## Abelian Categories and Module Categories

The blog’s been rather quiet lately, due in a large part to me being in research mode right now.  That also explains why my posts, when they occur, are mostly advanced.  I just don’t have much general-consumption math on the brain at the moment.

Today, I’d like to talk about some of the more basic things in the subjects I tend to work (as a compromise).  As you might recall, an abelian category is a category where the set of morphisms between any two objects, $Hom(A,B)$, is an abelian group; with some additional properties that make it nice enough to do things like homological algebra.  The classic example of an abelian category is $Mod(R)$, the category of finitely generated left modules of some ring $R$.

One can ask the question: how far is an arbitrary abelian category from being a module category?   One result in this direction worth knowing is the Freyd-Mitchell Embedding Theorem, which says that any abelian category has a fully faithful, exact embedding into some module category.  The principle use of this theorem is to make homological algebra proofs which assume the existance of ‘elements’ work.

But, how can we tell if an abelian category is equivalent to a module category?  As we will see, finding such an equivalence is the same as finding a sufficiently nice element in the category, called a compact progenerator.  It can also be interesting to find multiple progenerators, giving us non-trivial equivalences $Mod(R)\sim Mod(S)$, where we call $R$ and $S$ ‘Morita equivalent’.

We should start by considering a module category, and thinking about what we know aside from that it is an abelian category.  The first thing that comes to mind is that $Mod(R)$ has an object $R$, thought of as a left module over itself.  One thing we can do with $R$ is to look at maps from it into any other object, $Hom(R,M)$.  Of course, this is isomorphic to $M$ itself, with the left module structure induced by the right module structure on $R$ (being the first argument in a $Hom$ is dualizing, and so it swaps left and right actions).

Ok, so this is just the identity functor on $Mod(R)$, but now let’s pretend we are ignorant of the fact that our category is $Mod(R)$.  Picking this object $R$ still determines a functor $Hom(R,-)$, but where is this functor landing?  Well, $Hom(R,M)$ always has a left action by $Hom(R,R)$ (which is the ring $R$), given by composition.  Therefore, this functor maps to a concretely realized module category $Mod(Hom(R,R))=Mod(R)$, and we know by construction it is an equivalence.

Next, let’s take any abelian category $\mathcal{A}$.  If it is a module category, then there is some special object inside such that the “Hom out of” functor is an equivalence.  Therefore, let’s think about $Hom(O,-)$ for an arbitrary object $O\in \mathcal{A}$.  It determines a functor to $Mod(R):=Mod(Hom(O,O))$, and we want to know if there is an inverse equivalence going the other way.  However, our functor $Hom(O,-)$ does have a left adjoint $O\otimes_R -$, and we know that if a functor has both an adjoint and an inverse, they must coincide.  Therefore, it suffices to check if $O\otimes_R -$ is an inverse to $Hom(O,-)$.

Now, let’s assume that $Hom(O,-)$ and $O\otimes_R -$ are equivalences of categories.  Since they must be exact functors, $O$ is projective as both an object in $\mathcal{A}$ and a right $R$ module.

We also know that every object in $\mathcal{A}$ is isomorphic to one of the form $O\otimes_R M$, for some $R$ module $M$.  If we replace $M$ with a free resolution $R^{n_1}\leftarrow R^{n_2}\leftarrow$, then we get that $O^{n_1}\leftarrow O^{n_2}\leftarrow$ is a resolution of $O\otimes_R M$.  Hey, then every object in $\mathcal{A}$ has a resolution by sums of copies of $O$!  We say that $O$ generates the category $\mathcal{A}$.  A projective generator is often called a progenerator.

Technically, we have made the mild assumption that $O\otimes_R R^n=O^n$ even for infinite $n$, but this only needs that $O$ is a finitely presented $R$ module.  This requirement on $O$ is called being compact.

So what have we found so far?  A compact object $O$ for which $Hom(O,-)$ is an equivalence of categories is a progenerator.  Ah, but it’s not too hard to see that this is a sharp characterization.  This fact is usually called ‘Morita’s Theorem’:

Morita’s Theorem.  Let $O$ be an object in an abelian category $\mathcal{A}$, and let $R:=Hom_\mathcal{A}(O,O)$.  Then $Hom_\mathcal{A}(O,-)$ is an equivalence of categories between $\mathcal{A}$ and $Mod(R)$ with inverse functor $O\otimes_R-$ if and only if $O$ is a compact projective generator (‘progenerator’) in $\mathcal{A}$.  Furthermore, every equivalence between $\mathcal{A}$ and a category of modules arises in this way.

We haven’t completely justified every statement here, but we have given the rough outline.  To see why the last statement is true, that every equivalence arises in this way, assume the existance of the equivalence and let $O$ be the image of the ring $R$.

We can reap some immediate fruits from this theorem.  Let’s say we have a module category $Mod(R)$, and we want to know what other module categories $Mod(S)$ it is equivalent to.  This is a strong relationship between $R$ and $S$ called Morita equivalence.  We now know that $R$ and $S$ are Morita equivalent if and only if there exists a bimodule $_RT_S$ such that it is a compact progenerator of $Mod(R)$ and $Hom_R(_RT,_RT)=S$.  This first condition is the same as being finitely presented, projective and $R$ being a direct summand of $T^n$ for some $n$.

Well, we can think of some examples of these right away.  $T=R^n$, for instance, most certainly satisfies all the above requirements.  $Hom_R(R^n, R^n)$ is $M_n(R)$, the ring of $n\times n$ matrices with coefficients in $R$.  Thus, we have established the equivalence of categories between $Mod(R)$ and $Mod(M_n(R))$$M_n(R)$ and $R$ are the classic example of Morita equivalent rings, so this is comforting.

This idea, in particular the way I’ve presented it, beg to be generalized to derived categories.  After all, we replaced an element in $Mod(M)$ with a free resolution to see where it went in $\mathcal{A}$.  If we work with derived categories, we no longer need resolutions to go to resolutions, they can go to complexes; therefore, we no longer need projectivity.

In the context of derived categories, this process of finding an object which defines an equivalence to a derived module category is called ’tilting’, with the object called a ’tilting sheaf’.  The biggest success story in this direction is perhaps the Beilinson equivalence, which reveals the surprising fact that $D(Mod(\mathbb{P}^n))$ is equivalent to $D(Mod(Q_n))$, for a relatively simple quiver $Q_n$.

### 6 Responses to “Abelian Categories and Module Categories”

1. John Armstrong Says:

The principal use of this theorem is to make homological algebra proofs which assume the existance of ‘elements’ work.

Great post, but I just have to jump on one point here. Yes, this is a standard use of the Embedding Theorem, but there are simpler ways to do diagram chases as if you have elements. MacLane outlines one approach in CWM, and I ran through it last September.

2. Greg Muller Says:

That’s certainly a much better way to think of diagram chases, since the actual ring produced by Freyd-Mitchell is pretty horrible (if memory serves). However, Freyd-Mitchell is more useful as a short rebuttal to an audience member’s concerns about whether a given technique is valid, which is my principle use for it.

In this way, my relationship to Freyd-Mitchell is almost identical to my relationship with the theory of ‘Grothendieck Universes’, which I only think about when banishing set-theoretic concerns.

3. John Armstrong Says:

Theorems as tools of intimidation to keep the audience in line. I like it!

4. pritam Says:

Its really exlain short & nice way.

5. Evgeny Says:

I guess in the Morita theorem you should require that the abelian category has arbitrary direct sums.

Otherwise I can apply it e.g. to the category of finite-dimensional vector spaces, which is abelian and has a generator and get that it is equivalent to the category of all vector spaces.

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