# Why is the Artin-Rees lemma used here?

I am currently engaged in independent study of algebraic geometry, using Dan Bump's book. One of the exercises in it outlines a proof of the Krull Intersection Theorem, which [here] is the following:

Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$, and let $M$ be the intersection of all of the $\mathfrak{m}^n$. Then $M = 0$.

The hints direct me to use the Artin-Rees lemma to show that $\mathfrak{m} M = M$, then use Nakayama's lemma to show that $M = 0$ (this second step is easy). I showed this to a professor and he accused the book of using big machinery for no reason, arguing that

$$\mathfrak{m} M = \mathfrak{m} \bigcap_{n \ge 0} \mathfrak{m}^n = \bigcap_{n \ge 1} \mathfrak{m}^n = M.$$

Does this argument work? Does Bump apply Artin-Rees because that argument works in some broader context where the above argument fails?

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Don't believe everything that professors tell you. How did he show that $m\bigcap m^n=\bigcap m^{n+1}$? –  Robin Chapman Nov 4 '10 at 18:57
@Qiaochu: thanks –  user3120 Nov 4 '10 at 19:11
@user3120: no problem. It's clear that m times the intersection of the m^n is contained in the intersection of the m^{n+1}, but there is no reason to expect the reverse inclusion in general. I don't know a counterexample off the top of my head, though. –  Qiaochu Yuan Nov 4 '10 at 19:18
Incidenally, the Artin--Rees Lemma is not particlarly heavy machinery: it is a direct application of the Hilbert Basis Theorem (although not always explained this way), which makes it a pretty basic and fundamental fact about Noetherian rings. –  Matt E Nov 5 '10 at 4:03
$(x^2) \cap (x) = (x^2) \neq (x^3)$ which is the product of the ideals inside $\mathbb{Z}[x]$ –  Sean Tilson Dec 5 '10 at 5:01

See Theorem 2.5 in Chapter 6 of the CRing project for another elementary argument due to Perdry (American Math. Monthly, 2004) using only the Hilbert basis theorem. In fact, it shows more: if $R$ is a noetherian domain, $I \subset R$ a proper ideal, then the intersection of the powers of $I$ is trivial.