Talk:Graded ring

Graded school

I have removed the disambiguation link to Graded School since the latter is a nick-name for a school in Brazil, and is not referred to in any way as "Graded algebra". The article does not even mention the term algebra. The concept is unrelated to that of this article, and is unlikely to be a useful disambiguation. This disambiguation smacks of internal spam, and I need to see some clear evidence that some significant number of Wikipedia readers might be confused before I am willing to see this disambiguation added. For more details, I would encourage editors to see the Wikipedia disambiguation guidelines for when disambiguation should be performed. Silly rabbit (talk) 22:09, 9 February 2008 (UTC)[reply]

Homogeneous parts?

"... is homogeneous if for every element a ∈ , the homogeneous parts of a are also contained in..." Homogeneous parts need to be defined. Rschwieb (talk) 01:11, 1 November 2011 (UTC)[reply]

Yes, that is confusing. Is it clear now? ᛭ LokiClock (talk) 05:26, 11 November 2012 (UTC)[reply]

Yep, those edits were helpful. Rschwieb (talk) 21:15, 11 November 2012 (UTC)[reply]

Requested move

The following discussion is an archived discussion of the proposal. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the proposal was moved. While I'll see to double redirects and such, I'll leave the lede rewriting to you. It's Greek to me. --BDD (talk) 16:15, 25 July 2013 (UTC)[reply]

Graded algebra → Graded ring – A more standard, in my humble opinion. -- Taku (talk) 00:31, 18 July 2013 (UTC)[reply]

The above discussion is preserved as an archive of the proposal. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

Homogeneous Ideals

Shouldn't homogeneous ideal be a separate entry? Currently it redirects here. Wishcow 12:22, 8 September 2013 (UTC) — Preceding unsigned comment added by Wishcow (talk • contribs)

It's only defined for graded rings, so not unless the information found here about homogenous ideals outgrows the article. ᛭ LokiClock (talk) 14:50, 8 September 2013 (UTC)[reply]

The result is no consensus. -KAP03(^{Talk • Contributions • Email}) 00:08, 29 May 2017 (UTC)

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Proposed Merge

Supposedly there has been a proposal to merge Graded Ring and Supercommutative Algebra. I say "supposedly" since I see no evidence that any serious proposal was made. A serious proposal would include a clear argument for the change. Furthermore, the 2014 proposal seems to have resulted in a 2013 discussion: lacking time travel, there is no reason to take this seriously. My two cents is to vote NO. My reasoning is exemplefied in the ledes of the two articles: the first sentences of the GR lede:"In mathematics, in particular abstract algebra, a graded ring is a ring that is a direct sum of abelian groups R_i such that R_i R_j \subset R_{i+j}. The index set is usually the set of nonnegative integers or the set of integers, but can be any monoid or group." While the first sentence of the Supercommutative Algebra is:"In mathematics, a supercommutative algebra is a superalgebra (i.e. a Z2-graded algebra) such that for any two homogeneous elements x, y we have yx = (-1)^{|x| |y|}*xy." Clearly, the two concepts are related but not the same. Even if we grant that a "superalgebra" is identical to a Z_sub2 - graded algebra, imposing the requirement that its commutation properties obey the above equation seem to make it enough of a specialization as to merit its own article.imho72.172.11.204 (talk) 16:19, 14 October 2014 (UTC)[reply]

~~The superalgebra is a special case~~No, the algebra need not be a ring. --Mathmensch (talk) 07:03, 1 October 2016 (UTC). In my opinion there should be no merger~~, but the concept should be mentioned on the page regarding the graded ring, in the "examples" section~~. --Mathmensch (talk) 06:51, 1 October 2016 (UTC)[reply]

As I understand, a superalgebra is precisely a graded ring over the index set Z/2Z (as opposed to Z or N.) Note a graded ring in this article is not required to be commutative. Since the article "Supercommutative algebra" is short, it makes sense to merger it to this article. When the section becomes larger, we can alway split it off. -- Taku (talk) 00:25, 25 November 2016 (UTC)[reply]

A supercommutative algebra is not necessarily associative, see Superalgebra. Thus, if a merge should occur, it should be with Superalgebra. I suggest

adding here a section on Z2 grading, with a link to Superalgebra
merging Supercommutative algebra into Superalgebra; this is supported by the fact that notation of Supercommutative algebra is not defined in the article, but only in Superalgebra.
possibly merging Supercommutative algebra and Superalgebra into Lie superalgebra; this is motivated by the fact that is seems to me that the only superalgebras that have been studied under this name are Lie superalgebras.

D.Lazard (talk) 10:38, 25 November 2016 (UTC)[reply]

I know but they only discuss associative ones (e.g., exterior algebra) and not Lie algebra; so, in terms of materials, the merger makes. I do agree we should have a more extensive treatment of Z/2-case. -- Taku (talk) 21:27, 27 November 2016 (UTC)[reply]

Ok, the main issue seems that there is no article on "graded non-associative algebra"; I don't know if there should be such an article. -- Taku (talk) 23:00, 27 November 2016 (UTC)[reply]

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

I'm probably being thick...

I'm probably being thick but I don't see how a ring can be defined to be a direct sum of abelian groups. That takes care of the ring addition, but what of its multiplication? 86.185.161.165 (talk) 18:17, 1 March 2020 (UTC)[reply]

Clarified. D.Lazard (talk) 18:57, 1 March 2020 (UTC)[reply]

Grading of the product

This article does not breath a sigh about the grading of the (associative) product itself. I'm not sure why. This becomes an issue when the algebra has two distinct products: the "ordinary" associative product, and also a Lie bracket or a Poisson bracket which may or may not be super-brackets. For these cases, a careful distinction is then made, so as to differentiate between a Poisson superalgebra and a Gerstenhaber algebra. Both of these algebras have a degree-zero ordinary (associative) product:

|ab|=|a|+|b|

where $|a|=\deg a$ is the grading of a pure element $a$ . Sounds good, right? Most commonplace products behave like that: e.g. the plain old ordinary tensor product, or the plain-old alternating product of differential forms.

The two algebras above differ in how they treat their Lie-bracket product: in one case, the bracket itself has degree zero:

|[a,b]|=|a|+|b|

In the other, it has degree -1:

|[a,b]|=|a|+|b|-1

and that's fine, too. The second form is used in defining the Gerstenhaber algebra, used in both BRST quantization and in the Batalin–Vilkovisky formalism. No problem here, either. However, it is pretty easy to get cross-eyed while walking through all the super-grading wiki pages, and the sore spot for this article is that the ring product in this article can be validly understood as either the "normal associative product" or the "Lie-bracket product", and when there's two, it can be either, and then the grading of the product itself begins to matter. What I'm asking for is some clever words that can say something like "products can have gradings, too, and it is not necessarily the case that the grading is zero, and here are examples." Of course, lack of references on my part, and lack of patience, prevents my tackling this.

Perhaps I'm being childish; the article already says that

R_{m}R_{n}\subseteq R_{m+n}

which can be interpreted as saying $|ab|\leq |a|+|b|$ because $a\in R_{m}$ and $b\in R_{n}$ and $ab\in R_{m+n}$ and so nothing is violated. The issue is, I guess, notational conventions across ,ul;tiple articles, and examples where $\subseteq$ is a strict $\subset$ rather than strict equality.

Also FYI, take a look at Talk:Graded Lie algebra where it is explained that gradings can be abelian algebras and not just integers. Thanks to Deligne. 67.198.37.16 (talk) 22:57, 24 May 2024 (UTC)[reply]

This article is about graded rings and graded associative algebras; this was ambiguous in section § Graded algebra, and I just fixed the first sentence of the setion, which was nonsensical if "algebra" includes algebras that are not rings. The above post is about structures that have several products, one of them being a Lie algebra structure. So, this discussion should appear either in Graded structure or in Lie algebra. D.Lazard (talk) 09:00, 25 May 2024 (UTC)[reply]

In fact, the opening post seems to be more about filtered algebra that are graded as vector spaces, and not about graded algebras (strictly speaking). There are several possible definitions of a graded algebra; they may be graded vector spaces with bilinear product that is either "graded" or "filtered". More precisely, a graded vector space is a direct sum of vector space that is indexed by the naturel numbers:

A=\bigoplus _{n=0}^{\infty }A_{n}.

Such a graded vector space is also a filtered vector space, filtered by the

F_{n}=\bigoplus _{i=0}^{n}A_{i}.

I say that a product is "graded" if

A_{m}A_{n}\subseteq A_{m+n}

and "filtered" if

A_{m}A_{n}\subseteq F_{m+n}.

Both imply that

F_{m}F_{n}\subseteq F_{m+n}.

The above formula

|[a,b]|=|a|+|b|-1

suggest that it concerns a graded vector space with a filtered product that has the further property that the Lie bracket decreases the degree. As such a structure is rather common, it has certainly a name, but I ignore it. Certainly, Wikipedia should have an article that discusses the different sorts of graded or filtered algebras. This could be in a section of Algebra over a field or in a stand alone article possibly name Graded and filtered algebras. In any case, the redirect Graded algebra must be retargetted to this new content. D.Lazard (talk) 14:42, 25 May 2024 (UTC)[reply]

OK, thanks. Err, um, "yes but". Multiple issues.

There are just enough different kinds of gradings, between Lie superalgebra and Poisson superalgebra and another 3-5 super-articles, that one can get easily lost as to which grading is which, and it takes effort to find one's bearings. I feel like some super-overview is needed for all the different gradings.
All of these different articles link to graded algebra and that redirect lands here, so its mildly disappointing to find "ring" in the title. There are an assortment of physics articles that all land here. See Special:WhatLinksHere/Graded_algebra.
Yes, one can go from filtered to graded and vice-versa, but linking to filtered feels like overkill if one is just lost in the super-gradings. Yes, filters and flags are adjacent to all this, but it kind-of adds more noise to an already noisy picture. Though, I suppose a good article could be gracefully written.
FYI, it gets worse, and I know you don't want to hear it, but the folks that visit places like geometric algebra are working hard to wreck things for the rest of us. They have two gradings, one from Clifford algebra and another that is supersymmetric $(\varepsilon ^{2}=0)$ and they are proud of using that formula without using the prefix "super-". They point at deformation theory instead. Woo-hah. (In case you don't know, that's the friggin tip of the iceberg called "quantum".)
Graded structure is a veritable cornucopia!

I'm being Chatty Cathy here. My apologies. I have no plans to fix grading-anything just right now. My continuing mission is to explore strange new worlds, to seek out new life forms and to boldly go where no man has gone before. Specifically, Talk:Supersymmetric theory of stochastic dynamics. It hasn't fired it's phasers at graded algebra only by dint of pure luck. 67.198.37.16 (talk) 08:04, 30 May 2024 (UTC)[reply]

Confusion regarding homogeneous prime ideals

Zariski topology#Projective varieties links to Homogeneous Ideal which redirects here, which means Projective variety#Variety structure mentioning "homogeneous prime ideals" also goes here.

One property of a prime ideal is:

If $a$ and $b$ are two elements of $R$ such that their product $ab$ is an element of $P$ , then $a$ is in $P$ or $b$ is in $P$

However, in Projective variety#Variety structure it is only shown how scalar multiples of polynomials are equivalent. Consider scalar multiples of a homogeneous polynomial of degree m times n, which can be factored into homogeneous polynomials of degree m and of degree n. These smaller polynomials are not scalar multiples yet they have to be in the prime ideal. So I clearly don't understand something. AltoStev (talk) 12:55, 28 September 2024 (UTC)[reply]