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In this proof I know since R is noetherian it can be written as descending sequence of ideals which stabilizes after finite steps. Also I know since R is noetherian implies every ideal is finitely generated then I have some doubt;

1) Why $\cap \mathfrak m^n$ is an ideal of R and why $\mathfrak m\cdot \cap \mathfrak m^n=\cap \mathfrak m^n$ is unclear. Please help regarding this.

user26857
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maths student
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1 Answers1

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It should be clear that the intersection of ideals is an ideal, so that $\mathfrak{a}=\bigcap_n\mathfrak{m}^n$ is an ideal.

It really isn't obvious that $\mathfrak m\mathfrak a=\mathfrak a$. One way to prove this is to invoke the Artin-Rees lemma. A special case of this is that for any ideal $\mathfrak b$ then $\mathfrak m^{n+1}\cap\mathfrak b =\mathfrak m(\mathfrak m^n\cap\mathfrak b)$ for all large enough $n$. Taking $\mathfrak b=\mathfrak a$ gives $\mathfrak a=\mathfrak m\mathfrak a$.

Angina Seng
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  • A direct proof can be found in theorem 2.1 of this document – Kenny Lau Sep 23 '18 at 19:40
  • @Lord Shark can you please explain me proof of theorem 2.1 sent by kenny lu and give it as seperate answer if possible for you. It is not clear to me. – maths student Sep 23 '18 at 19:45
  • I think the method employed there is originally due to Herstein. A classic reference is Kaplansky's Commutative Rings, theorem 73. Both that presentation and the one given by @Kenny Lau are quite clear if you are willing to work through them. – Badam Baplan Sep 24 '18 at 00:30