Let $A,B$ be commutative rings with unit, $\phi:A\to B$ a ring homomorphism and $I$ an injective $A$-module. Prove that $\text{Hom}_A(B,I)$ is an injective $B$-module.
Here is my attempt:
Let $M,N$ be $B$-modules, $g:M\to\text{Hom}_A(B,I)$ and $f:M\to N$ with $\ker f=0$. Our goal is to construct $h:N\to\text{Hom}_A(B,I)$ with $h\circ f=g$.
Let $\phi^*:B\text{-Mod}\to A\text{-Mod}$ be the functor of restriction of scalars (notation: $X^*:=\phi^*(X)$ for any object or morphism $X$ in $B$-Mod).
That way, we get $f^*:M^*\to N^*$ with $\ker f^*=0$ and, for $b\in B$, we may define $g_b:M^*\to I$ with $g_b(m)=g(m)(b)$. Since $I$ is injective, there exists $h_b:N^*\to I$ with $h_b\circ f^*=g_b$.
What I want to do is to define $h:N\to\text{Hom}_A(B,I)$ as $h(n)(b)=h_b(n)$. But in order to $h$ be well-defined, I'd have to show that $h_{b+c}(n)=h_b(n)+h_c(n)$, which I still can't figure out.
Is it possible? How? If not, how to I approach this problem?
The structure of $B$-module on $\operatorname{Hom}_A(B,I)$ is defined by $$ bh\colon x\mapsto h(bx) $$ for $h\colon B\to I$. In particular, $bh(1)=h(b)$.
You can use Baer’s criterion. Suppose $\mathfrak{b}$ is an ideal of $B$ and you are given $f\colon\mathfrak{b}\to\operatorname{Hom}_A(B,I)$.
Define $\hat{f}\colon\mathfrak{b}\to I$ by $\hat{f}(b)=f(b)(1)$, which is a homomorphism of $A$-modules. This extends to a homomorphism of $A$-modules $g\colon B\to I$, since $I$ is injective.
Now we can show that $f(b)=bg$. Indeed, for $x\in B$, $$ bg(x)=g(bx)=f(bx)(1)=(xf(b))(1)=f(b)(x) $$ because $f$ is $B$-linear.