Krull intersection theorem for modules

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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

Results used

Applications

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.

References

Textbook references