An explicit imbedding of $(R\mathbf{-Mod})^{op}$ into $S\mathbf{-Mod}$ 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.
 A: 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 $\mathsf{f.g.Mod}(R)^{op}$.
A: 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).
A: 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.
