Krull intersection theorem for modules: Difference between revisions

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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=j=1IjM. Then, IN=N
  2. There exists rI such that (1r)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:=1IjM

Now consider the filtration:

MIMI2M

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:

NIMNI2MN

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

NNN

This being I-adic forces that IN=N.

Finding the element r

Since IN=N, we can find an element rI such that (1r)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