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…
Best Answer
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.