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?