Let $I$ be an ideal of $R$ and $M$ an $R$-module. When do we have the equality:
$\bigcap_{n \ge 1} (I^nM) = (\bigcap_{n \ge 1} I^n) M$
I was hoping "always" would be the answer, but this seems too good to be true, and I'm not quite sure how to prove it. I can certainly see some special cases where it's true, but I'm trying to find the most general conditions, maybe some constraint on $R$ if necessary.
Even for relatively "nice" $R$ and $M$, this equality can fail. For example, if $R = \mathbb{Z}$, $I = (m)$, $M = \mathbb{Z}/n\mathbb{Z}$, where $m$ and $n$ are relatively prime integers, the left side is all of $M$, while the right side is 0.
Some cases where equality holds: Assume $R$ is Noetherian and $M$ is finitely generated. Then by Krull's Intersection Theorem (see e.g. Eisenbud, Corollary 5.4), there exists an $r \in I$ such that $(1-r)(\bigcap_{n \ge 1}(I^nM)) = 0$. Now if $M$ is torsion-free, or if $R$ is a local ring and $I$ is the maximal ideal (which forces $1-r$ to be invertible) then $\bigcap_{n \ge 1}(I^nM) = 0$ and hence both sides are 0 (since the right side is clearly contained in the left).