cap N=s(tM\cap N)
$$
holds?


Or, at least, can one find an element $t\in S$ such that the above equality holds for a cofinal subset of elements $s\in S$ (in the divisibility preorder)?


A straightforward attempt to argue similarly to the Hilbert Basis Theorem proof of the Artin-Rees lemma does not seem to work, as the ring $\bigoplus\_{s\in S}sR$ does not have to be graded Noetherian when the multiplicative set $S$ is not finitely generated.



Artin-Rees lemma for multiplicative subsets?
This is straightforward . Define the submodule $P$ as $N\subset P\subset M$, the set of all elements $m\in M$ such that $sm\in N$ for some $s\in S$. Then, choose $t\in S$ such that $tP\subset N$.


Now for what you need, clearly the right hand side is contained in the left. So, let $x\in stM\cap N$ for $s\in S$. Then, $x=stm$ and then, $m\in P$. Then $tm\in N$ by our choice of $t$. So, $tm\in tM\cap N$ and thus $x\in s(tM\cap N)$.


