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Let $M$ and $N$ be respectively right and left $R$-modules and suppose that $N$ is also a right $S$-module. Then it is known that $M\otimes_R N$ has a right $S$-module structure - you can just take the usual construction for $M\otimes_R N$ and define the $S$-action explicitly.

Is there a way to do this just using the universal definition of $M\otimes_R N$, rather than using an explicit representative?

edit: I guess in this case the $S$-action is simply a linear map $n:N\otimes_\mathbb Z S\to N$, in which case we can define $(M\otimes_R N)\otimes_\mathbb Z S\to M\otimes_R N$ as $\text{id}_M\otimes n$, which should work. This works because the $\mathbb Z$ tensor product matches the balanced tensor product of $\mathbb Z$ - this follows from $\mathbb Z$ being commutative and then left and right modules are the same thing.

My question was born in a more general context of $R$-algebras and bimodules over them, and in this case the balanced tensor product and the tensor over $R$ product might differ...

Daniel Teixeira
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  • See my answer here. I didn’t use any construction of the tensor product. – azif00 Sep 20 '22 at 19:24
  • Maybe read Yuan’s excellent answer and a very detailed blog in https://math.stackexchange.com/questions/3932465/mathrmhom-sets-in-categories-with-finite-biproducts-have-the-structure-of-a and then decipher Wolsey’s very abstract answer in https://math.stackexchange.com/questions/4326788/adjunction-in-abelian-categories too. Once I wanna to do it universally too but at the end give it up and think the concrete verification is much easier. Nevertheless, I can feel it definitely has something to do with the Tensor-Hom adjunction – onRiv Sep 21 '22 at 04:35

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It follows directly from functoriality. That is, the universal property implies that $M \otimes_R N$ is a functor in the second variable, so it canonically inherits an action of $\text{End}(N)$. Similarly for the first variable, and since the tensor product is in fact a bifunctor (which also follows from the universal property) these actions commute.

So if $M$ is an $(S_1, R)$-bimodule then the tensor product inherits a left action of $S_1$ and if $N$ is an $(R, S_2)$-bimodule then the tensor product inherits a right action of $S_2$, and in the presence of both bimodule structres these actions commute. Altogether we get that the tensor product of an $(S_1, R)$-bimodule and an $(R, S_2)$-bimodule is an $(S_1, S_2)$-bimodule.

Qiaochu Yuan
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