Let’s start applying to some specific situations now. Suppose is a Noetherian local ring with maximal ideal . Let be an -primary ideal. Let be a finitely-generated -module, and a stable -filtration of .

Don’t panic from the set-up. I think I haven’t talked about filtrations. All the stable -filtration means is that we have a chain of submodules such that for large .

The goal for the day is to prove three things.

1) has finite length for all .

Define and . We have a natural way to make into a finitely-generated graded -module. The multiplication in the ring comes from the following. If , then let the image in be denoted . We take . This does not depend on representative.

We’ll say is the n-th grade: . Now is an Artinian local ring and each is a Noetherian -module annihilated by . Thus they are all Noetherian -modules. So by the Artinian condition we get that each is of finite length. Thus .

2) For large , is a polynomial of degree where is the least number of generators of .

Suppose generate . Then in generate as an -algebra. But is an additive function on the filtration, so by last time we saw thatfor large there is some polynomial such that , and each has degree 1, so the polynomial is of degree .

Thus we get that . So from two posts ago, we get for large that is some polynomial of degree .

3) Probably the most important part is that the degree and leading coefficient of depends only on and and not on the filtration.

Let be some other stable -filtration with polynomial . Since any two stable -filtrations have bounded difference, there is an integer such that and for all . But this condition on the polynomials says that and , which means that . Thus they have the same degree and leading coefficient.

That seems to be enough for one day. Unfortunately, I haven't quite got to the right setting that I want yet.

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November 4, 2009 at 8:55 pm

Some typos:

1. The definition of stable q-filtration. I think you have for all $n$ and equality holds only for big . Otherwise there is only one stable filtration as opposed to your third proprosition.

2. I guess in the setting part you want to have to be your noetherian ring. (Since you are using thereafter)

3. In the proof of 1, you want instead of .

November 5, 2009 at 2:19 pm

Thanks. I’m really trying to hold to this notation (since that is how I started forever ago) even though the literature tends to favor , so slipping is bound to occur.