Tensor product, Artin-Rees lemma and Krull intersection theorem I asked another question about tensor product, but can't conclude from the answer, so here is another more concrete question.

Let $(A,m)$ be a local ring then by Artin-Rees Lemma $m^k \bigcap I \subset m^sI$ for $s$ fixed and $k$ big enough. So now, is it true that for any module $M$ over $A$ we have
$$(m^k \cap I) \otimes_A M \subset m^s(I\otimes_A M)$$
for $s$ fixed and $k$ big enough?

 A: When I arrived here, this question is 5 years old. The way it was stated suggests that the author of the question had read the proof of Theorem 4.2 in Section 15.4.1, the CRing Project (this is also the reason I found this question). I have to find the answer on my own, and I'd like to write an answer here, for anyone comes to this place with the same problem in the future.
The statement of Theorem 4.2 is as follow:

Let $(R,\mathfrak{m}), (S,\mathfrak{n})$ be noetherian local rings, $S$ is an $R$-algebra (the morphism $R\to S$ is local, i.e. $\mathfrak{m}S\subset \mathfrak{n}$). Then a finitely generated $S$-module $M$ is flat over $R$ if and only if
$$\text{Tor}_1 (k,M)=0$$
where $k=R/\mathfrak{m}$.  In this case, $M$ is even free.

In the proof, the authors proved that the map $I\otimes M\to M$ is injective for any ideal $I$ of $R$ (so $M$ is flat). They did that by showing the kernel $K$ of $I\otimes M\to M$ belongs to the image of the maps
$$(\mathfrak{m}^t\cap I)\otimes M\to I\otimes M$$
for all $t$. They claimed that $K$ belongs to $\mathfrak{m}^t(I\otimes M)$, which happened if for all $s$, there exists $k$ such that

$$\text{im}\left((\mathfrak{m}^k\cap I)\otimes M\right)\subseteq \mathfrak{m}^s(I\otimes M)$$

If this is true, then the proof is done by Krull's Intersection Theorem, and I think this is what OP had in mind.
The solution is to look at the simple tensors
$$\begin{aligned}
(\mathfrak{m}^t\cap I)\otimes M &\to I\otimes M\\
r\otimes m &\mapsto r\otimes m
\end{aligned}$$
By Artin-Rees lemma, for all $s$ there exists $k$ such that $\mathfrak{m}^{k}\subseteq \mathfrak{m}^sI$. Hence if $r\in \mathfrak{m}^{k}\cap I$, there exists $r'\in \mathfrak{m}^s,a\in I$ such that $r=r'a$. It follows that $\text{im}(r\otimes m)=r\otimes m=(r'a)\otimes m\in \mathfrak{m}^s(I\otimes M)$. Now we have
$$K\subseteq \text{im}\left((\mathfrak{m}^k \cap I)\otimes M\right)\subseteq \mathfrak{m}^s(I\otimes M)$$
for all $s$.
