Proof. Let $N = \bigcap _{n \geq 0} I^ nM$. Then $N = I^ nM \cap N$ for all $n \geq 0$. By the Artin-Rees Lemma 10.50.2 we see that $N = I^ nM \cap N \subset IN$ for some suitably large $n$. By Nakayama's Lemma 10.19.1 we see that $N = 0$. $\square$

Comment #3856 by Alice on

While minor, It would be better to simply say $N=IN$, insteady of the roundabout $N\subset IN$. Also a note, the only point where locality is used is to guarente $I \subset \mathop{rad} (R)$ for all I for nakayama's lemma, so this can be generalised very easily if so desired.

Comment #3940 by on

Of course the application of Artin-Rees 10.50.2 does not immediately give what was said in the proof of the lemma or what you say. What I mean is that you have to think about it! So we can leave it as is I think for now.

The point about non-local cases is made in Remark 10.50.6 for the module equal to the ring. But yeah, we could discuss the case of a module there too.

Further discussion: The interesting part of the Artin-Rees lemma, and the way most people probably think about it, is that $I^nM \cap N$ is contained in $IN$ for $n \gg 0$. The fact that in the Artin-Rees lemma, as stated in the Stacks project and many texts, you get an equality is a bit of a red herring and rarely useful. This is why you'll see most often the Artin-Rees lemma used by saying $I^nM \cap N \subset IN$ for some $n > 0$.

This whole section can probably be improved upon and extended. For example, do lemmas 10.50.2 and 10.50.3 really cover all possible reformulations of Artin-Rees? Etc.

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