# Mod-$R$, Mod-$S$ and Mod-$R \otimes S$

Let $R,S,T$ be commutative rings and assume that $R,S$ are $T$-algebras.

In an answer to this question, Pierre-Yves Gaillard gives an example of an $R \otimes_T S$-module that cannot be written as the tensor product of an $R$-module and an $S$-module (there, $T=k$, $R=S=k^2$ where $k$ is a field).

I'm interested in the relation between the module categories Mod-$R$, Mod-$S$ and Mod-$R \otimes_T S$. Is there some kind of general operation (a "tensor product") on abelian categories that takes Mod-$R$, Mod-$S$ and Mod-$T$ (or $T$ itself) and produces Mod-$R \otimes_T S$?

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Dear Sebastian: If $M$ is an $R$-module and $N$ is an $T$-module, then there is a unique structure of $R\otimes_TS$-module on $M\otimes_T N$ satisfying $$(r\otimes s)(x\otimes y)=rx\otimes sy$$ for all $r\in R,s\in S,x\in M,y\in N$. – Pierre-Yves Gaillard Jan 31 '12 at 15:05
@Sebastian: Well, by the general theory of Morita equivalence, $\textrm{Mod-}R$ is equivalent to $\textrm{Mod-}S$ if and only if $R \cong S$ as rings, so in theory, yes, it should be possible if you are willing to restrict your attention to commutative rings. But the general case looks more doubtful, as the passage from $R$ to $\textrm{Mod-}R$ loses information. – Zhen Lin Jan 31 '12 at 17:22

Yes, Deligne's tensor product, Kelly's tensor product, as all as the coproduct of cocomplete $\otimes$-tensor categories. They all have the property $\mathsf{Mod}(R) \otimes_k \mathsf{Mod}(S) \simeq \mathsf{Mod}(R \otimes_k S)$. See Schäppi's work for generalizations to schemes and quasi-coherent sheaves.