Integrally Closed Noetherian Domain

I don’t feel like an all out post, but I did need to use this fact today and it relates to what I’ve been doing, so may as well post the proof. Let be an integrally closed Noetherian domain. Then .

We start with a Lemma: Every prime divisor of a non-zero principal ideal has height 1.

Suppose is a prime divisor of . Then there is an element such that . Define . So . Thus and .

If we had then the standard determinant trick (see a proof a Nakayama’s Lemma) would give us that is integral over contradicting being integrally closed. So and hence and so is a DVR which means dimension of is 1 which means . This gives the lemma.

We will prove that for with we have for all height 1 primes means that is actually in .

Let be the prime divisors of and let be a primary decomposition of where is -primary. But each has height 1 by the Lemma above, so . Thus .

Maybe a short discussion could be useful at this point. The use of this fact came up when trying to prove something about a normal point. This says that at a normal point on a variety, if you want to check if a function regular, you only need to check that it is regular on all codimension 1 subvarieties through that point. This almost immediately gives that if you have a regular function on () where is a normal point. Then the function actually extends to be regular on all of . A function on a dim variety cannot blow up at just a single normal point.

Discrete Valuation Rings

Maybe this title isn’t exactly what the post is about, but today will mostly be a hodgepodge attempt to get some more out there. I’m not sure what else to do with regular local rings (other than systems of parameters which I’m not too excited to post on), so I’ll move on. The next set of theorems in Hartshorne (that are not proved in the text) has to do with Noetherian local domains of dimension 1.

Before this is stated we need quite a bit of terminology. Let be a field and a totally ordered abelian group. A valuation is a map such that and .

We form a subring of by taking the set which is called the valuation ring of . This ring is local with maximal ideal .

Mostly we care about discrete valuation rings (DVR). These are the ones whose value group is the integers. Now we can state and prove the Theorem stated in Hartshorne:

Let be a Noetherian local domain of dimension 1. Then the following are equivalent:

1) is a discrete valuation ring
2) is integrally closed
3) is principal
4)
5) Every non-zero ideal is a power of
6) There exists such that every non-zero ideal is of the form , .

Proof: We’ll just go in the standard cyclic order for proof. If is a DVR, then we consider an integral element . If then we are done. If , then the claim is that . This is simply because . Since , and $v(x)=-v(x^{-1})$, we get that . Thus if we get by multiplying by $x^{1-n}$ that and every integral element is in the ring.

For the next, assume is integrally closed. Let and . Since is the only non-zero prime ideal, . Thus there is some integer such that such that . Now let . Let . Now since , we cannot reduce to a form , so . By integrally closed, we get that is not integral over . If , then would be a finitely generated (as an module) faithful -module, and hence would be integral. Thus . Clearly, , so . Thus is principal.

Now if is principal, we have . But since by the Noetherian condition, we get that .

Suppose . Thus is principal. Let be any non-zero ideal. Since all ideals are -primary we again get that for some n. Since is Artinian we get that for some r. Thus .

Suppose every non-zero ideal is a power of . By Noetherian we have , so we can pick . So there is some such that . Thus or else we’d have a prime chain longer than 1. Now given any ideal, .

Suppose such that every non-zero ideal has the form . Again, we must have . So by Noetherian . Thus if is non-zero, there is a well-defined such that . Naturally we get a discrete valuation and extend in the obvious way to the rest of the field by . By putting everything in reduced form, we see that something in that is not in has negative valuation, and hence is the valuation ring of .

Regular Local Rings I

We have defined and used the associated graded ring . Now we want to see how it behaves under completions.

By the last post, we have , so we immediately get that .

A great theorem that I’ll skip proving is that if is Noetherian, and is any ideal, then the completion with respect to the -adic topology is Noetherian. As a corollary we get that for any Noetherian ring , is Noetherian by noting that the completion of the Noetherian ring with respect to the -adic topology is .

After this brief excursion, we’ll come back to the dimension theory we left off from. The next natural place to go is to regular local rings. A local ring is regular if . (Recall that it is always true that ).

Suppose we have a Noetherian local ring such that . Then the following are equivalent definitions of regular: or can be generated by elements.

The first condition implies that , so it implies regular. Regular implies that can be generated by elements, by this post. Lastly, if can be generated by (if you’ve seen the term, this is a system of parameters), then we have a surjective map of graded rings with kernel . So it is an iso. This finishes up the equivalences.

Last time we saw without proof that (for Noetherian local rings) is regular if and only if is regular. Now we can prove it.

By the equivalent definition of regular, is regular iff , but we proved that , so but this happens iff is regular.

We’ll wrap up today with trying to keeping the geometric picture in mind. Regular means non-singular geometrically. So we see that passing to the completion doesn’t introduce any singularities. But since the dimension of the local ring at a point equals the dimension of the variety we actually get that completion of where is non-singular is where is the dimension of the variety.

So if we interpret completion of as the “analytically local” picture, then we see that locally all non-singular points on a variety are analytically isomorphic.

Properties of Completions

First note that taking an inverse limit is a functor. I won’t need the functorial properties in the immediate future, but it would be good to state some of them. First off, the functor is not exact, but it is left exact. So given an exact sequence of inverse systems (it is exact at each level and all the diagrams commute) we get an exact sequence .

It turns out that if the first system has the property that the homomorphisms are surjective, then the inverse limit is exact. So in our case with completions, this always happens.

The properties I’d really like to prove are the ones listed in Hartshorne without proof. Suppose for the rest of this post that is a Noetherian local ring and is its completion with respect to the -adic topology. The numbers will refer to Hartshorne numbering:

5.4A(a) is a local ring with maximal ideal and there is a natural injective homomorphism .

We already discussed the second part, since the kernel of the hom is just . Using right exactness of tensoring and exactness of completions, we get that for any finitely generated -module , is an iso (if we remove Noetherian on R, we only get surjective). This gives us that is an isomorphism and since the image is we get the first part of the statement.

Now we need that it is a unique maximal ideal. But applying the above result to any ideal (which is finitely generated since R is Noetherian) we get that . Thus . Taking inverse limits gives that and hence is a maximal ideal since the quotient is a field. But for any , we can define an inverse for formally by . Since we are in the completion, this converges in and hence . But a maximal ideal means that it is the Jacobson radical and hence the unique maximal ideal.

5.4A (b) If is a finitely generated -module, its completion is isomorphic to . Well, I needed this to prove (a) and briefly described how it would go, but since I didn’t prove the exactness properties, it seems needlessly detailed to do a full proof using them. For more details, see posts at delta epsilons.

5.4A (c) .

Let’s use some of the machinery we developed. The dimensions are equal to the degree of the Hilbert polynomial, but says precisely that . So the polynomials are actually the same.

5.4A (d) is regular if and only if is regular.

We’ll hold off on this until I cover regularity (which will either be next time or the time after).

Completions II

We will call a topological group complete if is an isomorphism.

The case that we are particularly concerned with is when our group is a ring and we take for our inverse system some ideal and . The topology that this determines is the “-adic topology”. This makes into a topological ring.

If we take the completion , then the continuous ring homomorphism has kernel .

Now we can also do all this with -modules by taking the group to be and the inverse system . The topology determined by this system is called the -topology on M. If we take the completion with respect to this topology (i.e. w.r.t this system), we get which is a topological -module meaning the action is continuous.

Rephrasing the motivating example from last time in this language we see that the -adic integers are formed as the completion of the ring with respect to the -topology where is the ideal .

The other really important example is to form the completion of with respect to the -adic topology. The completion is the ring of formal power series. Recall that by definition the inverse limit are all sequences such that . This just says that each is a polynomial, and it has to agree with the one before it up to the coefficient. So we can write each sequence where is the coefficient on the of the polynomial . And for any power series we get a sequence in this way.

Recall our notion of -filtrations. We had a chain such that , and if equality held for all large , then we called the filtration stable. Well, in our new language, these filtrations are inverse systems of modules, and hence determine a topology on . A few posts ago we used the fact that any stable -filtrations have bounded difference. In this new language, this says precisely that all stable -filtrations determine the same topology on M, moreover this is the -topology.

Lastly, if we convert the Artin-Rees Lemma to this language, we get that if is Noetherian, an ideal, a f.g. -module, and a submodule of , then the -topology on is actually just the subspace topology from the -topology on .

We should probably do some properties of completions next time.

Completions I

Today we’ll start a new section, but only because it is a tool we need when we come back to the stuff we just finished. We will look at completions.

To motivate the process take a Hausdorff abelian topological group . Suppose there is a countable local basis at 0 (which implies countable basis, since the neighborhoods of 0 determine the entire topology). Since we assumed Hausdorff we have the usual notion of Cauchy sequences, so we can define the completion of to be completion in the usual sense . In particular, if , then .

Now suppose we have a local basis about 0 of subgroups (this rules out ), say . If we are in this situation, then our topology is actually determined by a sequence of subgroups, so we will want to try to define the completion solely in terms of algebra.

Take any Cauchy sequence . If we fix k, then at some , is constant for all . Note that really does depend on . Set the limit .

If we make what we mod out by bigger, namely we go from to , then projection maps . Thus our Cauchy sequence determined a “coherent sequence” , where .

Conversely, we can define a Cauchy sequence corresponding to any coherent sequence by just picking an element in the equivalence class at each step. So we can now define the completion to be the set of coherent sequences with group structure given entry-wise by the quotient group. The standard example here is the -adic integers, where the group is and our fundamental system is . Coherent sequences are where .

Whenever we have in general a sequence of groups and homomorphisms this is called an inverse system. The group of all coherent sequences is called the inverse limit of the system written . Thus our definition of completion can be written succinctly as .

Next time we’ll transfer this to module language and get to a few results.

Some Corollaries

Today will just be some quick results we get from this build up.

First, if we localize a polynomial ring at a maximal ideal, say at , then . This is because has Poincare series so the order of the pole is which is the dimension by the last post.

This one will be really useful later: . Let such that are a basis for the vector space. Then by Nakayama’s Lemma the generate . Thus .

This one is also useful in algebraic geometry. If is Noetherian, and , then every minimal ideal belonging to has height . Unfortunately, we cannot push this to equality. Geometrically the example is that if is the twisted cubic, then has height 2, but cannot be generated by less than 3 elements.

Lastly, we’ll do the famous Principal Ideal Theorem. If is Noetherian and is neither a zero-divisor nor a unit, then every minimal prime ideal of has height 1. By the last paragraph we know that . If then it belongs to . Thus every element of is a zero-divisor which is a contradiction since .

Finishing up Dimensions

We are now on the last inequality: . Recall we’re supposing is Noetherian and local. Let , then the inequality is saying we can find an ideal, that is an -primary ideal and is generated by elements: .

Let’s construct these elements inductively. The way we want to do it is so that at each step any prime ideal containing has height bigger than or equal to to force the dimension of to be big.

Suppose the have been constructed in the given way. Let be the minimal prime ideals of with height exactly . But we have that for any since . Thus (there is a well-known fact that if any ideal , then in fact for some ).

Now pick some element , and let be a prime ideal containing . We definitely have that , since contains a minimal prime ideal of , say . If for some , , then since , we have strictly increasing the height. If for any , then since are all minimal primes of height , we have .

All that is left is to show that is -primary. Let be any prime ideal of . Then if we have that . So this is impossible and we have . Thus is -primary.

Over the last couple of posts we have finally completed the first goal. We have . In other words, for Noetherian local rings we have an equivalence between the maximum length of chains of prime ideals, the degree of the Hilbert polynomial, and the least number of generators of an -primary ideal.

Next time I'll derive some results directly from this including the Principal Ideal Theorem. Then we'll move on to something different (only for awhile, then we'll return).

The Next Inequality

Considering it has been at least a post removed, I’ll bring us back to our situation. We have a local Noetherian ring . Our notation is that is the least number of generators of an -primary ideal (which was shown to be independent of choice of ideal here). The goal for the day is to show that .

Suppose is -primary. We’ll prove something more general. Let be a finitely generated -module, a non-zero divisor in and . Then the claim is that .

Since is not a zero-divisor, we have an iso as -modules: . Define . Now take . Since is a stable -filtration of , by Artin-Rees we get that is a stable -filtration of .

For each we have exact.

Thus we get . So if we let , we get for large : .

But is also a stable -filtration of , since . We already showed that the degree and leading coefficient of depends only on and and not on the filtration. Thus and have the same degree and leading coefficient, so the highest powers kill eachother which gives .

In particular, we will need that as an -module gives us .

Now we prove the goal for today. For simplicity, let . We will induct on . The base case gives that is constant for large . In particular, there is some such that for all . But we are local, so and hence by Nakayama, . Thus for any prime ideal , we have for some , so take radicals to get . Thus there is only one prime ideal and we actually have an Artinian ring and hence have .

Now suppose and the result holds for . Let be a chain of primes. Choose . Define and be the image of in .

Note that since is an integral domain, and is not 0, it is not a zero-divisor. So we use our first proof from today to get that .

Let be the maximal ideal of . Then is the image of , so which is precisely . Plugging this into the above inequality gives .

So by the inductive hypothesis, . Take our original prime chain. The images form a chain in . Thus . Since the chain was arbitrary, .

A nice corollary here is that the dimension of any Noetherian local ring is finite. Another similar corollary is that in any Noetherian ring (drop the local) the height of a prime ideal is finite (and hence primes satisfy the DCC), since which is local Noetherian.

The Artin-Rees Lemma

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: .