$R$ is commutative ring, $S$ is commutative $R$-algebra and $M$ is $R$-module.
So we have the S-module isomorphism $$S\otimes_R \mathrm{Hom}_R(M,N) \simeq \mathrm{Hom}_S(S\otimes_R M, S\otimes_R N)$$ given that $S$ is flat over $R$ and $M$ is finitely presented. Even though $S$ is not flat or $M$ is finitely presented, at least there exists $S$-homomorphism between the modules.
What I am considering is some restriction of the isomorphism. Let $\phi \in \mathrm{End}_R(M)$. $1\otimes_R \phi$ will denote endomorphism of $S\otimes_R M$. $$ S\otimes_R R[\phi] \simeq S[1\otimes_R \phi]$$ Will the above isomorphism hold without the condition that $S$ is flat over $R$ or that $M$ is finitely presented?
Given that $S$ is flat over $R$ and $M$ is finitely presented, $S\otimes_R R[\phi]$ is submodule of $S\otimes_R \mathrm{End}_R(M)$ and the isomorphism in consideration easily follows.
That the induced homomorphism is surjective is obvious but I cannot quite verify whether it is injective.
- Does the isomophism hold without flatness or finitely-presentedness?
- If the isomorphism holds, what is this structural difference between $R[\phi]$ and the entire $\mathrm{End}_R(M)$ that makes the isomorphism possible on one and not on the other?