# Krull intersection theorem for modules

This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
View more results involving Noetherianness or Read a survey article on applying Noetherianness
This fact is an application of the following pivotal fact/result/idea: Artin-Rees lemma
View other applications of Artin-Rees lemma OR Read a survey article on applying Artin-Rees lemma
This fact is an application of the following pivotal fact/result/idea: Cayley-Hamilton theorem
View other applications of Cayley-Hamilton theorem OR Read a survey article on applying Cayley-Hamilton theorem

## Statement

Let $R$ be a Noetherian ring and $I$ be an ideal inside $R$. Suppose $M$ is a finitely generated module over $R$. Then, we have the following:

1. Let $N = \bigcap_{j=1}^\infty I^j M$. Then, $IN = N$
2. There exists $r \in I$ such that $(1 - r)N = 0$

## Proof

### The intersection equals its product with $I$

We first show that the intersection equals its product with $I$. This is the step where we se the Artin-Rees lemma.

Let:

$N := \bigcap_1^\infty I^jM$

Now consider the filtration:

$M \supset IM \supset I^2M \supset \ldots$

this is an $I$-adic filtration and the underlying ring is Noetherian, hence by the Artin-Rees lemma, the following filtration is also $I$-adic:

$N \supset IM \cap N \supset I^2M \cap N\supset \ldots$

Since each $I^jM$ contains $N$, the filtration below is the same as the filtration:

$N \supset N \supset N \supset \ldots$

This being $I$-adic forces that $IN = N$.

### Finding the element $r$

Since $IN = N$, we can find an element $r \in I$ such that $(1 - r)N = 0$. This is an application of the Cayley-Hamilton theorem: we first find the Cayley-Hamilton polynomial, then observe that $1$ is a root of the polynomial, and then take the negative of the sum of all coefficients of higher degree terms.