For the purpose of this discussion, all rings are commutative with a unit.

Any time we have a ring morphism $f: A\rightarrow B$, we can view $B$ as an $A$-module. Moreover, if $N$ is a $B$-module, we can give it the structure of an $A$-module by $a\cdot n:=f(a)\cdot n$. It seems that people often say in this case that we can 'view $N$ as an $A$ module'. If I understand correctly we say this because $A$-module and $B$-module structures are compatible in the sense that $a\cdot_A(b\cdot_B n)=(a\cdot_A b)\cdot_B n$ (where '$\cdot_A$' represents the multiplication from the $A$-module structure).

Its obviously not true that $M$ viewed as a $B$ module is the same as $M$ as a $M$ as an $A$-module, take $f:A\rightarrow B$ to be the zero map for example. But there are some nice things that are true:

For example exercise 2.15 in Atiyah-Macdonald gives us an associativity over tensor products

If $P$ is an ($A$-$B$)-module and $M$ (resp. $N$) is an $A$ (resp. $B$)-module then $(M\otimes_A P) \otimes_B N \cong M\otimes_A (P \otimes_B N)$

I find this very confusing. My questions are:

  • Is there a notion for $M$ and an $A$-module being isomorphic to $M$ as $B$ module?
  • If $f:A\rightarrow B $ is a ring morphism, does this induce a functor from $B$-modules to $A$-modules.
  • It seems there is a lot more we could say about such things. For example, if $M$ and $N$ are $(A,B)$-modules via $f:A\rightarrow B$, is $M\otimes_A N$ the restriction of $M\otimes_B N$?
  • Is there a good reference where I could read more about these modules, or some good exercises to get used to this notion?
  • 1
    $\begingroup$ For your obvious counterexample, you're forgetting that a ring-homomorphism of commutative rings,by convention, maps the unit element onto the unit element. $\endgroup$
    – Bernard
    Feb 3, 2021 at 19:57
  • $\begingroup$ It seems that your understanding about $\textit{bimodules}$ is not correct. When we say two structures are compatible we mean $a(nb)=(an)b$. See here. $\endgroup$
    – MathEric
    Feb 4, 2021 at 20:28
  • $\begingroup$ I am using the term in the sense it is used in Atiyah-Macdonald. I should have made that clear as it does seem the term bimodule is more commonly used in the sense you describe. $\endgroup$ Feb 4, 2021 at 20:43

1 Answer 1


I think that you should think of it as follows: Given $f:A\to B$ and a $B$-module $(M,+,\cdot_B)$, there is an induced $A$-module structure on (the underlying abelian group of) $M$, i.e. there is an $A$-module structure $(M,+,\cdot_A)$ in the way you are familiar with. I can't speak for other people but I typically think of the $B$-module structure on $M$ and the induced $A$-module structure as distinct things, i.e. I don't bother to think of it as a bi-module structure, etc., and you don't want to ask questions like whether these are isomorphic (because these are modules over different rings).

Let's go through some of your questions:

  • Yes, this induces a functor from $B$-modules to $A$-modules. Given a $B$-module homomorphism $\varphi:M\to M'$, it is straightforward to check that the same map $\varphi:M\to M'$ is an $A$-module homomorphism when $M$ and $M'$ are equipped with their induced $A$-module structures, and since we are using the same set-theoretic map it is trivial that this preserves the identity, composition, whatever else you need for a functor.

  • No, you should not expect $M\otimes_A N$ to be the restriction of $M\otimes_B N$. If you take $\mathbb R\to\mathbb C$ and $M=N=\mathbb C$ you get a counterexample, proved here.

  • I don't have a good idea of where you can find exercises dealing with this specific construction; I say just keep reading Atiyah-Macdonald and working with it when it comes up and soon you will find it a comfortable thing to work with.


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