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If $R$ and $S$ are commutative rings, then does the category $R \oplus S$-modules encompase the category of $(S,R)$-bimodules?

I was thinking we can accomplish this by defining the action to be: $(r,s)\cdot x:=r\cdot (x\cdot s)$ and then doing something accoring for the morphisms.

Am I overlooking something or is this indeed true?

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    $\begingroup$ The category of $(S, R)$-bimodules is equivalent to the category $S \otimes R^{op}$-modules. The category of $S \times R$-modules is equivalent to the direct sum of the category of $S$-modules and the category of $R$-modules. $\endgroup$ – Qiaochu Yuan Mar 22 '16 at 3:51
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The direct product of rings is not really the right way to go. You cannot get suitable additive properties (as Eric Wofsey describes.) An $R,S$ bimodule structure is equivalent to an $R\otimes S^{op}$ module structure. You can prove this beginning with what you have already written. You just need to apply the properties of the tensor product.

If $R$ and $S$ are central simple $K$ algebras, then $R\otimes_K S^{op}$ is central simple too, but $R\times S$ is not, so there will be differences in the module categories.

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  • $\begingroup$ Right, I forgot about this. Thanks :) $\endgroup$ – AIM_BLB Mar 22 '16 at 3:39
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This doesn't work. If you define $(r,s)\cdot x=rxs$, then for this to give you an $R\times S$-module structure you would need to have $(r,s)x+(r',s')x=(r+r',s+s')x$, or $rxs+r'xs'=(r+r')x(s+s')$, but this is usually not true. For instance, if $s'=0$, this is saying that $rxs=(r+r')xs$ so $r'xs=0$ for all $r'\in R$ and all $s\in S$, which is almost never true.

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  • $\begingroup$ Doh, you beat me to filling this in. This is indeed the big flaw. $\endgroup$ – rschwieb Mar 22 '16 at 3:18

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