TITLE: An explicit imbedding of $(R\mathbf{-Mod})^{op}$ into $S\mathbf{-Mod}$
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: 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.