We have a somewhat bumpy road to traverse today. I’ll start with the Artin-Rees lemma and see if we can get to a use of it to continue our set of inequalities we’re trying to prove.

First we’ll need some new ideas. Suppose is any old ring (in particular, we are dropping graded and Noetherian assumptions). Then if is an ideal, we can form a new ring which by construction is graded. Now for any -module, say and an -filtration we can form a graded -module, .

Note that if is Noetherian in the situation above, then , so , so by Hilbert Basis Theorem, we get is Noetherian.

We’ll need that in the situation above the following two statements are equivalent: is finitely generated as an -module, and that the filtration is stable.

Proof: Each is finitely generated, so is finitely generated for all . Let’s form . We have that each is finitely generated as an -module, so we get that is finitely generated as an -module.

Clearly, , so since is Noetherian we get that is finitely generated iff the ascending chain terminates iff for some and for all iff the filtration is stable.

Now we can prove the Artin-Rees Lemma which says that if is a Noetherian ring, an ideal, a finitely generated -module, a stable -filtration and a submodule of , then is a stable -filtration of .

The situation is fairly simple from the previous fact. Note that . So we do indeed get a filtration. But is a graded -submodule of , so it is finitely generated. Now by the equivalence of finitely generated and stable we are done.

There are two important corollaries (both get referred to as the Artin-Rees Lemma as well). In the special case we get that the stable filtration condition says that there is some integer such that for all .

The other result uses the bounded difference result from last time. Since and are both stable -filtrations, they have bounded difference, so the -topology of coincides with induced topology from the -topology on .

I think that is sufficient for today. Next time I’ll go ahead and knock off the next step of the inequalities: .