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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 $M$.

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I forgot: if there exits $M$ such that ${\rm\mathop Mod}(M)={\rm\mathop Prod}(M)$ then it is a test module for projectivity, that is ${\rm\mathop Ext}^1_R(P,M)=0$ iff $P$ is projective. Note that such modules are relatively rare! –  George C. Modoi Jun 21 '11 at 9:09
    
Just to clarify, $R$ (left) pure semisimple means it has pure global dimension zero, right? So you know that all flat (left) modules are projective and that every (left) $R$-module is a direct sum of finitely generated $R$-modules, right? I'm just trying to understand the terminology. Is there a better way to think of pure semisimple? –  David White Jul 20 '11 at 21:18
    
Yes, pure semisimple means pure global dimension zero, i.e. every module is pure projective and pure injective. I found another characterisation of pure semisimplicity in the cited paper by Stovicek, namely $R$ is left pure semisimple iff there is an $R$-module $M$ such that every left $R$-module is a direct summand of a direct sum of copies of $M$. –  George C. Modoi Jul 22 '11 at 16:42

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