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The Freyd-Mitchell embedding theorem states the following:

Let $\mathcal{A}$ be a small abelian category. There exists a unital ring $R$ and a full, faithful and exact functor $F\colon\mathcal{A}\rightarrow R\mathcal{Mod}$, where $R\mathcal{Mod}$ is the (abelian) category of (left) $R$-modules.

I am searching for (elementary and non-elementary, useful and non-useful) applications of this theorem. I am also interested in applications, for which one does not necessarly need the embedding theorem.

The only "real" application I know at the moment is all about diagram chasing:

Checking exactness of diagrams in arbitrary abelian categories can be done by checking those in the categories of $R$-modules, where one can apply element-wise diagram chasing.

(This can be made precise, what is e.g. done in Freyd's "Abelian Categories - An Introduction to the Theory of Functors".)

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    $\begingroup$ In my (inexpert) opinion, it is indeed mostly of "psychological" use, giving an a priori explanation of why certain contructions that are obvious and intuitive in $R$-Mod can be performed in any abelian category. Since I for one find the basic formalism of abelian categories somewhat tedious, I do appreciate the result for this reason. $\endgroup$ Apr 21, 2013 at 1:00
  • $\begingroup$ math.stackexchange.com/questions/361351 $\endgroup$ Apr 22, 2013 at 13:28
  • $\begingroup$ I would not call it just psychological. People wanted to understand how much like module categories a general abelian category must be. This was a good answer though it is not particularly useful. $\endgroup$ Aug 31, 2013 at 3:44

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An interesting class of examples comes from the following construction, also due to Freyd. Let $\mathcal{T}$ be a small triangulated category. Form a new category $\mathcal{A}$, with one object $I(u)$ for each morphism $u$ in $\mathcal{T}$. The morphisms from $I(u:A\to B)$ to $I(v:C\to D)$ are the pairs $(f:A\to C,g:B\to D)$ such that $gu=vf$, modulo those for which $gu=vf=0$. There is a functor $J:\mathcal{T}\to\mathcal{A}$ given by $J(X)=I(1_X)$. One can show that $\mathcal{A}$ is abelian, that $J$ is full and faithful, and that the essential image of $J$ is the subcategory of projective objects, which is the same as the subcategory of injective objects.

I think that the embedding theorem is much less obvious for these abelian categories than it is for the more usual examples.

As a special case, we can take $\mathcal{T}$ to be the category of finite spectra in the sense of stable homotopy theory. The embedding theorem, combined with the above construction, tells us that $\mathcal{T}$ embeds in the category of $R$-modules for some $R$ (which is not unique). Freyd also conjectured more specifically that in this case the stable homotopy functor gives a full and faithful embedding of $\mathcal{T}$ in the category of modules over the ring of stable homotopy groups of spheres (and he proved that many interesting consequences would follow from that). This conjecture is still wide open half a century later.

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  • $\begingroup$ What is multiplication on stable homotopy groups of spheres? (To make them ring? ) $\endgroup$ Apr 21, 2013 at 12:12
  • $\begingroup$ Would you mind pointing out a reference to this construction? Are there any more properties that $\mathcal{A}$ have, such as Frobenius? $\endgroup$
    – Aaron Chan
    Apr 21, 2013 at 15:13
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    $\begingroup$ Take $\alpha \in \pi_i^s$, $\beta \in \pi_j^s$. THey can be represented by maps $\alph : S^{i+j+k} \rightarrow S^{j+k}$ and $\beta : S^{j+k} \rightarrow S^k$ so that the composition is defined $\alpha \circ \beta : S^{i+j+k} \rightarrow S^k$ which then determines an element of $\pi_{i+j}^s$. From a more general perspective $\pi_*^s = \mathbb{S}^*(pt)$---the cohomology of a point where $\mathbb{S}$ is the sphere spectrum. The sphere spectrum is a ring spectrum so cohomology groups have a product structure. $\endgroup$ Apr 21, 2013 at 15:27
  • $\begingroup$ @Aaron: see ams.org/mathscinet-getitem?mr=209333 and the other papers by Freyd in the same proceedings volume. Unfortunately I do not think that the full text is online anywhere. $\endgroup$ Apr 21, 2013 at 15:50
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    $\begingroup$ The entire La Jolla volume is now online on Springer and can be accessed here (modulo subscription). The quality of the scans is pretty good. (@AaronChan) $\endgroup$
    – commenter
    Sep 1, 2013 at 15:36

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