1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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, $\endgroup$ – Hoot Feb 7 '15 at 18:11
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.