Talk:Module (mathematics)

What is a semimodule?
The word redirects here but is not used in the article. Equinox (talk) 06:58, 11 September 2015 (UTC)


 * In a nutshell, a semimodule is just like a module, except that the underlying abelian group is replaced with an abelian semigroup, so the elements do not necessarily have inverses. For example, the set of natural numbers is a semiring, and it is a semimodule over itself just as any ring is a module over itself. I think an argument could be made to redirect this to "semiring" instead of "module." Rschwieb (talk) 13:59, 11 September 2015 (UTC)

Add a Link Please
In the second paragraph the term "abelian" should be linked to our relevant article for ease of use. I'm not a skilled editor, so request someone else add this link. Thx — Preceding unsigned comment added by 97.125.86.137 (talk) 23:12, 14 June 2016 (UTC)

Proposition: Latex conversion
Proposition. Latex should be preferred as the default for mathematical typography on this page because:

1. Usage of multiple templates or raw unicode leads to a LOT of rendering variability and browser interaction, whereas Latex focuses efforts of mathematical typography and rendering onto preferred community libraries. This makes usability very hard to test.

2. Latex provides many fallback options, including rendering to SVG or PNG.

3. It's easier for amateur Wikipedians to copy-paste Latex, and it's easier to follow along when the community has consistent style.

4. Maintenance becomes easier with uniformity. Across different math pages one may find raw unicode, a no-wrap template, a variables template, or a generic Math template.

5. Editing by source becomes very ugly with multiple styles.

6. Mathematical typography should be consistent at least within-page even if not between pages.

7. More popular peer math pages prefer this style, such as Linear Algebra.

SirMeowMeow (talk) 14:35, 22 January 2021 (UTC)


 * This has been widely discussed in Wikipedia, see MOS:MATH. The current consensus on WP:WikiProject Mathematics is to use templates math and mvar for simple inlines formulas, and otherwise. If you want to change this consensus, please, discuss it on WT:WPM.


 * For those links, where do you see the language which suggests there is consensus to use math and mvar? There is also language to suggest that encoding and typography should be locally decided, rather than decided on a Wikipedia math-wide level. We also have this statement: "This essay offers a comparison of different encodings and presentation of mathematical formulae. The three principal ones are the tag, raw wiki (or HTML) code, and "texhtml" templates. The  and "texhtml" encoding may have different presentations for registered users, depending on user preferences and personal styles." This would suggest that mvar is not among the top 3 choices for encoding. There is a LOT of rendering variability which occurs when we use multiple templates, whereas rendering with Latex is more battle-hardened and continuously improved upon. Some of the rendering is typographically AMBIGUOUS.


 * When you use something other than, you are losing mass-browser support, huge amounts of accessibility work, and multiple-fallback solutions. Doing so ought be a deliberate choice made for conscious tradeoffs.  SirMeowMeow (talk) 14:21, 22 January 2021 (UTC)


 * Again, there is no point in making your case here. Suggested changes to the consensus should be discussed at WT:WPM as suggested above. Rschwieb (talk) 15:11, 22 January 2021 (UTC)
 * MOS:FORMULA says
 * MOS:VAR says
 * and
 * So, please, respect wikipedia rules. Thanks. D.Lazard (talk) 15:15, 22 January 2021 (UTC)
 * So, please, respect wikipedia rules. Thanks. D.Lazard (talk) 15:15, 22 January 2021 (UTC)


 * (1) MOS:FORMULA says not to proceed BOLDLY and to build consensus on the talk page. That suggests that local talk pages are precisely the right place to build consensus.
 * I also refer to this passage Manual of Style/Mathematics, which suggests that while large-scale changes may be controversial, one should refer to clear improvements: "Proposed changes should generally be discussed on the talk page of the article before implementation. If there is no positive response, or if planned changes affect more than one article, consider notifying an appropriate Wikiproject, such as WikiProject Mathematics for mathematical articles."
 * (2) The citation to the Wikipedia Arbitration Committee is about style. It's difficult to believe that an author's style is to surrender control for more rendering variability, where the author is less aware of whether their formulas will even appear correct. Will the identity matrix look more like the number 1 depending on browser interactions? Is that a question of style? And is the homogeneity of encoding a matter of screen reader accessibility (heterogeneous markup is a death sentence to screen readers)? What is this concept of style to a screen reader?
 * (3) It is not obvious why there will be edit wars over a matter of encoding or rendering, especially as the rest of the world outside of Wikipedia has very strong consensus for Latex and MathML. Is there actually a problem of people converting Latex into a myriad of math templates?
 * (4) I refer to this passage from Manual of Style/Mathematics: "Even for simple formulae the LaTeX markup might be preferred if required for uniformity within an article."
 * (5) I refer to the page Help:Displaying a formula, which is almost entirely full of Latex and MathML examples. This was the start of many people's on-ramping for learning to encode math on Wikipedia.


 * This is a question of whether UNDERNEATH the text there will be one kind of element tag versus another, and how that impacts site accessibility. I'm here to argue why there are net positive reasons for local style consistency on this page, but so far I've been getting no response to my actual arguments.


 * Improved site accessibility for screen readers, reduced variability of rendering due to browser interactions, massive technological support and great fallback solutions, etc. are all well known arguments about why there is non-trivial BENEFIT. Where does focus on style fit in all of this?


 * SirMeowMeow (talk) 17:47, 22 January 2021 (UTC)

Examples - K is not defined
The examples refer to K but that is not defined anywhere. — Preceding unsigned comment added by Gcsfred2 (talk • contribs) 13:13, 3 July 2021 (UTC)
 * The sentence "If K is a field" defines K. More precisely, it specifies K sufficiently for giving sense to what follows. D.Lazard (talk) 13:35, 3 July 2021 (UTC)

Is it reasonable to say that modules generalize abelian groups?
Usually when we generalize something, we drop unneeded axioms and make something work in a more general setting.

However, when comparing modules and abelian groups, modules contain strictly more axioms than abelian groups.

Wouldn't it therefore make sense to say that abelian groups generalize Z-modules, and not the other way around?

Of course, this "generalization" is rather superficial, since of course dropping the scalar multiplication is hardly a generalization at all.

What's actually going on is that any abelian group can be blessed with a useful scalar multiplication operation, and when you do that, the axioms of a module are satisfied. To me, that does not mean that modules generalize abelian groups, but rather that there is a useful construction you can make for every abelian group that turns the group into a module. SchwaWolf (talk) 19:31, 30 May 2024 (UTC)


 * A concept A is a generalization of a concept B when B is a special case of A. Here, the abelian groups are exactly the modules over the integers, and the vector spaces are exactly the modules over a field. So, it is correct to say that vector spaces and abelian groups are special cases of modules, and so modules generalize both vector spaces and abelian groups. D.Lazard (talk) 21:03, 30 May 2024 (UTC)
 * To go from a vector space to a module means to drop the requirement that scalars must be invertible.
 * To go from an abelian group to a module means to define an additional scalar multiplication operation.
 * Therefore I think modules generalize vector spaces while abelian groups (trivially) generalize any R-module.
 * Every vector space (less general) induces an R-module (more general) but not every R-module induces a vector space.
 * Every R-module (less general) induces an abelian group (more general) but not every abelian group induces an R-module.
 * To put it yet another way,
 * abelian group -> module -> vector space
 * forms a sequence of algebraic structures where each structure (more or less) contains the same axioms as the previous one plus some additional axioms. Therefore they are sorted from most general to least general, and in particular abelian groups generalize modules, not the other way around.
 * To really emphasize why modules contain more axioms than abelian groups, consider that an abelian group imposes the existence of ONE set with certain criteria, while a module imposes the existence of TWO sets (one of which having the same abelian group axioms). SchwaWolf (talk) 14:39, 1 June 2024 (UTC)
 * It is true that every module is an abelian group. But one has also that every abelian group is a module over the integers, with the scalar multiplication defined naturally by $$nx=x+\cdots+x$$ ($n$ summands) for positive $n$, and $$(-1)x=-x.$$ So, one passes from abelian groups to modules by dropping the requirement that the ring is the ring of integers. A witness that this is a relevant generaliztion is that the Fundamental theorem of finitely generated modules over a principal ideal domain is a direct generalization of the Fundamental theorem of finitely generated abelian groups (almost the same statemrnt and same proof). D.Lazard (talk) 15:14, 1 June 2024 (UTC)
 * Let us consider a world where we have a new algebraic structure "jury module" M that is in all other respects a module, except we additionally require that there exists some function f : M -> Z where Z is the set of integers such that for every m in M, f(m) maps to the decision of a well-established jury (an integer from 1 to 10) where the jury decided how appealing that particular element is. If the module is infinite then the default rating is 6 and the jury decides their rating manually for some amount of elements, based on how they feel that day.
 * Do these jury modules generalize modules?
 * If they do (the argument being that "there are more jury modules than modules" since there may be several well-established juries and we may agree that if f(m)=6 for all m in M then M is a jury module automatically, meaning every module induces a jury module in an obvious way but there may be different ways to turn a module into a jury module) then I argue that the word "generalize" has completely lost its meaning and we are not on firm mathematical ground anymore.
 * If they don't, then I urge you to think about how this jury function is any different than imposing a scalar multiplication function (and a set of scalars) for some abelian group. After all, according to you, the action of doing that generalizes abelian groups. So why should jury modules not generalize modules?


 * EDIT: It took me a while but I finally understood what you meant. If we were to treat abelian groups as Z-modules in the way that you described, then in that case YES, modules WOULD be a generalization of abelian groups (instead of having the set Z and a particular function, you have a general ring R and any function you choose that satisfies some criteria that the previous function satisfied). In fact, I think this is an excellent idea, since in practice you often do need the scalar multiplication when working with abelian groups, and that's why I've seen many authors define it as a "notational shorthand". Would it make more sense to include it in the definition? In my opinion, absolutely! However, that's unfortunately not the current world we live in, and at present an abelian group is a very barebones set of axioms that, as far as I can see, is just an R-module with some of the axioms dropped out (and thus a generalization).


 * SchwaWolf (talk) 15:45, 1 June 2024 (UTC)
 * The notion of module is a generalisation of that of abelian group because (as explained by D. Lazard) every abelian group has a structure of a module (of the ring ZZ) which allows to recover its structure as an abelian group. The converse is clearly not true (for instance a free abelian group of rank 2 has infinitely many non-isomorphic ZZ[X]-module structures).
 * Modules generalise abelian groups because the former allow to deal with many more objects. D. Lazard's example of the structure theorem for f.g. modules over PIDs is an excellent illustration of why this is useful: it shows that the concept of module allows you to fit the structure theorem for fg abelian groups, and the classification of conjugacy classes of matrices (among many other results) under the same umbrella.
 * On the other hand, the "forgetful" construction that you mention can be useful only insofar as you are interested in properties of modules that follow from their mere additive structure. To illustrate further why your conception of "generalisation" is fraught, let me say that all groups, rings, etc. are "generalisations" of sets as they are just sets with added structure: but this will not get you far if you want to prove results about groups or rings.
 * You can probably express all of the above in categorical language if you fancy that kind of thing. Cheers, jraimbau (talk) 07:17, 2 June 2024 (UTC)
 * An abelian group does not have the structure of a module because an abelian group does not take the existence of scalar multiplication (let alone the existence of a distinct set of scalars) as an axiom.
 * The converse is somewhat true because a subset of the axioms of a module forms the axioms of an abelian group.
 * You note that a free abelian group of rank 2 has infinitely many non-isomorphic modules associated with it but if anything, doesn't that just solidify the point that an abelian group is a more general set of axioms?
 * Unfortunately I can't comment on the structure theorem nor a category theory approach because that is beyond the scope of my current understanding.
 * I disagree with the idea that all groups and rings are generalizations of sets. They are sets with additional axioms imposed on them, so evidently they are not generalizations, and I would not really call them "special cases" of sets either because a group, for example, is not just a set, but a tuple also containing an operation defined over the elements of that set. In particular, it's not a set with some of the set theory axioms dropped out, which would in my view make it a generalization.
 * I don't understand how you're simultaneously claiming that my reductio ad absurdum view of generalizations is both fraud, and yet you're using the exact same notion to claim that modules generalize abelian groups. Could you try to enlighten me on where the analogy breaks down? SchwaWolf (talk) 13:10, 4 June 2024 (UTC)
 * This is easily explained : i made a mistake while typing my answer, your contention regarding groups vs. sets would be that sets are generalisations of groups (not the reverse). The rest stands. jraimbau (talk) 16:19, 4 June 2024 (UTC)
 * So you think that sets generalize groups because groups contain more axioms than a set?
 * Okay... wouldn't that mean that abelian groups generalize R-modules because an R-module contains more axioms than an abelian group? SchwaWolf (talk) 13:13, 6 June 2024 (UTC)
 * It would. That's why the second proposition is useless.jraimbau (talk) 14:12, 6 June 2024 (UTC)
 * Sorry, what do you mean by "the second proposition"? Do we in fact agree that abelian groups generalize R-modules? SchwaWolf (talk) 08:14, 7 June 2024 (UTC)
 * I mean that "abelian groups generalise R-modules" is a completely useless proposition, as is "sets generalise groups". Sorry for not being clearer. jraimbau (talk) 08:19, 7 June 2024 (UTC)
 * But certainly either A generalizes B or B generalizes A or neither. Certainly it can't be that A generalizes B and B generalizes A, or what do you think?
 * So if abelian groups generalize R-modules, then in particular it means that R-modules do NOT generalize abelian groups. That's certainly not a useless result as is evident by the length of this comment thread. SchwaWolf (talk) 12:31, 7 June 2024 (UTC)
 * Abelian groups DO NOT generalise R-modules in any useful sense. The converse is true as illustrated by the example given by D. Lazard. jraimbau (talk) 14:03, 7 June 2024 (UTC)
 * Do you, or do you not, agree with the idea that A cannot generalize B while B already generalizes A? If you agree with that idea, then you can't have an additional remark like "in any useful sense". If abelian groups generalize R-modules in any sense whatsoever then we can deduce that R-modules do not generalize abelian groups. SchwaWolf (talk) 14:23, 12 June 2024 (UTC)
 * R-modules are abelian groups and sets, exactly as do rings and fields. Similarly, all are sets. "Generalizing" does not mean dropping some axioms. It means enlarging the context in a way that allow applying some properties to new situations. Here, almost all important properties of abelian groups remain true for modules over a principal ideal domain. Similarly, principal ideal domains are generalizations of integers, in the sense that the fundamental theorem of arithmetic and many other properties of integers can be extended to them. D.Lazard (talk) 14:55, 7 June 2024 (UTC)
 * If you define integers as a certain kind of principal ideal domain then I agree that principal ideal domains generalize integers because then, a principal ideal domain is just the same requirements that you'd require for the integers except you don't require that the positive elements are well-ordered.
 * There's something weird going on because I would never say that an R-module is an abelian group or that an abelian group is a set. An abelian group, to me, is a tuple of both sets and operations, while a set is just a set. So how could they be the same thing in any reasonable sense?
 * Of course one has to be careful with what equality really means, since in a pedantic sense you could argue that the natural number 1 is not the same as the real number 1 on the basis that they have different set-theoretic constructions. But that's obviously not a helpful notion of equality.
 * Here, on the other hand, I don't think it's pedantic at all to say that a set and a group are two different things and neither equals the other, I think that's just common sense.
 * "It [Generalizing] means enlarging the context in a way that allow applying some properties to new situations"
 * How is that different from saying that a vector space generalizes R-modules because we enlarge the context in a way that we can apply some properties to new situations, like for example we can now invert scalars and thus show that each basis has equal cardinality? SchwaWolf (talk) 14:21, 12 June 2024 (UTC)
 * You would never say that a group is a set, but most mathematicians do. For example, the article Group begins with "In mathematics, a group is a set with an operation that satisfies the following constraints: ...". Wikipedia must follow the common usage of mathematicians rather than your own thoughts.
 * Also you discuss on the meaning of "generalization" as if it would be a mathematical concept subject to a formal definition. This is not the case here. The word is herefor informally explain the relatioship between the topic and other related concepts. In other words, it is used here as jargon that is useful for explaining the context. So, the common use of the word by mathematicians is more important than a formal definition. If you don't like this use, replace it in your mind with "strongly related", this will do the job. D.Lazard (talk) 15:02, 12 June 2024 (UTC)


 * SchwaWolf seems to have essentially two objections to the idea that "module" is a generalisation of "Abelian group":
 * 1) Modules contain more axioms than Abelian groups, but a generalisation has fewer axioms. However, there is a confusion of two senses of "more axioms" here. "Field" has more axioms than "Abelian group", in the sense that it has an extra operation, and that extra operation is defined by axioms. "Abelian group" has more axioms than "group", not in the sense of having extra axioms defining an extra operation (or any other kind of extra structure) but in the sense that there is an extra axiom placing a restriction on the existing operation. It is in the second sense that a generalisation has fewer axioms, not the first sense.
 * 2) An abelian group does not have the structure of a module because an abelian group does not take the existence of scalar multiplication as an axiom. That doesn't matter: it has a scalar multiplication; that scalar multiplication is not included in the axioms because it follows naturally from the existing axioms, not because it isn't there. Suppose I choose to define something, let's call it a JBW group, with exactly the same axioms as an Abelian group, but with the additional requirement "There exists an operation called "scalar multiple" such that  if x is any element of the JBW group and n is any integer, then the scalar multiple  nx of n  and x is the element of the JBW group such that if n is positive then nx = x +  x + ... + x with n terms, and ... [continuation of the definition to cover n ≤ 0]". It logically follows from SchwaWolf's position that a module is a generalisation of what I have defined, as I have added an extra axiom. However, what I have defined is simply the concept of "Abelian group"; I have done so in an unnecessarily verbose way, stating something which would have been true whether I had stated it or not, but it is exactly the same, with the extra axiom actually adding nothing whatever to the concept. It would be totally unhelpful if we were to use the word "generalisation" in such a way that something is a generalisation of something else if we define it in one way, but not if we define exactly the same thing in different words.
 * started this discussion by saying "Usually when we generalize something, we drop unneeded axioms", which is of course perfectly true, but that is just a property of how we commonly define generalisations, not a definition of what "generalisation" means. A correct definition is the one which gave in answer to that opening post: "A concept A is a generalization of a concept B when B is a special case of A" or, equivalently, A concept A is a generalization of a concept B when every example of B is an example of A, but not necessarily vice versa. In the sense which D.Lazard then  went on to explain, the concept "module" is clearly a generalization of the concept "Abelian group". Endlessly trying to argue that that is not the case is really not constructive. JBW (talk) 18:53, 13 June 2024 (UTC)
 * "concept A is a generalization of a concept B when every example of B is an example of A, but not necessarily vice versa"
 * An abelian group is a generalization of an R-module because every example of an R-module induces an abelian group in a natural way but not necessarily vice versa. Consider, for instance, an F-module where F is the set of real numbers and the module is F^n (the n dimensional euclidean vector space). Now elements of this set form an abelian group in a natural sense, but what is the natural way to extend the abelian group F^n into an R-module for some ring R? It could be the vector space (F, F^n) or it could be the "spiced up abelian group" (Z, F^n) where Z are integers.
 * There's lots of possibilities in this direction, because you're ADDING axioms and you have different ways to satisfy those new axioms. When you generalize something, you drop some axioms, and the resulting structure is highly obvious, which is not the case here.
 * "that scalar multiplication is not included in the axioms because it follows naturally from the existing axioms, not because it isn't there"
 * I disagree with this statement. If it's not included in the axioms, then from the point of view of comparing different sets of axioms, it isn't there.
 * If we have two people Alice and Bob living in very restricted logical worlds where Alice's world is governed by the axioms of abelian groups and Bob's world is governed by the axioms of R-modules, then every true statement that Alice comes up with will correspond very naturally to a statement in Bob's world, however a statement in Bob's world does not necessarily carry over to Alice's world. What if Bob makes a statement about the ring R? Alice's world doesn't even have the language to talk about two different operations between elements! To me, this means that Alice's world (of abelian groups) is more general than Bob's world (of modules).
 * EDIT: I will add that many mathematicians attempt to generalize something as much as they can, and often by generalizing something, you end up with more stuff to do. But this isn't always the case. Sometimes if you drop some axiom, you actually lose very interesting results, and sometimes having a very specific set of axioms is important for studying a new kind of interesting scenario. Dropping all axioms ultimately leads you to be able to prove absolutely nothing. Just because something is more general, does not mean it's more "powerful", in fact it's quite the opposite! The more general a theory is, the weaker it becomes. But on the flipside you can apply it to more and more different scenarios. So that's what's going on. Abelian groups are WEAKER but MORE GENERAL than modules, however the generality is very superficial because you could define some trivial scalar multiplication (such as the one with integers) and suddenly any results about modules (which could potentially be stronger statements) also apply to your new structure that used to be an abelian group. SchwaWolf (talk) 14:02, 16 June 2024 (UTC)

i want to chime in here as well, because i am tired of people in "physics" acting like their gymnastics qualify as mathematics. analogous to the discussion above is this belief that semi-positive-definiteness is somehow a generalisation of positive-definiteness. the relaxation of the positive definiteness property has tradeoffs. using a statement from the poster above:


 * "concept A is a generalization of a concept B when every example of B is an example of A, but not necessarily vice versa"

well, in this case, that fails per Horn and Johnson's Matrix Analysis (2013):

"Corollary 7.1.7. A positive semidefinite matrix is positive definite if and only if it is nonsingular."

i'm tired of these physicists abusing numbers and equations under the guise of mathematics. stop calling positive-semidefinite ANYTHING a generalisation of positive-definiteness. it's not!!

it relaxes the set of solutions big time, and i can't see how the properties we appreciate for positive-definiteness simply carry over to the former "just because" we extended the solution set to include zero, which means any psd matrix is NECESSARILY NOT pd because its distinguishing attribute is also one that introduces singularity!! — Preceding unsigned comment added by 162.157.84.254 (talk) 03:17, 18 June 2024 (UTC)


 * Despite being a slightly off-topic rant, I found your comment interesting and I agree with you entirely! Thank you for the good thought-provoking post.
 * Any theorem about positive definite matrices carries over seamlessly to a statement about positive semi-definite matrices, so therefore positive definite is the more general setting, despite positive semi-definite matrices being the larger set with respect to inclusion.
 * If people somehow managed to internalize this idea then the original debate would also be settled. SchwaWolf (talk) 21:34, 23 June 2024 (UTC)