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
Contents
Statement
Let be a Noetherian ring and
be an ideal inside
. Suppose
is a finitely generated module over
. Then, we have the following:
- Let
. Then,
- There exists
such that
Results used
Applications
- Krull intersection theorem for Jacobson radical, also covers the case of a local ring
- Krull intersection theorem for domains
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
- ''Dimensionstheorie in Stellenringen by Wolfgang Krull, 1938
Textbook references
- Book:Eisenbud, Page 152