Let $A$ and $B$ be rings. Show that the right adjoint of restriction-of-scalars functor along any ring homomorphism $f: A \to B$ preserves injectives and show that the unit of this adjunction is always a monomorphism.
By the restriction-of-scalars functor, I assume that the question means the functor $F: B-\text{mod} \to A-\text{mod}$ defined by sending a $B$-module $M$ to the same set $M$ but with the operation $a \cdot m := f(a)m$. I believe that the right adjoint should be $B \otimes_A -$, where $B$ is given the right $A$-module structure as $b \cdot a := bf(a)$. Please correct me if I am mistaken. Then I am to prove that $\text{Hom}_{A-\text{mod}}(-,B \otimes_A I)$ is exact if $\text{Hom}_{B-\text{mod}}(-,I)$ is exact, if $I$ is an exact $B$-module. But $\text{Hom}_{A-\text{mod}}(-, B \otimes_A I) \cong \text{Hom}_{B-\text{mod}}(F(-),I)$ by adjointness, and $F$ is exact since any exact sequence in the category of $B$-modules is in particular an exact sequence of sets and functions between the sets, which is sent to itself. Since composition of exact functors is exact, we have that $\text{Hom}_{A-\text{mod}}(-,B \otimes_A I)$ is exact, i.e. $B \otimes_A I$ is injective. However, I am not sure how to show that the unit of this adjunction is always a monomorphism. I would be grateful for your insight.
Thank you so much!