Krull intersection theorem for modules

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 $$(1 - r)N = 0$$

Results used

 * Artin-Rees lemma
 * Cayley-Hamilton theorem

Applications

 * Krull intersection theorem for Jacobson radical, also covers the case of a local ring
 * Krull intersection theorem for domains

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.

Textbook references

 * , Page 152