Hopefully today we can finish this topic off. We’ll jump right in. Let be Noetherian and finite, then is primary if and only if consists of a single element.

We’ll use the second formulation of primary. Suppose . Then by last time , so we have . Suppose is a zero divisor for . Then from AP I we get that . Thus by the second formulation, is a primary submodule.

For the reverse, suppose is a primary submodule and . Then every element of is a zero-divisor for . So . Thus . By definition of associated prime we get , and primes are radical so taking radicals of both sides get the other inclusion and so . i.e. the only element of is .

We now note that is actually a primary ideal. Let and suppose but that . Then but . So is a zero-divisor for which gives . Thus the ideal is -primary.

Thus we make the definition for modules that if then the submodule is -primary (sometimes called a primary submodule belonging to ).

Note that the intersection of any two -primary submodules is again -primary. This is seen by embedding .

We call a submodule reducible if it can be written as such an intersection and irreducible otherwise (this is a property on submodules, not to be confused with the notion of irreducible for modules).

Any submodule of a Noetherian module can be written as a finite intersection of irreducible submodules. This is seen by applying Zorn’s lemma to the set of submodules having no such representation.

We are about to the point where we can define a primary decomposition. Of course there is not going to be a unique way of doing it, but we’ll make some contrived definitions to get it as unique as possible.

We’ll call an intersection irredundant if none of the components of the intersection can be omitted (in particular this will prevent unnecessary repetition sort of like multiplying by a bunch of 1’s in a prime factoring).

A decomposition of a submodule is an expression of the submodule as an intersection of a finite number of submodules, and if each component is irreducible, then we say it is an irreducible decomposition. Likewise, we define primary decomposition if each component is primary.

Now suppose we write as an irredundant primary decomposition with . Since the intersection of any finite number of -primary submodules is again -primary we can group that intersection together and consider it as just a single submodule. In this way we get a decomposition in which when . This makes the decomposition as short as possible.

To wrap up we need to prove that everything behaves the way we want (note that we’ve been assuming Noetherian ring and finite module).

I) An irreducible submodule of is a primary submodule. Suppose is not primary. We can assume without loss of generality by replacing with . Then by the first theorem in this post we get that has at least two elements . i.e. contains submodules isomorphic to , so which means is reducible.

II) Now for an important one. We want to be able to read off the associated primes from the decomposition, so If we have an irredundant primary decomposition of a proper submodule , then where .

We’ll again assume WLOG that . Thus is isomorphic to a submodule of . i.e. .

Now by being irredundant . So pick a non-zero element . Then . But we have is primary belonging to , so for some . Thus which gives that for some we have and . Choose a non-zero element . Then .

But note that , so we have . Since the submodule is primary so giving . If we do this for all the other we get that giving equality of sets.

III) Lastly we want the existence and uniqueness. Every proper submodule has a primary decomposition. This is just because we know that there is an irreducible decomposition, so apply (I) to each irreducible component.

Uniqueness is a little trickier. We must restrict our attention to minimal primes. Suppose is a minimal associated prime of , then the -primary component of is where . Thus it is uniquely determined given the data and . Non-minimal are not unique: Take the ring , then .

Today we’ll continue towards a primary decomposition for modules. First, I’ll list two facts without proof that may come up (they are quite straightforward to prove if you want to try). If R is any ring and is an exact sequence of -modules, then . Secondly, if is a Noetherian ring and a non-zero finite -module, then there is a chain of submodules of such that for each we have with .

I don’t remember, but I may have even proved that second one when talking about Artin-Rees. Now let be Noetherian and a finite -module.

I) is finite. We induct on the length of the chain in the second fact. Suppose this is true for all having a chain of the above form of length . If is a finite module with a chain of length , then since is Noetherian and is a submodule it is also finite. So by the inductive hypothesis, is finite. Now consider the exact sequence . By the first fact . But the chain has the condition that for some . Since we have that the cardinality of can increase by at most one from the cardinality of which was finite.

II) . Suppose . Then contains a submodule isomorphic to (it is just the image of the hom and apply first iso theorem). So is exact, so when we localize we still have an exact sequence . Since , which means .

III) The set of minimal elements of coincides with the minimal elements of . Well, (II) gave one inclusion so suppose is a minimal element. Then since by the last post we get that . But we also figured out a formula for this set . Thus by non-emptyness we must have .

Recall when we working in the Zariski topology on . We have an operator on ideals , and the Zariski closed sets of are precisely those sets that are of the form for some .

So by definition of this operator, if are the minimal elements of , then . Another property of the topological space is that a subspace is irreducible if and only if it is for some minimal prime . So if we think of as a closed subspace of , then the irreducible components are precisely . We call the the isolated associated primes of M. The other associated primes are called embedded primes.

Due to the above geomtric interpretation of what isolated and embedded primes are, the terms make sense. An isolated prime gives you full irreducible component of whereas an embedded prime gives some embedded subspace of the component generated by the prime it lies over.

I’ll finish with the new definitions. Suppose is a submodule. Then we call a primary submodule if for all and we have the condition and for some .

The above condition is equivalent to the condition: if is a zero-divisor for , then . Showing these are equivalent is immediate when you write out what the definitions of all these things are. This shows that the property of being primary is dependent only on the quotient module .

Sorry to end on some definitions, but I think if I do another theorem this post will become too long.

I’d like to go over associated primes in general rather than just the Noetherian ring form of primary decomposition of ideals. The natural generalization is to modules, since ideals are sub-modules over the ring treated as a module over itself. We’ll need to define a few things first.

Let M be an R-module. Let P be a prime ideal of R. Then we call this an associated prime of M if for some . The set of associated primes is denoted (since R will be understood, we’ll just drop that from here on).

Now suppose an ideal. Then the elements of are called the “prime divisors” of I.

Now we’ll get some basics out of the way. Suppose that R is Noetherian. First off, when . We show this by showing that any maximal element of the family is an associated prime. This is an important fact on its own.

Note that all we really are trying to show is that a maximal element of is prime as an ideal, since it will already be of the form . Let be a maximal element. Suppose and that and that . Then but . Thus . But by maximality, . Thus which means . Thus is prime and hence an associated prime of M.

The other fact we need is that the set of zero-divisors for M is precisely the set .

If , then this just says there is some such that and hence is a zero-divisor. The reverse inclusion just uses the previous fact. Let be such that for some . So we have . Then take a maximal element containing . By the last fact and hence .

For the theorem of the day, you may need a refresher on the spectrum of a ring, and on localization.

We no longer assume R is Noetherian. Let be multiplicative, and an -module. Viewing , then we have . In general, if is Noetherian, then for any R-module M, we have .

Let . Then . This is just because the elements of R that kill x, are just the fractions that kill x that are “actually” in R. This immediately gives us one inclusion, since if then .

Now suppose . Then there is some such that . Thus giving . Thus with . This proves the first statement.

We now show the second statement about M. Suppose . Thus we again get that and for some non-zero . Suppose that that in . Thus there is some such that in . But we’ve already noted that and , thus by primality . So . Thus giving one inclusion.

For the reverse, suppose . By clearing the denominator we can assume that for a non-zero we have . Let . Then . We have that is finitely generated since is Noetherian, so there is some such that . Thus which gives the reverse inclusion.

A nice little corollary is that for Noetherian rings a prime ideal if and only if .

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.

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 .

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.

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).

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.

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.

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 .