# Understanding viewing $M$ as both an $A$ and $B$ module

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?
• For your obvious counterexample, you're forgetting that a ring-homomorphism of commutative rings,by convention, maps the unit element onto the unit element. Feb 3, 2021 at 19:57
• 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. Feb 4, 2021 at 20:28
• 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. Feb 4, 2021 at 20:43

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).
• 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.