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This is related to another question of mine. Suppose you met someone who was well-acquainted with the basic properties of rings, but who had never heard of a module. You tell him that modules generalize ideals and quotients, but he remains unimpressed. How do you convince him that studying modules of a ring is a good way to understand that ring? (In other words, why does one have to work "external" to the ring?) Your answer should also explain why it is a good idea to study a group by studying its representations.

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To explain the usefulness of modules, claiming it helps to understand the ring does not strike me as the right initial rationale. For example, is the main use of real vector spaces to give you insights into the real numbers? Nope. Similarly, I think a better motivation for caring about modules is to point out that they're the setting for linear algebra over a ring. Then illustrate why the module-over-a-ring viewpoint is useful in examples tailored to the other person's background (e.g., canonical forms for linear operators via modules over F[x], analogous to f.g. abelian groups). –  KConrad Mar 25 '12 at 4:04
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In short, I'd tell your friend: "If you believe a ring can be understood geometrically as functions its spectrum, then modules help you by providing more functions with which to measure and characterize its spectrum."

Elements of a module over a ring $R$ are like generalized functions on $Spec(R)$. We can talk about the support of a module element, or its vanishing set. More concretely, think of how global sections of a line bundle can act as functions you can use to define map into projective space.

When you glue together a module on open sets of a spectrum or a scheme, you get to glue using maps which are module isomorphisms, which are more flexible than the ring isomorphisms required to glue together a scheme. Borrowing intuition from smooth manifold land, the twist in the Moebius band (as a line bundle on the circle) is formed by gluing a copy of the reals to itself via multiplication by $-1$, a module map, not a ring map. This allows us to think of functions like $\cos(\theta/2)$ as being globally defined: as a map to the Moebius band.

In the same vein, when you have a representation $V$ of a group $G$, each element $v\in V$ gives you a nice evaluation map from $G$ into $V$, so lurking everywhere we've got these morphisms from our object of interest into a known object, which are nicely related to each other via the group laws. A fortiori, this certainly doesn't capture the full utility of group representations, but a priori I think it's a decent justification.

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The "categorical" approach: Understand an object not via intrinsic properties (rather ignore them, they are not part of the categorical data) but via the morphisms.

By choosing a category to work in, we choose the class of properties that are interesting. If we choose to work in the category of groups and group homomorphisms, then two groups with different underlying sets but having a group isomorphism between will not be distinguished.

A group representation is a group morphism G -> End(M) and this way we get properties about the group itself.

A module M over a ring R ist nothing else than a morphism R -> End(M) and so we get information about R from this morphism (or, better: from the collection of all these morphisms).

Then, you could continue asking "why should we do this and not study the ring/group intrinsically"?

My answer is: what is a group, to you? For me it is a set satisfying the group axioms - and two groups are the same if there is a group isomorphism between them.

The "intrinsic" study makes sense only if you want to distinguish different groups/rings that are isomorphic. But then you could also choose a different category, with less isomorphisms :-)

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Sure, but my point is why distinguish sources of the form End(M)? –  Qiaochu Yuan Nov 12 '09 at 20:16
    
In other words, your answer does not tell me why representations of groups are more interesting than group actions. –  Qiaochu Yuan Nov 12 '09 at 20:17
    
Yes, group actions are interesting in their own right. Representations are just the linear ones, thus easier to understand. The same holds for End(M) above. You could say to the unimpressed mathematician "you don't think linear algebra is useful?". –  Konrad Voelkel Nov 12 '09 at 20:20
    
Oh wait, maybe you're asking "how to convince somebody to use linear algebra". Then my answer would be: Don't try to. If they're happy with difficult math, let them do it. –  Konrad Voelkel Nov 12 '09 at 20:21
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@Konrad: I dont' think you have addressed Qiaochu's question. I guess what he's looking for, is why the study of all such endomorphisms from R -> End (M) (in the modules case, let's say) should give information about R, i.e. study of these endomorphisms as a whole. Your answer seems to be focusing on each such endomorphisms, thus not quite to the point. –  Ho Chung Siu Dec 5 '09 at 4:48
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I think the main reason is the flexibility of working in the category of $R$-modules rather than just with the ring $R$. For instance suppose we stick to rings - we have some ways of building new rings like localization and taking factor rings and limited ways of "building new things" - basically linear combinations of elements and maybe taking limits if $R$ is complete with respect to some topology.

At the level of module categories we still get all of this, torsion theories deal with localization (and make it clear that this is really an "internal" concept), instead of a quotient map we get useful adjoints, and we can still add and compose endomorphisms of $R$. But we also have lots of other structure to work with. We have all limits and colimits, possibly a tensor product, injective modules (which can have a lovely structure theory), duality, etc... So not only can we build a lot more objects but one can prove that just the existence of certain objects gives us a lot of information about the ring.

In a sense (which one can make precise) the category of $R$-modules is the same as $R$ (which is the same as $D^{\mathrm{perf}}(R)$ with its tensor product for $R$ commutative with unit) so the distinction between a ring and its modules shouldn't really exist!

I thought I would add the following quote which sprang to mind when I read the question:

"Grothendieck would later describe each sheaf on a space T as a “meter stick” measuring T."

taken from McClarty's article The Rising Sea: Grothendieck on simplicity and generality I (which can be found here).

In the commutative with unit case this can be interpreted literally, however, I still think it is relevant (suitably adjusted) in the case of noncommutative rings.

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I always thought one should regard issues like ring theory/module theory or theory of (abstract) groups/ representation theory of groups in an analogous manner to theory of abstract manifolds/embeddings of manifolds. So you can disentangle "mixed" notions and work out the concepts more clearly. It's not like embeddings of manifolds were "more interesting" than the theory of manifolds - on the contrary, the gist is distinguishing both.

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If the ring $R$ is ${\mathbb R}$, then the modules are vector spaces. Vector spaces are more interesting and more useful. Modules over an arbitrary ring, then are analogous to vector spaces. Moreover, the places where the analogies break down form the interesting parts of the theory.

For groups and representations, I think the question is different. A group can be thought of as a set of symmetries --- the original meaning. Symmetries of what? Well, for example a particular vector space, or rather, for a family of vector spaces. The relations among those spaces tells you about the group.

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This relates to an earlier question you asked:

What representative examples of modules should I keep in mind?

My answer is the same: look at the frontispieces of Miles Ried's book undergrad commutative algebra, which visualize this central fact: Modules to rings are what bundles are to varieties.

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Would now be a good time to admit that I also have no intuition for why bundles are useful? –  Qiaochu Yuan Nov 13 '09 at 0:31
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A better time was when the original question asked, but now is a good time too. One important feature with bundles/sheaves (depending on the category) you can cook a lot of numerical invariants. E.g. if you want to prove that a scheme is general type, one standard technique is to represnt the canonical sheaf as effective plus ample; many times this reduces to a lot of computations in the Chow ring, which is much simpler then finding many sections of the pluri-canonical family. –  David Lehavi Nov 13 '09 at 6:05
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Cuz otherwise it'd be like getting to know a bicycle without riding it.

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I want to answer your question twice: first with a "top-down" approach and second with a "bottom-up" approach. Let me limit myself to the first answer here and see how I do.

I claim the following analogy:

abstract groups : group actions on sets :: abstract rings : linear actions of rings on abelian groups (= modules)

I will take it as mostly self-evident that it is desirable to study groups acting on sets. If you are not doing this -- but thinking of groups only as sets with a certain law of composition -- then you are thinking about groups "in the wrong way". You are missing out not only on powerful tools for studying abstract groups (e.g. the Monster was constructed as the automorphism group of a certain algebra), but also, even more importantly, on why groups are important and interesting to mathematics: they come up as automorphisms of things, not (or rather, rarely) abstractly.

The way to think about a group action is that you have a set S, and it has an automorphism group, Sym(S), the group of all bijections from S to itself. Then an action of G on S is simply a homomorphism of groups G -> Sym(S). More generally, if x is any object in a category C, then it has an automorphism group Aut(x), and one can think of a homomorphism from an abstract group G to Aut(x) as a group action on x.

Now in place of a set, we take an abelian group M. This has more structure -- apart from a group Aut(M) of Z-linear automorphsims, it also has an endomorphism ring End(M): the ring of Z-linear maps from M to itself. Note that End(M) is in general noncommutative, so this construction is more general than any "ring of functions" construction in (commutative!) algebraic geometry.

So given a ring R, the analogy is completed by considering ring homomorphisms R -> End(M). As for rings, this provides a bridge between the abstract notion of a ring and the "real world" notion of endomorphisms of an abelian group. Moreover, just as the notion of a symmetry group of a set generalizes to that of an automorphism group of an object in a category, similarly, any object x in an abelian category C has an endomorphism ring, and hence one can consider "ring actions" of an abstract ring R on x, via homomorphisms R -> End(x).

In the first part of the analogy, a distinguished role is played by the category of [left, or right] G-sets for a particular group G. In particular, this provides a way for every group G to be the precise automorphism group of some object in a category: it is the automorphism group of itself. It is of course not true that any abstract group is equal to the full symmetry group Sym(S) of a set S, so this is an important construction.

In the second part of the analogy, a distinguished role is played by the abelian category of [left, or right] R-modules. In particular, End_{R-Mod}(R) = R, so every ring is the precise endomorphism ring of some object in an abelian category. (I am pretty sure that not every ring is isomorphic to the full endomorphism ring of an abelian group, although this is less obvious than the other case. It might make a good question in its own right...)

This is I think the right "general" answer to the question "Why is studying modules of a ring a good way to understand that ring?" A different kind of answer would give instances in commutative algebra when theorems about rings are proved using modules. I'll try that at some future point.

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Thanks Pete! Could you recommend a book about ring/modules/group-theory with this view as its driving force? I think I am lacking this kind of insight. –  Jose Brox Nov 16 '09 at 22:17
    
I like this answer. Thanks! –  Qiaochu Yuan Nov 16 '09 at 22:27
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The introduction to section 12.A in Isaacs Algebra: A Graduate Course talks a little about this view. –  David Dynerman Mar 28 '11 at 19:29
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+1. Beautifully written. –  Dejan Govc Mar 24 '12 at 23:36
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First, category of commutative ring is opposite category of commutative affine schemes(whether in classical algebraic geometric sense or presheaf view point sense)

Second, considering an affine scheme (Spec(R),O),where R is a commutative ring. Qcoh(SpecR,O) is R-mod (category of R-modules). You should know, we can reconstruct the affine scheme (SpecR,O) from this R-module up to isomorphism. So study the ring and module category over it is the same. (Actually this is from Grothendieck view point: to do geometry, we just need category of sheaves on "would be space", this view point is supported by Gabriel-Rosenberg reconstruction theorem).

But, the module theory is just the theory of linear algebra. It is much easier to deal with than theory of ring. In fact, if we study quasi compact and quasi separated scheme. According to Barr-Beck's theorem(in particular,Grothendieck flat descent),all the algebraic geometry turns into linear algebra.(for affine scheme, we deal with module over monad, for non affine schemes, we deal with comodule over comonad, for all semiseparated scheme(all algebraic varieties), we even have simpler form for comonad). But I have to remind you. I assume you are talking about unital ring.

For non-unital ring, we can still consider R-module as category of associative action to some linear vector space. This is just the quasi coherent sheaves on quasi affine scheme(which is called cone). In fact, this is also not very difficult to deal with. A good reference for general framework is "Des categories Abéliennes" or more recent, O.Gabber's work on almost ring theory.

For your question about studying representation theory of G. The reason is the same. We can use Tannaka formalism to reconstruct locally compact group from its representations. Moreover, if G is an algebraic group. Then all algebraic geometry on group scheme can be turned into study on Rep(G) which is a symmetric monoidal category with linear algebra.

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This is a bit late, but here is one example I like:

Theorem. A localization of a regular local ring at a prime ideal is still regular.

One way to prove this is to deduce it from

Theorem. Let $R$ be a local ring. Then the following are equivalent:

  1. $R$ is regular

  2. Every $R$-module has a finite length projective resolution

  3. The residue field has a finite length projective resolution.

(To use it, let $P$ be the prime ideal. Since $R$ is regular, $R/P$ has a finite length projective resolution. Now localize--this is exact, so we get a finite length $R_P$-projective resolution of $(R/P)_P$, which is the residue field of $R_P$)

This stuff is in Chapter 19 of Eisenbud's Commutative Algebra.

It's not clear to me how one would try to prove the first theorem from the definitions of regular.

Edit: fixed some mistakes

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One way to convince the guy would be to make him list interesting questions he can make about rings, and show him which can be solved by looking only at the category of modules.

Somewhere in the middle of that conversation, one should make the point that it may very well happen that two different rings have equivalent categories of modules, so in answering any such question we can change the ring.

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In homage to Serge Lang, I might suggest that your friend pick up any book on Morita theory and solve all the exercises.  Less facetiously, I might point out that many interesting ring-theoretic properties can be characterized, sometimes unexpectedly, by properties of the (right) module category over the ring.  For instance, consider what it means for all right $R$-modules to be injective, or what it means for all right $R$-modules to be projective, or what it means for all right $R$-modules to be flat, or what it means for all direct sums of injective right $R$-modules to be injective, or what it means for all direct products of projective right $R$-modules to be projective, or what it means for all flat right $R$-modules to be projective, or ...

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@Greg: +1. Indeed my own commutative algebra notes (as well as the class currently being taught out of them) take this perspective a lot. For instance, just today I proved that for a commutative ring $R$, all modules are projective iff $R$ is a finite direct product of fields. To my taste, a lot of the standard texts don't emphasize these sort of "inverse problems" as much as they should (e.g. where is this fact in Atiyah-Macdonald or Matsumura? perhaps some exercise, but I don't know where). –  Pete L. Clark Mar 28 '11 at 23:59
    
This is what is usually called 'the homological study' of ring theory –  Jose Brox Mar 29 '11 at 3:00
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This is closely related to Pete Clark's answer, but stated in a slightly different way that I personally find helpful. I think it's not too hard to convince people that when studying an abstract object, it helps a great deal to be able to "write it down concretely," i.e., to consider a representation of the object. It's not an accident that the word "representation" is used for a homomorphism of a group into the group of matrices over a field; matrices over a familiar field like $\mathbb{C}$ are, intuitively, "more concrete" than an arbitrary abstract group and hence can be thought of as providing a concrete representation of the group. When you write something down concretely, often some structure will emerge that is not immediately evident from the original abstract definition. For example, the character table is an important invariant of a finite group. I don't know how you would come up with this invariant without considering representations of the group.

Similarly, modules are representations of rings. I think the Artin-Wedderburn theorem is a good illustration of the usefulness of considering representations of rings. Even if your interest is only in rings themselves, it's clearly an important result that you can classify all (Artinian) semisimple rings as products of matrix rings over division rings. If you possess the concept that a ring can be represented as a matrix ring, then it is not too shocking (and may even seem natural) that something like the Artin-Wedderburn theorem should be true, and moreover you can even see that to prove it you should somehow construct the matrix rings by having the original ring act on something. Without the concept of a representation (or equivalently a module), I don't know how you would proceed; it seems like a difficult and clumsy task at best.

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I guess they are dozens of points of view. Here are the one that I found most useful (for group representations) :

  • Studying the linear representations of a group is a way to linearize the problem. You thus get all the powerful tools of linear algebra (diagonalization, trigonalization...). The simplest (yet striking) example I know of that principle : you can show that every finite commutative group is a direct product of cyclic groups (essentially by decomposing its regular representation).

  • Studying the complex representations of a group $G$ is equivalent to studying modules over $\mathbb{C}[G]$ (you can adapt the algebra depending on the setting). So you may view it as analysis on $G$. A well known example : roughly, the theory of Fourier Series gives you a decomposition of $L^2(\mathbb{R}/\mathbb{Z})$ as a sum of irreducible representations of $\mathbb{R}/\mathbb{Z}$.

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A simple example: If you have a representation of a group presented by generators and relations, you have an explicit algorithm to solve the word problem for this group. Another example: the proof of Burnside's theorem.

A representation of an algebraic structure gives, generally speaking, additional tools for understanding the structure. See for instance the topological proof of the Nielsen-Schreier theorem.

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I'm afraid I don't fully understand what you are driving at in your first sentence, since we know by Cayley's theorem that every group acts canonically on the underlying set of itself. –  Todd Trimble Jun 26 '13 at 0:49
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If your friend knows well rings and ideals, he has certainly encountered quotient like $I/J$, hasn't he? What are they for him? sets? abelian groups? This point of view is wanting. You can formulate basic and very useful results, such as (for $A$ local, noetherian): if $I/mI$ has a set of $n$ generators, so has the ideal $I$ (Nakayama), only if you can see $I/J$ as a module...

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Well, $I/J$ is a ring, so that argument is not going to convince the extremists :-) –  Mariano Suárez-Alvarez Jun 11 '13 at 0:17
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Mariano: a quotient of two ideals is not a ring, just as an ideal in a ring (other than the whole ring) is not a subring. At least not in commutative algebra. –  KConrad Jun 11 '13 at 4:27
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