# Do categories of bimodules have enough projectives?

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.

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

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.