# Artin-Rees lemma for multiplicative subsets?

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.

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 $$x\in s(tM\cap N)$$.