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.

  • 8
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.