Background: For a ring $R$, we denote by ${\rm\mathop Mod}(R)$ the category of all (say right) $R$-modules. If $R$ is pure semisimple, then it is known that ${\rm\mathop Mod}(R)={\rm\mathop Add}(M)$, for some $M\in{\rm\mathop Mod}(R)$, where by ${\rm\mathop Add}(M)$ we understand the full subcategory consisting of all direct summands of arbitrary direct sums of $M$ (see Proposition 2.2 in Stovicek's paper: Locally well generated homotopy categories of complexes). My question is about a sort of dual: if we denote ${\rm\mathop Prod}(M)$ the class of all direct factors(=direct summands) of arbitrary products of $M$, when is it true that ${\rm\mathop Mod}(R)={\rm\mathop Prod}(M)$? It is clear that if $R$ is pure semisimple as above, then the module $M$, for which holds the equality ${\rm\mathop Mod}(R)={\rm\mathop Add}(M)$, is product-complete, since ${\rm\mathop Add}(M)$ is closed under products, thus the equality ${\rm\mathop Mod}(R)={\rm\mathop Prod}(M)$ holds too. But otherwise?
My question is related to a question about the homotopy category of complexes: it is shown in the paper Brown representability often fails for homotopy category of complexes by Stovicek and myself, that the homotopy category of complexes over ${\rm\mathop Mod}(R)$ satisfies Brown representability if and only if $R$ is pure semisimple, and does not satisfy Brown representability for the dual, provided that ${\rm\mathop Mod}(R)\neq{\rm\mathop Prod}(M)$ for some $M$.
