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It is well-known and easy to prove (assuming the axiom of choice) that the category $R\text{-}\mathrm{Mod}$ has enough projectives for any ring $R$: Let $M$ be any $R$-module, and let $P$ be the free left $R$-module with basis given by the set $M$. Define a map $\pi:P\to M$ using the universal property of free objects by sending the basis element $m$ in the set $M$ to the corresponding element of the module $M$. The module $P$ is free and therefore projective, and clearly $\pi$ is surjective.

Does the category $R\text{-}\mathrm{Mod}\text{-}S$ of $(R,S)$-bimodules have enough projectives? It seems that the above argument doesn't generalize. Does the category $R\text{-}\mathrm{Mod}\text{-}S$ even have free objects? I'd love to see a proof, counterexample, or resource one way or the other.

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    $\begingroup$ A $(R,S)$-bimodule is the same as a left $R\otimes S^{op}$ module $\endgroup$ – Maxime Ramzi Feb 27 '19 at 22:17
  • $\begingroup$ Of course. Thanks! Feel free to post as an answer and I'll accept it. $\endgroup$ – Doeke Feb 27 '19 at 22:55
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A $(R,S)$-bimodule is the same thing as left $R\otimes S^{op}$-module, so you have enough projectives, and in fact as you mentioned we have free objects.

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