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: Nakayama's lemma
View other applications of Nakayama's lemma OR Read a survey article on applying Nakayama's lemma

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
• Nakayama's lemma

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:

$N:=\bigcap _{1}^{\infty }I^{j}M$

Now consider the filtration:

$M\supset IM\supset I^{2}M\supset \ldots$

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:

$N\supset IM\cap N\supset I^{2}M\cap N\supset \ldots$

Since each $I^{j}M$ 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.

Proving the intersection is trivial for an integral domain

If $R$ is an integral domain, and we set $M=R$ in the above, we find an element $r\in I$ such that $(1-r)(\bigcap _{j}I^{j})=0$. Thus $1-r$ is a zero divisor on the intersection. Since $R$ is an integral domain, one of these must hold:

• $1-r=0$. This forces $1\in I$, contradicting the assumption that $I$ is a proper ideal
• $\bigcap _{j}I^{j}=0$

This completes the proof.

Proving that the intersection is trivial for a local ring

In this case, $I$ is contained inside the unique maximal ideal ${\mathfrak {m}}$, which is the Jacobson radical. As above, let $N=\bigcap _{j}I^{j}$. Then, we have $IN=N$. Since $I$ is contained in the Jacobson radical, this forces $N=0$.

References

• ''Dimensionstheorie in Stellenringen by Wolfgang Krull, 1938

Textbook references

• Book:Eisenbud, Page 152