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Gradings, filtrations, and gr August 18, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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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 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 {<n}. 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.

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Comments»

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 [...]


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