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I am taking a course next term in homological algebra (using Weibel's classic text) and am having a hard time seeing some of the big picture of the idea behind homological algebra.

Now, this sort of question has been asked many times on forums such as mathoverflow (viz. here and here) so, let me try to make more specific what I'd like to understand, which isn't covered in either of the responses to these questions. I have a (with no small amount of help from the first link I posted) a fairly good intuition for the uses of homology in topology, what I am more interested in is the intuition for the use of homological algebra in "pure algebra". I want to understand better why studying what we study in homological algebra gives us valuable information about the ring over which we are concerned--what information does it exactly give us? Of course, for commutative rings the question is answered definitively by Morita's theorem which tells us that $R\text{-}\mathbf{Mod}$ is categorially equivalent to $S\text{-}\mathbf{Mod}$ implies that $R$ and $S$ are isomorphic as rings. Fine, I can see why this obviously gives us motivation to study some of the things we study in homological algebra, but the only problem is I have no intuition for why Morita's theorem should be true. Can anyone elucidate this, in the most simple terms possible?

While this very well may be equivalent to what I have asked in the above paragraph (if so, feel free to concatenate answers) I was wondering if someone could more fully explain Eisenbud's analogy that homological algebra is to ring theory as representation theory is to group theory (in your own opinion, I know you don't know what he was thinking).

Lastly, for me, to get a basic motivation/intuition it is necessary for me to see how powerful a subject $X$ can be, in the sense that it can answer questions which (ostensibly!) have nothing to do with subject $X$. The classic example being that Burnside's theorem has nothing directly to do with representation theory (there is no use of character theoretic language in its statement) yet the only "simple" proof uses character theory. Unfortunately, in the realm of pure algebra I have been able to find very few examples of such uses of homological algebra--the only exception being the Schur-Zassenhaus theorem. So, any (as elementary possible) applications of homological algebra to problems in more elementary algebra (group theory, module theory, ring theory, and to some extent [but preferably less so] commutative algebra) where the statements would seem to suggest that the proof could be self-contained, yet realistically requires homological algebra would be great.

Thank you very much friends, help with any of these questions would go a LONG way to helping a very excited, and eager learner of homological algebra.

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Dear Alex, the enthusiasm expressed at the end of your question is quite refreshing and forces me to try to give an answer! –  Georges Elencwajg Nov 11 '11 at 23:59
    
@GeorgesElencwajg I have some more in-depth questions I wanted to ask you--is it possible that I could email you? Thanks so much! –  Alex Youcis Dec 18 '12 at 2:55

7 Answers 7

up vote 17 down vote accepted

I certainly won't try to give a general philosophical answer to your question, but I'll mention a success story that persuaded specialists that homological algebra was an amazingly powerful tool in commutative algebra.

Auslander, Buchsbaum and Serre proved that a local noetherian ring is regular if and only if it has finite global dimension.
From this it is easy to deduce that the localization at a prime ideal of a regular local ring is still regular.
The statement of that result has nothing to do with homological algebra but since nobody had managed to prove it before, without homological algebra, this duly impressed algebraists.

Optional technicalities
Let me give some relevant definitions here.
A noetherian local ring $(R,\mathfrak m)$ is called regular if its maximal ideal can be generated by $dim (A)$ (=Krull dimension of $A$) elements.
This definition, due to Zariski, is a purely algebraic way of ensuring that an algebraic variety has no singularities.
The global dimension of the ring $A$ is the supremum of the projective dimensions $pd_AM$ of its modules $M$.
And $pd_AM$ is the infimum of the lengths of resolutions $0\to P_n\to...\to P_0\to M\to0$ of $M$ by projective $A$-modules $P_i$.

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Georges, thank you very much for this response! I am just starting homological algebra and so notions of global dimension and projective dimension are a little beyond me at this point (although, I shall learn them soon enough). That said, even with my minimal commutative algebra background (Atiyah-Macdonald) I can definitely understand your example. In that sense (despite the fact that its a little more advanced than I had wished) this is PRECISELY the kind of "success story" I was looking for! Thanks again! –  Alex Youcis Nov 12 '11 at 0:06

Another success of homological algebra is the standard proof of the First Brauer-Thrall conjecture:

Theorem. (Roiter 1968) A finite dimensional algebra over an algebraically closed field is either representation-finite or admits indecomposable modules of arbitrary dimension.(1)

This statement is, in principle, quite independent of homological algebra, yet we now prove it using what's called Auslander-Reiten theory, which is at its heart homological algebra (I don't know what Roiter's argument was)

(1) This is not the most general statement known to be true

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That should be Auslander-Reiten theory, not Artin-Reiten theory. –  mt_ Mar 3 '12 at 11:09
    
@mt_, indeed! Corrected. Thanks! –  Mariano Suárez-Alvarez Mar 3 '12 at 19:27

Rings appear in nature, most often, through their representations, that is, their modules. Modules are what you see of a ring experimentally.

If two rings have equivalent categories of modules, then they are indistinguishable in that you cannot tell them apart by looking at their modules. Most interesting things that you want to know about rings are things that you can learn about it from looking at their modules: it is therefore quite natural that two Morita equivalent rings share, for the most part, lots of properties.

In particular, you can compute the center of a ring from the knowledge of its modules alone (and the maps between them) and one can easily check then that two Morita equivalent rings have isomorphic center. In the special case in which the two rings are commutative, so that they coincide with their centers, they are then isomorphic, as you say.

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Hat trick!${}{}{}$ –  Alexei Averchenko Nov 12 '11 at 2:00
    
Mariano, thanks for your replies! I was well-aware of the applications of homological algebra to topology, but was unaware of the result of Roiter--it's a nice result! Thanks also for this post I'm commenting on, it definitely starts to make me realize the similarity of homological algebra to representation theory. I, in one of those embarrassing cases of right-in-front-of-your-nose-missing, never made quite the connection I should have between studying $R$-modules and studying $G$-modules. Thanks again! –  Alex Youcis Nov 12 '11 at 2:41

A famous application of homological algebra is to algebraic topology. It plays a key role there, where in fact it arose originally. The most elementary instance of this is that the Tor and Ext functors appears conspicuously in the Universal Coefficient Theorems, but homological algebra is really essential to topology.

McCleary's book on spectral sequences is a great place to find information about this —it does get into lots of the nasty, difficult technical details, but one can always get by by reading the first section of each chapter!

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Suppose you have a variety $V=V(f_1,\dots,f_n)$. One possible question to ask on $V$ is if you can define it by less equations? Equivalently, are there polynomials $g_1,\dots,g_m$ with $m<n$ such that $\sqrt{(g_1,\dots,g_m)} = \sqrt{(f_1,\dots,f_n)}$?

If there is such a set of polyonmials you can of course try to find them. But what if there isn't? How can you prove that?

Local cohomology is your answer:

Let $A$ be a noetherian ring, and let $I\subseteq A$ be an ideal. If $I$ can be generated by $n$ elements then for any $A$-module $M$ and for any $i>n$ you have that $H_I^i(M)=0$. Thus, to prove that an ideal cannot be generated by $n$ elements, it is enough to find a module $M$ such that $H_I^{n+1}(M) \ne 0$.

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I have no intuition for why Morita's theorem should be true. Can anyone elucidate this, in the most simple terms possible?

As Mariano says, this is a special case of a more general result, namely that the center $Z(R)$ of a ring can be computed from its category of modules. First we need the following general definition: if $C$ is a category, its center $Z(C)$ is the monoid of natural transformations from the identity functor $\text{id}_C : C \to C$ to itself. Written out more explicitly, the elements of the center are collections of morphisms $\eta_x : x \to x$ ($x$ an object of $C$) such that for every morphism $f : x \to y$, we have $$\eta_y \circ f = f \circ \eta_x.$$

In particular, each $\eta_x$ must be an element of $Z(\text{End}(x))$, and this necessary condition is sufficient when $C$ has only one object. So the center of a category is a commutative monoid that generalizes the center of a monoid, and if we think of rings as one-object categories enriched over abelian groups $\text{Ab}$ (this just means that the Hom-sets are abelian groups and composition is bilinear), then the center of an $\text{Ab}$-enriched category is a commutative ring that generalizes the center of a ring.

Summarizing, the center of a category consists of natural families of endomorphisms of its objects, and naturality implies in particular that each endomorphism is central.

Now, it turns out that the center of $R\text{-Mod}$ is still $Z(R)$, from which it follows that the center of a ring is a Morita invariant. It should be clear that every element of $Z(R)$ defines a natural family of endomorphisms of $R$-modules, so the only thing to show is the other inclusion, and this follows from the fact that an element of the center of a category is determined by what it does to a generator; for a full argument see this blog post.

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Koszul duality is a nice topic where classical constructions and homological algebra intertwine themselves. Consider, for example, the symmetric algebra $A=S(X^{*})$ and the exterior algebra $B=\wedge(X^{*})$ over a finite dim. vector space $X$; let us consider $\mathbb K=\mathbb R, \mathbb C$ as ground field for simplicity. $A$ and $B$ are very different object as vector spaces over $\mathbb K$ (this is our "classical level"); if we introduce homological algebra instead, we can find something really interesting.

In fact, from the homological algebra point of view $A$ and $B$ are augmented differential bigraded associative algebras with natural augmentations given by evaluation at zero. The differential is equal to zero for both $A$ and $B$.

The beautiful duality (called Koszul duality) given by the isomorphisms of associative algebras $A\simeq Ext_{B}(\mathbb K,\mathbb K)$ $B\simeq Ext_{A}(\mathbb K,\mathbb K)$ is very well known and it is the prototype of a wider class of examples. In general one can move to the level of categories with the Bernstein-Gelfand-Gelfand correspondence, to the realm of $A_{\infty}$-structures and deformation quantization (with a generalization of the classical Morita duality) or operadic structures simply starting from the classical triple $(A,\mathbb K, B)$.

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