Bourbaki has a whole chapter in *Commutative Algebra* devoted to “graduations, filtrations, and topologies,” which indicates the importance of these concepts. That’s the theme for the next few posts I’ll do here, although I will (of course) be more concise.

In general, all rings will be commutative.

**Gradings **

The idea of a graded ring is necessary to define projective space.

Definition 1Agraded ringis ring together with a decompositionsuch that . The set is said to consist of

homogeneous elementsof degree .

Note that is a subring (containing ) and an -algebra. Also, in many cases we actually have , so the negative elements don’t matter, but to talk about localization, we want the greater generality.

The example to keep in mind here is the polynomial ring associated to any commutative ring . Here the homogeneous elements of degree are the monomials of degree , or multiples of .

More generally, consider the polynomial ring and let the homogeneous elements of degree be the polynomials which are sums of monomials of degree . In this way, homogeneous elements correspond to homogeneous polynomials. This is the example that leads to projective space.

Naturally, we can form a category of graded rings, but we need the appropriate morphisms:

Definition 2Let be graded rings. Then ahomomorphism of graded ringsis a ring-homomorphism such that for all , in other words preserves homogeneous elements and degrees.

To keep up with the theme from my previous posts, let’s do the standard test for when a graded ring is Noetherian:

**Theorem 3** *If is a graded ring with for , then is Noetherian if and only if is Noetherian and is a finitely generaed -algebra. *

One direction is the Hilbert basis theorem. Conversely, suppose is Noetherian. First I claim is Noetherian. Indeed, otherwise given ideals , we consider

note that strict inclusion holds, by considering the components of degree (which are just the ideals ). This is a contradiction, so is Noetherian.

Now we need to check is a finitely generated -algebra. For this, consider the -ideal

and choose generators . By splitting them into components, assume each is homogeneous of degree .

I claim that . To prove this, we will check inductively that

which will imply the claim. This is clearly true for . Assume it true for , and let . We can write

and by taking the -th homogeneous component, we may assume each is actually homogeneous of degree . Then by (complete) induction each , so the same is true for .

Now we can look at graded *modules*:

Definition 4If is a graded ring, agraded -moduleis an -module together with a decompositionsuch that .

So in particular is a graded -module. In the same vein, there is a category of graded -modules with homomorphisms preserving the grading. This is all essentially a repetition of what was already said.

**Filtrations **

Filtrations are a more general concept. Basically, you don’t have a notion of “degree ,” but you instead have a notion of “degree .”

Definition 5Afiltered ringis a ring together with subgroups with , , and

For simplicity I am only looking at filtrations for nonnegative integers. It is as usual possible to define filtered modules (we want subgroups with and ) and homomorphisms preserving filtrations.

As an example, if is graded, we can let

This is a filtration. A more interesting example comes from the theory of Lie algebras. If is a Lie algebra over a field with a basis, then the Poincaré-Birkhoff-Witt theorem (which I hope to discuss, eventually) states that products of the form

are a basis of the enveloping algebra . So we can filter this ring by letting

Nevertheless, this is a noncommutative ring in general, so we have slightly violated our conventions.

Conversely, we can get from a filtered ring a graded ring as follows:

Definition 6If is a filtered ring, we define theassociated graded ringby

where . To define the product of with , lift to , and take the image of .If is a filtered -module, define the

associated graded modulesimilarly.

It is easy to check the above definition is legitimate.

Similarly, we can define a *descending filtration* on a ring (resp. module) by reversing the inclusions: thus (resp. ). There is a similar definition for the associated graded ring

(resp. ). An important example of this is obtained as follows. If is an ideal, then the filtration is called the **-adic filtration**.

As before, there is a category of filtered rings, and given a filtered ring , a category of filtered -modules. [*Edit:* *For some reason I forgot to add the next comment in the post at first. AM] *Then is a functor from filtered rings to graded rings, or filtered -modules to graded -modules.

As an aside, Anirudha Balasubramanian suggested during a talk (to a non-mathematical audience) a clever analogy: a descending filtration (in his example a lower central series) is like a Matryoshka doll, or a rock with many layers.

So, next will be some discussion of topologies, the Artin-Rees lemma and its applications, and completions. Today we finished the basics.

August 18, 2009 at 12:01 pm

In definition 5, it’s said that A_i are subsets. Then what does A_i/A_{i-1} mean in definition 6?

August 18, 2009 at 12:29 pm

The condition that implies that is a subring, and the condition that implies that each is a -module. The quotient in definition 6 is the module quotient.

August 18, 2009 at 1:24 pm

Qiaochu was right. Thanks for the correction!

August 19, 2009 at 10:04 pm

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