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Given a ring $R$ consider $(R\mathbf{-Mod})^{op}$, the opposite category of the category of left $R$-modules. Since it is the dual to an abelian category and the axioms of abelian categories are self-duals, it is an abelian category itself and thus, by the Freyd-Mitchell Imbedding Theorem, has to be a full subcategory of $S$-Mod, for some ring $S$.

Is it possible to describe $S$ and the embedding in a particular nice form? At least for some special rings, I would like to see a construction of $S$ and the embedding which is as concrete as possible.

Dominik
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    If you restrict to finitely generated modules, then Morita duality is a very nice answer. I think it may also be nice enough to answer your question: Take S=Z, the embedding is Hom_R(-,E((+) S_i)) where S_i ranges over the simple modules. – Jack Schmidt Jan 06 '14 at 19:24
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    I retract my second comment. That embedding does not have image a full subcategory. – Jack Schmidt Jan 06 '14 at 19:38

3 Answers3

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This is not an answer, but a comment which should not be overlooked.

The Freyd-Mitchell embedding theorem does not apply to arbitrary abelian categories. It only applies to small abelian categories.

It is possible that $\mathsf{Mod}(R)^{op}$ has no fully-faithful exact embedding into some $\mathsf{Mod}(S)$.

The easiest example should be that of a field $R=K$; then $\mathsf{Vect}(K)^{op}$ is equivalent to the category of linear compact topological vector spaces over $K$, with continuous linear maps. This gives a faithful exact embedding $\mathsf{Vect}(K)^{op} \to \mathsf{Vect}(K)$, namely $V \mapsto V^*$. But this embedding is not full. And I really cannot imagine any fully-faithful exact embedding, because this would mean that we can encode continuous linear maps by abstract linear maps between certain modules (of course this is not a proof). Note that, however, $\mathsf{FinVect}(K)^{op}$ is an (essentially) small abelian category, which by the above functor becomes equivalent to $\mathsf{FinVect}(K)$, which is a full exact subcategory of $\mathsf{Vect}(K)$.

Now for general $R$, there is an injective cogenerator in $\mathsf{Mod}(R)$, for example $G:=\hom_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$. This means that $\hom_R(-,G) : \mathsf{Mod}(R)^{op} \to \mathsf{Mod}(R^{op})$ is faithful and exact. But again it is not full.

Perhaps one might hope for an embedding of $\mathsf{f.g.Mod}(R)^{op}$.

Community
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Martin Brandenburg
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    PS: I tried to show that $\mathsf{Vect}(K)^{op}$ doesn't embed fully-faithfully into some $\mathsf{Mod}(R)$, but this seems to be a hard problem. The image of $K$ "cogenerates" a subcategory of $R$-modules with endomorphism ring $K$. Which $R$-modules have as endomorphism ring a field? Notice that, for every cardinal $\kappa$, there is an embedding of $\mathrm{Vect}_{\dim \leq \kappa}^{op}$ (since this is essentially small). So the proof will somehow have to rely on long chains of infinite-dimensional vector spaces ... – Martin Brandenburg Jan 09 '14 at 14:40
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I think that if Vopěnka's principle is true, then $\mathsf{Mod}(R)^{op}$ can't fully embed in $\mathsf{Mod}(S)$ for any non-zero $R$ and $S$ (all the references that follow are to "Locally Presentable and Accessible Categories" by Adámek and Rosický): If it did, then $\mathsf{Mod}(R)^{op}$ would be bounded (Theorem 6.6), and since it is also complete it would be locally presentable (Theorem 6.14). But if a category and its opposite are both locally presentable, then the category is equivalent to a complete lattice (Theorem 1.64).

Jeremy Rickard
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An obvious comment that nobody has mentioned: If you pass to a higher universe, then $\mathrm{Mod}(R)^\mathrm{op}$ is small, and so the embedding theorem tells us that it embeds exactly into some $\mathrm{Mod}(S)$. The catch is that (with respect to the original universe) $S$ will be a large (i.e. proper-class-sized) ring.

You don't actually need the embedding theorem to see this, though. You can just take $S$ to be a class-sized power of an injective cogenerator in $\mathrm{Mod}(R)$.

EDIT Oh this is actually correct modulo not making sense. See the comment by Eric Wofsey below.

tcamps
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  • The last line is almost correct--you just need to take $S$ to be the endomorphism ring of a class-sized power of an injective cogenerator. – Eric Wofsey Jul 11 '16 at 16:38
  • But don't you have the same problem I had on my question (link for the benefit of bystanders), where this embedding need not end up being full? – tcamps Jul 11 '16 at 16:47
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    No, that's what the class-sized power does for you. If you know that every object has a monomorphism into the injective cogenerator $I$, then fullness is automatic. Indeed, in that case a monomorphism $A\to I$ will generate $\operatorname{Hom}(A,I)$ as an $\operatorname{End}(I)$-module, and so you can lift any $\operatorname{End}(I)$-linear map to a map in the original category by just looking where this generator needs to go. – Eric Wofsey Jul 11 '16 at 16:56
  • Oh right, I was being stupid. Let $\kappa$ be the size of the original universe, and $\lambda$ be the size of a bigger universe. The point is that $\mathrm{Mod}\kappa(R)$ embeds into $\mathrm{Mod}\lambda(R)$, by taking the powers of an injective cogenerator, you get an injective object such that every object of $\mathrm{Mod}\kappa(R)$ embeds into into it, and then there's a theorem telling you that this object is codense with respect to $\mathrm{Mod}_\kappa(R)$. It's something nice about abelian categories that lets you jump from being colimit-dense to being actually dense. – tcamps Jul 11 '16 at 17:01