What is wrong with this proof of Krull Intersection Theorem? Let $M$ be a module over a ring $R$ and $I$ an ideal. $I^\infty M= \bigcap_{k=1}^\infty I^kM$. Then $I \cdot I^\infty M=\bigcap_{k=2}^\infty I^k M.$ Clearly we have $\bigcap_{k=2}^\infty I^k M \supseteq \bigcap_{k=1}^\infty I^k M$. But $IM \supseteq I^2M$, so $\bigcap_{k=2}^\infty I^{k}M = \bigcap_{k=1}^\infty I^k M$, and therefore
$I \cdot I^\infty M = I^\infty M$.
This proof avoids Artin-Reese, which seems to be frequently used to prove the Krull intersection theorem. I'm not sure why you would avoid using this proof, but the textbooks seem to.
 A: Products don't necessarily commute with intersections.

So you can't claim $I \cdot I^\infty M=\bigcap_{k=2}^\infty I^k M$.

Note that in your proof attempt, you didn't specify any conditions on the ring $R$ or on the module $M$ (e.g., $R$ is noetherian, $M$ is finitely generated), and without at least some additional conditions, there are known counterexamples to the claim of the Krull Intersection Theorem.

For example, see the following thread . . .

$\qquad$https://mathoverflow.net/questions/71699

With regard to the issue of whether products commute with intersections . . .

In general, if 


*

*$R$ is a ring.$\\[4pt]$

*$I$ is an ideal of $R$.$\\[4pt]$

*$M$ is an $R$-module.$\\[4pt]$

*$S$ is a collection of sub-modules of $M$.


then we get
$$
I
\left(
\bigcap_{M\in S}M
\right)
\subseteq
\bigcap_{M\in S}IM 
$$
but the reverse inclusion need not hold.

In fact, products don't necessarily commute even with finite intersections.

As an example, let 


*

*$R=\mathbb{Z}$.$\\[4pt]$

*$I=(2)$.$\\[4pt]$

*$G$ be the finite abelian group $Z_4\times Z_6$.$\\[4pt]$

*$A$ be the $4$-element cyclic subgroup of $G$ generated by $(1,0)$.$\\[4pt]$

*$B$ be the $4$-element cyclic subgroup of $G$ generated by $(1,3)$.


Then 


*

*$A\cap B$ is the $2$-element cyclic subgroup of $G$ generated by $(2,0)$, hence $I(A \cap B)=0$.$\\[4pt]$

*$IA$ and $IB$ are both equal to the cyclic subgroup generated by $(2,0)$, hence
$IA\cap IB \ne 0$.


Thus, for this example, $I(A \cap B)$ is a proper subset of $IA \cap IB$.
