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In most books, a graded ring/module/algebra means either a $\mathbb{N}$- or $\mathbb{Z}$-graded ring/module/algebra. But often, different gradings appear: doubly graded (spectral sequences) = $\mathbb{N}^2$- or $\mathbb{Z}^2$-graded; multigraded = $\mathbb{N}^n$- or $\mathbb{Z}^n$-graded; superalgebra = $\mathbb{Z}_2$-graded; etc. It is desirable to have a definition which is general enough, and still retains most properties of usually graded objects. One such definition (I think) is the following:

Let $G$ be any monoid, written additively, though commutativity of $G$ is not assumed. A $G$-graded ring is a ring $R$ together with a chosen direct sum decomposition into additive subgroups $R\!=\!\bigoplus_{i\in G}\!R_i$, such that the ring multiplication satisfies $R_iR_j\!\subseteq\!R_{i+j}$. A $G$-graded $R$-module is an $R$-module $M$ that has a direct sum decomposition into $R$-submodules $M\!=\!\bigoplus_{i\in G}\!M_i$. A $G$-graded $R$-algebra is an $R$-algebra $A$ that has a direct sum decomposition into $R$-submodules $A\!=\!\bigoplus_{i\in G}\!A_i$ such that $A_iA_j\!\subseteq\!A_{i+j}$. In the case when $R$ is a $G$-graded ring, there are more general definitions of $G$-graded modules and algebras.

Question 1: Do these definitions apply only to commutative rings/algebras? Are gradings used in noncommutative algebra?


I wish to have a general enough definition of a filtration on a ring/module/algebra as well:

  1. index set not necessarily $\mathbb{N}$ or $\mathbb{Z}$;
  2. preferably the definition for rings/modules/algebras are compatible when a ring $R$ is viewed as a $R$-module and a $\mathbb{Z}$-algebra;
  3. preferrably it is defined for noncommutative rings and algebras as well.

Most books define a filtration as a sequence. There are some generalizations:

In Grillet's Abstract Algebra, p. 462: enter image description here

In Pete Clark's Commutative Algebra notes, p.50: enter image description here

In Stacks Project, 10. Homological Algebra, 13. Filtrations: enter image description here

Question 2: What is the right definition of a filtration of a ring/module/algebra in regard to 1,2,3? If $R$ is noncommutative, do we need to take left or two-sided ideals to obtain a filtration? Does the index set need to be totally ordered, or a monoid, or a group, or a directed set?

I realise the answer is probably going to be "depends what you want to do with a filtration", and since I don't yet know what is or can be done with filtrations, or even what their main use is, I'll just say that my motivation comes from Algebraic Topology, Homological Algebra, and Commutative Algebra. I would simply like to have a definition, general enough so that I am not confused when I take 10 different books and each has a bit different definition of a filtration.

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@YACP: Do rings/algebras form an abelian category? What are subobjects? So there is never any need to consider filtrations with more general index sets? –  Leon Feb 21 '13 at 22:37
    
Filtrations can be considered indexed over a totally ordered group. Why do you need an abelian category to define a filtration on a ring? This is probably required for defining filtrations on categories. –  user26857 Feb 21 '13 at 22:45
    
Well, the 'Stacks Project' definition defines it for objects from an abelian category. Why is it so necessary that the index set is totally ordered? –  Leon Feb 21 '13 at 23:05
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Well, because we need to have some inclusions between the objects of a filtration. –  user26857 Feb 21 '13 at 23:08
    
@LeonLampret: in the definition of $G$-graded $R$-modules, you certainly want $R_iM_j\subseteq M_{i+j}$. For the filtrations, I think people wants to work in an abelian categories to be able to take quotients $F^nA/F^{n+1}A$. –  user18119 Feb 22 '13 at 17:09

2 Answers 2

up vote 3 down vote accepted

Very short answer: Gradings and filtrations are used everywhere for associative algebras.

Not very detailed exposition (I expect you don't know most of the terms, but I am just listing things I know that form examples to the answer of your questions)

(Q1) Finding a good grading on (usually finite dimensional) associative algebra has become one of the main trend in recent years in representation theory. The gradings people are looking for are usually the Koszul grading, i.e. a grading so that the algebra is Koszul. People who are more DG(differential graded)-oriented may need more than just a positive Z-grading. This comes under the name of Adams grading (still a group-graded theory anyway)

Other use of grading I know of: grade the preprojective algebra of finite type and take a smash product with it, one would arrive to a mesh algebra.

In a nut shell, as you have said it yourself, it really depends on what you want at the end before you actually start looking for a grading.

(Q2) Again I can only give examples that I know of. If $A$ is quasi-hereditary algebra, then there are certain objects called the standard modules, which exhibit certain nice (homological) properties. The category of the objects which can be filtered by standard modules have very interesting homological properties (this actually happens not only in quasi-hereditary algebras), such as, they are contravariantly finite.

If $A$ is Koszul algebra, then the Koszul dual $A^! := Ext_A^\bullet(A_0,A_0)$ in fact "determines" the category of modules which are filtered by the 0-th graded piece $A_0$. When people try to generalise Koszul theory, one thing they would study is that if the algebra $Ext_A^\bullet(M,M)$ can "determine" the category of objects filtered by $M$, etc. etc. A good exposition on this direction is Keller's Introduction to $A_\infty$ algebras and modules.

Filtration, sometimes, comes very close to determining some sort of useful/interesting resolutions (for Koszul algebras, the filtration gives you the projective resolution). Whether you want a filtration by two-sided ideal, or one-sided ideal, or submodule, is entirely up to you. As you have expected, it really depends on what you want to do.

EDIT: In response to OP comments. I think you need to understand filtrations and gradings are two different concepts and should not to be confused the two. I myself fell into this trap before. This is understandable, because, most of the time people talk about positive $\mathbb{Z}$-grading, so then you can get a filtration $\bigoplus_{i\geq 0} A_i \supset \bigoplus_{i\geq 1} A_i \supset \cdots$. However, as the comments to the question already tell you, in order to have 'filtration', you need a concept of inclusion for the objects in your category (warning: this is not really true, but I should not confuse you further), and you usualy need "quotient" for objects so that you can talk about things like "filtered by a family of objects". The first definition you wrote down in the comment looks more like a grading thing rather than a filtration thing (sorry for bad wording). The subscript appearing in a grading has significance in the study; but the subscript appearing in a filtration, is merely used to label the position of an object in the chain (at least I would choose to think about them this way).

TBH, I don't understand why you need to be so precise on whether you should choose one-sided or two-sided ideal for a ring (for me, I look at algebras). You should start by thinking what you want at the end; or if you don't know what you want at the end, why don't you try fiddle with some examples first and see which kind of filtration would give something interesting? May be you want to define a type of filtration on something you are studying, I guess you can, for example, write down the definition you had in your mind first, and go around asking people and see what they think?

Again, take quasi-hereditary algebras as example; such algebra is required to possess a chain of inclusions of some special two-sided ideal (a filtration of the algebra). On the other hand, you can successively take the (Jacobson) radical for an algebra, and this gives also gives you a filtration of the algebra. $A \supset Rad A \supset Rad^2 A \supset \cdots$. But these are two entirely different filtrations, and they have different uses.

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What is a filtration of a ring? A sequence of additive subgroups? Subrings? Left/right ideals? Two-sided ideals? And similarly for algebras. Furthermore, do the inclusions need to be strict? Some definitions say that it is a descending sequence of abelian subgroups $R_n$, such that $R_mR_n\!\subseteq\!R_{m+n}$. Other definitions say it is simply a descending sequence of ideals $\mathfrak{a}_1\!\supseteq\!\mathfrak{a}_2\!\supseteq\!\mathfrak{a}_3\!\supseteq‌​\!\ldots$. What definition is the right one (more frequent one)? –  Leon Feb 23 '13 at 6:02
    
Furthermore, are filtrations ever indexed by more exotic sets (direct or inverse systems, groups, $\mathbb{N}^n$, etc)? Do there exist multifiltrations? –  Leon Feb 23 '13 at 6:03
    
I edited the answer in response to your first comment. For your second comment: to upset you even further, people use filtration for entirely different meaning in triangulated categories (it stills resemble the `original' one and that's why the naming). I think filtration is defined to simplify statements; instead of defined in hope of generalising certain phenomenon. May be one can define filtration indexed by strange sets; I haven't seen any, and I think it is because there is no use to such things - you may want a grading instead of filtration in these cases. –  Aaron Feb 23 '13 at 14:41
    
Thank you for your answer and warning about different terminology. Regards! –  Leon Mar 1 '13 at 4:49

"Question 1: Do these definitions apply only to commutative rings/algebras? Are gradings used in noncommutative algebra?"

Yes, they are used. For example, see

Bahturin, Y.A.; Zaicev, M.V. Semigroup gradings on associative rings. Adv. Appl. Math. 37, No. 2, 153-161 (2006).

Bahturin, Yuri A.; Parmenter, Michael M. Group gradings on integral group rings. (English) Giambruno, Antonio (ed.) et al., Groups, rings and group rings. Proceedings of the conference, Ubatuba, Brazil, July 26–31, 2004. Lecture Notes in Pure and Applied Mathematics 248, 25-32 (2006).

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@YACP: Certainly. But I wrote an example only, without history –  Boris Novikov Feb 21 '13 at 22:42

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