# 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

There is a proof due to Herstein that is "elementary" in the sense of avoiding the Artin-Rees lemma. Below is Kaplansky's presentation of this proof, from his "Commutative Rings". A proof using primary decomposition (as in Krull's original proof) can be found in Zariski and Samuel.

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See Theorem 3.6 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.

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Dear Akhil, Nice. Of course, my fondness for Artin--Rees forces me to point out that this is very close to just a reproof of that lemma in this particular case. Best wishes, – Matt E Jan 30 '11 at 11:59
@Matt: Dear Matt, I agree, given that -- as you observe -- the Artin-Rees lemma is essentially a statement about the noetherianness of a polynomial ring. – Akhil Mathew Jan 30 '11 at 14:10
Perdry's proof is also given in my commutative algebra notes: math.uga.edu/~pete/integral.pdf. This is one of the most remarkable instances of a quick, easy proof of a nontrivial theorem that I know. – Pete L. Clark Jan 30 '11 at 20:12
@Akhil: the end of your proof of Artin-Rees is rather unorthodox: "This is easy to check, but we're out of time. We'll talk about this more next week." – Pete L. Clark Jan 30 '11 at 20:20
By the way, I encountered this CRing project about a week ago while googling for other online notes in commutative algebra to post to the webpage for my commutative algebra course. It's quite impressive. I certainly take it as serious competition to my notes. It might be worth chatting at some point to carve out complementary paths... – Pete L. Clark Jan 30 '11 at 20:21