I'm working on the following question:
Let $R$ be a ring (commutative and unital), $f: R \rightarrow B$ be a $R$-algebra, $M$ an $R$-module and $R$ a $B$-module. Show that
$\text{Hom}_B (B \otimes_R M, N) \simeq \text{Hom}_R (M,N)$.
Now for a moment if we forget about the algebra $B$, I know that $\text{Hom}_R(R \otimes_R M, N) \simeq \text{Hom}_R (R, \text{Hom}_R (M, N))$ holds for $R$-modules. Also, I know that there is a canonical isomorphism $\text{Hom}_R (R, M) \simeq M$.
I could try to realize $B$ as a $R$ module, but I'm confused about $R$ being a $B$-module. Would that imply that I'm just realizing $R$ as a module over itself? How exactly does the subscript on $\text{Hom}$ play out in here?
Best Answer
$B$ is an $R$-algebra. It becomes an $R$-module if we define "$rb$" as $f(x)b$ for $r\in R$ and $b\in B$.
Likewise, a $B$-module $N$ becomes an $R$-module via $rn=f(r)n$.
Also $B\otimes_ RM$ is a $B$ module via $b\sum b_i\otimes m_i =\sum(bb_i)\otimes m_i$.
So here we are claiming that the $B$-module homomorphisms from $B\otimes_R M$ to $N$ (where $N$ is considered as a $B$-module) correspond to the $R$-module homomorphisms from $M$ to $N$ (where $N$ is considered as an $R$-module).
Roughly speaking if $\phi:M\to N$ is an $R$-module homomorphism then the corresponding $B$-module map takes $\sum b_i\otimes m_i$ to $\sum b_i\phi(m_i)$.
In the language of category theory, the functor $M\mapsto B\otimes_RM$ is a left adjoint to the forgetful functor from $B$-mod to $R$-mod.