# Tensorproduct of $R$-module and $R$-algebra

let $M$ be an $R$-module and let $S$ be an $R$-algebra through the ring homomorphism $\phi$.

I can make $M\otimes S$ into a $R$-module in several different ways. Either by defining

1. $r. (m\otimes s)=rm\otimes \phi(r)s$ or

2. $r. (m\otimes s)=rm\otimes s$ or

3. $r. (m\otimes s)=m\otimes \phi(r)s$

Here the last structure agrees with the structure given on $M\otimes _S S$ (multiplication in second component) by extension of scalars. What can be said about the relationships among these $R$-modules? In particular in my class it was stated that making $M\otimes _S S$ to an $R$-module by extension of scalars agrees with "the original structure", i.e 3 should be isomorphic with 1 or 2.

• Why do you use $\oplus$ in 1–3? Anyway, 2 and 3 are the same; the tensor product (I'm assuming over $R$) is an $R$-bilinear gadget and $\phi$ is giving $S$ its $R$-module structure. Really, if you just want to think about it as an $R$-module the fact that $S$ is an algebra over $R$ isn't necessary information. I don't think 1 gives you a module structure: you want $(r_1 + r_2)z = r_1z + r_2z$ and I dont think it works out, – Hoot Feb 7 '15 at 18:11

I'm assuming $R$ is commutative and in the center of $S$ (which is a standard assumption when you call something an algebra). I also assume $M \otimes S$ is tensoring as $R$-modules.
Numbers (2) and (3) are equivalent because elements of $R$ can move between the tensor factors, hence $rm \otimes s = m \otimes \phi(r)s$. This indeed is what you get if you take extension of scalars $M \otimes_R S$ (it really should be a subscript $R$ on the tensor, not $S$), which is an $S$-module via multiplication into the right factor, and then restrict it to an $R$-module.
Number (1) is not a well defined action. By definition of tensoring over $R$ we have $rm \otimes \phi(r)s = r^2m \otimes s$ so distributivity over addition isn't going to work correctly.