Basic question about how to form left adjoint of forgetful functor $U: R\text-\mathrm{Mod}\text-S \to R\text-\mathrm{Mod}$ as tensor product with $S$? My algebra is a bit rusty, and I'm having some trouble forming a certain tensor product of modules over a non-commutative ring. Not sure if there is a typo, or if I'm forgetting something obvious. On page 87 of Mac Lane's Categories for the Working Mathematician (CWM) he offers a table of examples of left adjoints to forgetful functors. One of row of the table offers this example:
$U: R\text-\mathrm{Mod}\text-S \rightarrow R\text-\mathrm{Mod}$
taking $(R,S)$-bimodules to left $R$-modules, with the left adjoint given by:
$F:A\rightarrow A\otimes S$
taking left $R$-modules to $(R,S)$-bimodules. Presumably $S$ is intended to be a ring containing $R$, so that $S$ can serve as an $(R,S)$-bimodule here. But wouldn't $A$ is need to be a right $R$ module, so that we can even form the tensor product $A\otimes_R S$? Even if that were the case, wouldn't $A\otimes_R S$ then be merely a right $S$ module, when what is required is actually an $R,S$ bi-module? I'm apparently missing something, because I can't get the "types" correct.

The tensor product should be over $\mathbb{Z}$. This doesn't require any relationship between $R$ and $S$ to be specified in advance.
If $R$ and $S$ are algebras over a common commutative ring $k$ and "$(R, S)$-bimodule" means "compatible $k$-module structures" then the tensor product should be over $k$.
