Krull intersection theorem for modules: Difference between revisions

Revision as of 18:59, 3 March 2008

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

Template:Aplpicationof

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:

Let $N=\bigcap _{j=1}^{\infty }I^{j}M$ . Then, $IN=N$
There exists $r\in I$ such that $(1-r)N=0$

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 $I$

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.

References

''Dimensionstheorie in Stellenringen by Wolfgang Krull, 1938

Textbook references

Book:Eisenbud, Page 152

@@ Line 1: / Line 1: @@
 {{Noetherian ring result}}
 {{applicationof|Artin-Rees lemma}}
-{{applicationof|Nakayama's lemma}}
+{{aplpicationof|Cayley-Hamilton theorem}}
 ==Statement==
@@ Line 12: / Line 12: @@
 * [[Artin-Rees lemma]]
-* [[Nakayama's lemma]]
+* [[Cayley-Hamilton theorem]]
+==Applications==
+* [[Krull intersection theorem for Jacobson radical]], also covers the case of a [[local ring]]
+* [[Krull intersection theorem for domains]]
 ==Proof==
@@ Line 42: / Line 47: @@
 Since <math>IN = N</math>, we can find an element <math>r \in I</math> such that <math>(1 - r)N = 0</math>. This is an application of the [[Cayley-Hamilton theorem]]: we first find the Cayley-Hamilton polynomial, then observe that <math>1</math> 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 <math>R</math> is an integral domain, and we set <math>M = R</math> in the above, we find an element <math>r \in I</math> such that <math>(1 - r)(\bigcap_j I^j) = 0</math>. Thus <math>1 - r</math> is a zero divisor on the intersection. Since <math>R</math> is an [[integral domain]], one of these must hold:
-* <math>1 - r = 0</math>. This forces <math>1 \in I</math>, contradicting the assumption that <math>I</math> is a [[proper ideal]]
-* <math>\bigcap_j I^j = 0</math>
-This completes the proof.
-===Proving that the intersection is trivial for a local ring===
-In this case, <math>I</math> is contained inside the unique maximal ideal <math>\mathfrak{m}</math>, which is the Jacobson radical. As above, let <math>N = \bigcap_j I^j</math>. Then, we have <math>IN = N</math>. Since <math>I</math> is contained in the [[Jacobson radical]], this forces <math>N = 0</math>.
 ==References==