## Gradings, filtrations, and gr August 18, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
Tags: , , , , ,

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.

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

Definition 1    A graded ring is ring ${A}$ together with a decomposition

$\displaystyle A = \bigoplus_{n=-\infty}^\infty A_n \ \mathrm{as \ abelian \ groups},$

such that ${A_i \cdot A_j \subset A_{i+j}}$. The set ${A_i}$ is said to consist of homogeneous elements of degree ${i}$.

Note that ${A_0}$ is a subring (containing ${1}$) and ${A}$ an ${A_0}$-algebra. Also, in many cases we actually have ${A = \bigoplus_{n=0}^\infty A_n}$, 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 ${R[x]}$ associated to any commutative ring ${R}$. Here the homogeneous elements of degree ${i}$ are the monomials of degree ${i}$, or multiples of ${x^i}$.

More generally, consider the polynomial ring ${R[x_1, \dots, x_m]}$ and let the homogeneous elements of degree ${i}$ be the polynomials which are sums of monomials of degree ${i}$. 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 2 Let ${R,S}$ be graded rings. Then a homomorphism of graded rings is a ring-homomorphism ${\phi: R \rightarrow S}$ such that ${\phi(R_i) \subset S_i}$ for all ${i}$, in other words ${\phi}$ 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 ${A}$ is a graded ring with ${A_n =0}$ for ${n<0}$, then ${A}$ is Noetherian if and only if ${A_0}$ is Noetherian and ${A}$ is a finitely generaed ${A_0}$-algebra.

One direction is the Hilbert basis theorem. Conversely, suppose ${A}$ is Noetherian. First I claim ${A_0}$ is Noetherian. Indeed, otherwise given ideals ${I_0 \subsetneq I_1 \subsetneq I_2 \subsetneq \dots \subset A_0}$, we consider

$\displaystyle I_0 A \subsetneq I_1 A \subsetneq I_2 A \subsetneq \dots;$

note that strict inclusion holds, by considering the components of degree ${0}$ (which are just the ideals ${I_i}$). This is a contradiction, so ${A_0}$ is Noetherian.

Now we need to check ${A}$ is a finitely generated ${A_0}$-algebra. For this, consider the ${A}$-ideal

$\displaystyle A_{+} := \bigoplus_{n=1}^\infty A_n$

and choose generators ${x_1, \dots, x_m}$. By splitting them into components, assume each ${x_i}$ is homogeneous of degree ${d_i>0}$.

I claim that ${A = A_0[x_1, \dots, x_m]}$. To prove this, we will check inductively that

$\displaystyle A_n \supset A_0[x_1, \dots, x_m],$

which will imply the claim. This is clearly true for ${n=0}$. Assume it true for ${n-1}$, and let ${x \in A_n}$. We can write

$\displaystyle x = \sum_{j=1}^m c_j x_j, \quad \mathrm{where} \quad c_j \in A$

and by taking the ${n-d_i}$-th homogeneous component, we may assume each ${c_j}$ is actually homogeneous of degree ${. Then by (complete) induction each ${c_j \in A_0[x_1, \dots, x_n]}$, so the same is true for ${x}$.

Now we can look at graded modules

Definition 4  If ${A}$ is a graded ring, a graded ${A}$-module is an ${A}$-module ${M}$ together with a decomposition

$\displaystyle M = \bigoplus_{n=-\infty}^\infty M_n$

such that ${A_i \cdot M_j \subset M_{i+j}}$.

So in particular ${A}$ is a graded ${A}$-module. In the same vein, there is a category of graded ${A}$-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 ${n}$,” but you instead have a notion of “degree ${\leq n}$.”

Definition 5

A filtered ring is a ring ${A}$ together with subgroups ${A_i, i \in \mathbb{Z}_{\geq 0}}$ with ${A_i \subset A_{i+1}}$, ${A_i \cdot A_j \subset A_{i+j}}$, and $\displaystyle A = \bigcup_i A_i.$

For simplicity I am only looking at filtrations for nonnegative integers. It is as usual possible to define filtered modules ${M}$ (we want subgroups ${M_j}$ with ${M = \bigcup_j M_j}$ and ${A_i M_j \subset M_{i+j}}$) and homomorphisms preserving filtrations.

As an example, if ${A}$ is graded, we can let

$\displaystyle A_i := \{ \mathrm{sums \ of \ homogeneous\ elements \ of \ degree } \leq i \}.$

This is a filtration. A more interesting example comes from the theory of Lie algebras. If ${\mathfrak{g}}$ is a Lie algebra over a field with ${x_1, \dots, x_m}$ a basis, then the Poincaré-Birkhoff-Witt theorem (which I hope to discuss, eventually) states that products of the form

$\displaystyle x_1^{\alpha_1}\dots x_m^{\alpha_m} \in \mathcal{U}(\mathfrak{g}), \quad \alpha_i \in \mathbb{Z}_{\geq 0}$

are a basis of the enveloping algebra ${\mathcal{U}(\mathfrak{g})}$. So we can filter this ring by letting

$\displaystyle \mathcal{U}(\mathfrak{g})_n := \mathrm{span}\left( x_1^{\alpha_1}\dots x_m^{\alpha_m} \right), \quad \sum \alpha_i \leq n.$

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 6

If ${A}$ is a filtered ring, we define the associated graded ring ${\mathrm{gr} (A)}$ by

$\displaystyle \mathrm{gr} (A) := \bigoplus_{n=0}^\infty A_n/A_{n-1},$

where ${A_{-1} := 0}$. To define the product of ${\bar{x} \in A_i/A_{i-1}}$ with ${\bar{y} \in A_j/A_{j-1}}$, lift to ${x \in A_i, y \in A_j}$, and take the image ${\overline{xy} \in A_{i+j}/A_{i+j-1}}$ of ${xy \in A_{i+j}}$.If ${M}$ is a filtered ${A}$-module, define the associated graded module ${\mathrm{gr} (M)}$ similarly.

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 ${A_n \supset A_{n+1}}$ (resp. ${M_n \supset M_{n+1}}$). There is a similar definition for the associated graded ring

$\displaystyle \mathrm{gr}(A) := \bigoplus_n A_n/A_{n+1}$

(resp. ${\mathrm{gr}(M) := \bigoplus_n M_n/M_{n+1}}$). An important example of this is obtained as follows. If ${I \subset A}$ is an ideal, then the filtration ${I^n, n \geq 0}$ is called the ${I}$-adic filtration.

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

1. soarerz - August 18, 2009

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

2. Qiaochu Yuan - August 18, 2009

The condition that $A_0 \cdot A_0 \subset A_0$ implies that $A_0$ is a subring, and the condition that $A_0 \cdot A_i \subset A_i$ implies that each $A_i$ is a $A_0$-module. The quotient in definition 6 is the module quotient.

Qiaochu Yuan - August 18, 2009

Akhil may have forgotten the condition that each $A_i$ is closed under addition.

3. Akhil Mathew - August 18, 2009

Qiaochu was right. Thanks for the correction!

4. Topologies and the Artin-Rees lemma « Delta Epsilons - August 19, 2009

[…] filtered rings, filtrations, I-adic filtration trackback Today I’ll continue the series on graded rings and filtrations by discussing the resulting topologies and the Artin-Rees […]