Let $R$ be a commutative ring with 1. Under what conditions are $R$-mod (the category of $R$-modules) and $R$-mod-$R$ (the category of $R$-$R$ bimodules) equivalent as categories?
When are $R$-mod and $R$-mod-$R$ equivalent?
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abstract-algebra
category-theory
modules
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0Yes, I mean $R$-$R$ bimodules. I'll edit the question to be more clear. – 2012-03-27
1 Answers
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The category of $(R, R)$-bimodules is equivalent to the category of $R \otimes_{\mathbb{Z}} R$-modules by the universal property of the tensor product (since there's no distinction between a left and a right action of a commutative ring). Moreover, it is possible to recover a commutative ring from its category of modules (ostensibly as an $\text{Ab}$-enriched category but actually you only need to know what it looks like as an ordinary category, although the argument I have in mind is slightly involved).
So a necessary and sufficient condition is that $R$ is isomorphic to $R \otimes_{\mathbb{Z}} R$. For example $R = \mathbb{Z}$ and $R = \mathbb{Z}[x_1, x_2, ... ]$ have this property.