Question about If $B$ is a $A$-algebra and $N$ is a $B$-module, then can $N$ naturally become a $A$-module?

Question

So I think the only point is to show how to build the $A$-module structure on $N.$ We know that $B$ is an $A$-algebra which is also an $A$-module where the $A$-module structure is given by the scalar multiplication. For $N$ we know that $N$ is a $B$-module so by definition it has a group action $b:B\times N\rightarrow N$ where $(b,n)\mapsto bn.$ We also know that the $A$-module structure of $B$ is given by homomorphism $f:A\rightarrow B,\ ab=f(a)b$(group action). So we can define the $A$-module structure on $N$ simply inherit the scalar multiplication for $B$ as $A$-module. So for $f:A\rightarrow B$ a homomorphism, we have group action $a\in A, a:A\times N\rightarrow N$ defined by $(a,n)\mapsto f(a)n.$ How we maps $f(a)n$ gives by $N$ as the $B$-module. Can anyone tells me if I think it right? Appreciate!

Answer 1

What you have done is right.

A more elegant way to do the same construction could be the following: a $A$-algebra $B$ is a ring homomorphism $A\to Z(B)$ (where $Z(B)$ denotes the center of $B$), a $B$-module $N$ is a ring homomophism $B\to \mbox{End}(N)$.
If you compose these two ring homomrphisms you get a ring homomorphism $A\to B\to \mbox{End}(N)$, which is exactly an $A$-module structure on $N$.
Note that this construction in fact coincides with yours, it is just a more elegant way to write what you have already done.

Edit: I've seen that you have tagged the question after "commutative algebra". In this case, note that if you are working only with commutative rings you have $Z(B)=B$.