Let $R\longrightarrow S$ be a map of commutative rings. Suppose $I$ is an injective $R$-module. Then, is $Hom_R(S,I)$ an injective $S$-module?
I am unable to prove this. Google led me to a paper in Proc AMS by L W Christensen and others (see here)that says in its abstract that this is a "well known fact" but gives no proof/reference. Could you give me a proof or a reference to a proof?
A slick way to see this is using the hom-tensor adjunction and recalling that $I$ is injective if and only if $\operatorname{Hom}_R(-,I) : \mathsf{Mod}_R\to\mathsf{Ab}$ is exact. Then you want to prove that the functor $$\operatorname{Hom}_S(-,\operatorname{Hom}_R(S,I)) : \mathsf{Mod}_S\to\mathsf{Ab}$$ is exact. But, by the hom-tensor adjunction, we have $$ \operatorname{Hom}_S(-,\operatorname{Hom}_R(S,I))\cong\operatorname{Hom}_R(-\otimes_S S,I)\cong\operatorname{Hom}_R(-,I). $$ Thus, given an exact sequence $0\to M'\to M\to M''\to 0$ of $S$-modules, applying $\operatorname{Hom}_S(-,\operatorname{Hom}_R(S,I))$ is the same as first viewing the modules $M', M,$ and $M''$ as $R$-modules via the map $R\to S,$ and then applying $\operatorname{Hom}_R(-,I).$ This gives another exact sequence, by injectivity of $I$!
For a more direct argument, see here.
Edit: To prove the hom-tensor adjunction, let $N$ be an $R$-module, and let $M$ be an $S$-module. One needs to show that $$\operatorname{Hom}_S(M,\operatorname{Hom}_R(S,N))\cong\operatorname{Hom}_R(M\otimes_S S, N)$$ (functorially in $M$ and $N$).
To that end, let $f : M\otimes_S S\to N$ be a morphism of $R$-modules. Define a morphism $\phi^f : M\to\operatorname{Hom}_R(S,N)$ of $S$-modules by \begin{align*} \phi^f : M&\to\operatorname{Hom}_R(S,N)\\ m&\mapsto\left(\phi^f_m : s\mapsto f(m\otimes s)\right). \end{align*} I leave it to you to check that this is indeed an $S$-module morphism.
Conversely, let $\phi : M\to\operatorname{Hom}_R(S,N)$ sending $m\mapsto \phi_m$ be a morphism of $S$-modules, and define $f_\phi$ by the following formula on simple tensors and extend linearly: \begin{align*} f_\phi : M\otimes_S S&\to N\\ m\otimes s&\mapsto\phi_m(s). \end{align*} Again, I leave it to you to check that this is a morphism of $R$-modules.
We must check that these constructions are mutually inverse. So, let $f : M\otimes_S S\to N$ be a morphism of $R$-modules. As defined above, $\phi^f(m) = \phi^f_m$ is the function which sends $s\in S$ to $f(m\otimes s).$ Then we have $$f_{\phi^f}\left(\sum m_i\otimes s_i\right) = \sum \phi^f_{m_i}(s_i) = \sum f(m_i\otimes s_i) = f\left(\sum m_i\otimes s_i\right),$$ which shows that the composition $$ \operatorname{Hom}_R(M\otimes_S S, N)\to \operatorname{Hom}_S(M,\operatorname{Hom}_R(S,N))\to\operatorname{Hom}_R(M\otimes_S S, N) $$ is the identity. Similarly, you need to check that the composition in the other order is also the identity (and if you want to be picky, that these maps are all functorial in $M$ and $N$), but I'll let you check the rest of the details yourself.