I originally posted this on math.SE (https://math.stackexchange.com/questions/3438528/concrete-examples-of-freyd-mitchell-embedding) but since it's been a few days I figured I would crosspost it here. If this isn't the right level I'm happy to delete.

By the Freyd-Mitchell Embedding Theorem, any Abelian category admits an exact embedding into the category of modules over some ring. I'm not (currently) hoping to learn a proof, but instead I want to know if there are specific cases of this, aside from the obvious ones, which we can work out explicitly.

For example: is there a reasonably explicit way to describe the category of sheaves of Abelian groups on some space as a category of modules? Or the category of chain complexes of Abelian groups? Let's say the threshold for explicit is, at minimum, that we can cleanly describe the ring over which everything is a module.

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    $\begingroup$ The Freyd-Mitchell theorem doesn't state that any abelian category admits an exact embedding into a module category. It states that any small abelian category does. The examples you give (although they may admit such embeddings) are not small, so the existence of embeddings isn't an immediate consequence of the Freyd-Mitchell theorem. $\endgroup$ – Jeremy Rickard Nov 23 '19 at 10:49
  • $\begingroup$ @JeremyRickard good point. Given what you've said, how is it that Freyd-Mitchell is usually taken to justify diagram chasing? I guess we work with some essentially small subcategory containing all the objects we care about, and then embed that one instead? $\endgroup$ – Spencer Dembner Nov 23 '19 at 18:15
  • $\begingroup$ See also mathoverflow.net/questions/128178 $\endgroup$ – Neil Strickland Nov 24 '19 at 10:05

For some abelian categories it is also very easy to describe such a ring quite explicitly if the category you start with is similar enough to a module category.

Let's say you consider $\mathsf{Ch}(A\mathsf{-mod})$ and that $A$ is a $\mathbb{Q}$-algebra. Then $\mathsf{Ch}(A\mathsf{-mod})$ embeds as a full subcategory into $B-\mathsf{mod}$ where $$B:=A \otimes \mathbb{Z}\langle z,d\rangle / ([z,d]=d, d^2=0)$$ by mapping a chain complex $(X_\ast,\partial)\in\mathsf{Ch}(A\mathsf{-mod})$ to $\bigoplus_{n\in\mathbb{Z}} X_n$ and defining the $B$-action by $$\forall x\in X_n: z\cdot x := nx, d\cdot x := \partial(x).$$

In particular $z$ acts diagonalisable on this $B$-module, has spectrum $\mathbb{Z}$ and $X_n$ is exactly the eigenspace of $z$ for the eigenvalue $n$. It is easy to verify that every chain-map defines a $B$-linear map and vice versa so that this construction gives an equivalence between $\mathsf{Ch}(A\mathsf{-mod})$ and the full subcategory of $B\mathsf{-mod}$ consisting of those $B$-modules on which $z$ acts diagonalisable with spectrum $\mathbb{Z}$.

One can remove the condition $\mathbb{Q}\subseteq A$ by using slightly more complicated algebras, like $$C:=A \otimes \mathbb{Z}\langle e_n, d | n\in\mathbb{Z}\rangle / (e_i e_j = \delta_{ij} e_i, d e_n = e_{n-1} d, d^2=0)$$ This also algebra also acts on $\bigoplus_n X_n$: $e_n$ is the projection onto the $n$-component and $d x_n =\partial(x_n)$ for all $x_n\in X_n$.

In this way $\mathsf{Ch}(A\mathsf{-mod})$ gets identified with the full subcategory of $C\mathsf{-mod}$ of all those modules that satisfy $M=\sum_{n\in\mathbb{Z}} e_n M$.

This second construction can be generalised: $C$ is a special case of a category algebra of an $\mathsf{Ab}$-enriched category $\mathsf{Ch}$: It has object set $\mathbb{Z}$ and $Hom(n,m)=\mathbb{Z}\partial$. The category of chain complexes of $A$-modules is nothing else than the functor category (of additive functors) $Fun_\mathbb{Z}(\mathsf{Ch},A\mathsf{mod})$.

Generally: Every category that happens to be the functor category (of $k$-linear functors) $Fun_k(\mathsf{C},A\mathsf{-mod})$ from some $k$-linear category $\mathsf{C}$ into a module category can be embedded into $(A\otimes_k k[\mathsf{C}])-\mathsf{mod}$ in a similar spirit.

Even more generally: If $\mathsf{A}$ is an abelian category which you already know how to embed into a module category, then you can adapt this construction to get an embedding of $Fun_k(\mathsf{C},\mathsf{A})$ into a module category as well.

Even more generally yet: Any category can be embedded into a free $\mathbb{Z}$-linear category (using the same objects but the free abelian group generated by the original Hom-sets as new Hom-sets). Using this construction one can embed the category of all functors $Fun(\mathsf{C},\mathsf{A})$ into a module category. In particular you can do this with the category of open sets $\mathsf{Ouv}_X$ of a topological space $X$ and get an embedding of the category of $A\mathsf{-mod}$-valued sheafs on $X$ into $A \otimes \mathbb{Z}[\mathsf{Ouv}_X]\mathsf{-mod}$.

  • $\begingroup$ Thanks, I really like this. Could you explain how the argument uses that A is a $\mathbb{Q}$-algebra? $\endgroup$ – Spencer Dembner Nov 23 '19 at 18:36
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    $\begingroup$ You need that the eigenvalues $n$ all act differently on $X$. If, say $X$ where $2$-torsion, then $X_1$ and $X_3$ are both contained in the $1$-eigenspace. You could of course weaken that and ask that $X$ is torsionfree as an abelian group for example, but then you would get the full category of all chain complexes. $\endgroup$ – Johannes Hahn Nov 23 '19 at 21:10
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    $\begingroup$ I added another, more general construction that works without this simplifying assumption. $\endgroup$ – Johannes Hahn Nov 24 '19 at 0:46

Let me ignore any ambitious statement and say that your abelian category $\mathsf{A}$ has a projective generator $p$, this means that $p$ is projective and separates arrows.

Notice that the $p$-points $\mathsf{A}(p,a)$ of an object $a \in \mathsf{A}$ admit a structure of $\mathsf{End}(p)$-module via the multiplication $\phi \cdot c = c \circ \phi.$

Now the representable functor $\mathsf{A}(p,-): \mathsf{A} \to \mathsf{End}(p)\text{-}\mathsf{Mod}$ is faithful (because $p$ is a generator) and exact (because $p$ is projective), as desired.

When $\mathsf{A}$ is the category of abelian groups and $p$ are the integers, this recipe produces the identity functor on abelian groups.

You might be disappointed because I assumed that there is a $1$-object generator, while very often we deal with a family of generators, to take care this problem, notice that when $\{p_i\}_{i \in I }$ is a generator, the object $\prod_{i \in I} p_i$ is a generator, and if $\mathsf{A}$ has enough projectives, its projective cover $Q \twoheadrightarrow \prod_{i \in I} p_i$ is a generator too and is projective. As a result if your abelian category has enogh projectives and products, one can reduce any family of generators to a projective generator.

Cor. Let $\mathsf{A}$ be a complete abelian category with enough projectives and a generator $\{p_i\}_{i \in I }$, then there is a faithful and exact functor in the category of modules over the endomorphism ring of the projective cover of the product of the objects in the generator.


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