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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.

Doeke
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1 Answers1

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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.

Maxime Ramzi
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  • Just a comment that the tensor is over $\mathbb{Z}$. Or, if both $R$ and $S$ are algebras over a common $k$, then we can take tensor over $k$. – equin Oct 01 '21 at 16:45
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    I take that back... if you tensor over $k$ you impose an additional condition on your module; namely that the copy of $k$ acts the same way in $R$ as it does in $S$. (One could call such a thing a $(R,S)$-bimodule over $k$). – equin Oct 01 '21 at 16:53
  • @equin : absolutely ! With this addendum, this is actually important (you could also say "bimodules in $k$-modules" ) – Maxime Ramzi Oct 01 '21 at 16:58