What is wrong with this proof of Krull Intersection Theorem?

Question

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.

Answer 1

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 $IA \cap IB$.