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

  1. Let . Then,
  2. There exists such that

Results used

Applications

Proof

The intersection equals its product with

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

Let:

Now consider the filtration:

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

Since each contains , the filtration below is the same as the filtration:

This being -adic forces that .

Finding the element

Since , we can find an element such that . This is an application of the Cayley-Hamilton theorem: we first find the Cayley-Hamilton polynomial, then observe that is a root of the polynomial, and then take the negative of the sum of all coefficients of higher degree terms.

References

Textbook references