A tensor identity - $\text{Hom}_{R}(A,B \otimes_S C) \cong \text{Hom}_{R}(A,B)\otimes _SC$

Question

Let $R,S$ be associative algebras over $\mathbb{C}$. Let $A$, $B$ and $C$ be, a left $R$-module, a $(R,S)$-bimodule, a left $S$-module, respectively. Assume that $B\otimes_S C$ is finite-dimensional.

Note that $\text{Hom}_{R}(A,B)$ and $B\otimes_S C$ are right $S$-modules and left $R$-modules in a natural way.

$\bf{My \ \ Question:}$

Prove that $\text{Hom}_{R}(A,B)\otimes _SC \cong \begin{align} \text{Hom}_{R}(A,B \otimes_S C) \end{align}$.

My idea is that the map $\Phi:\text{Hom}_{R}(A,B)\otimes _SC \rightarrow \begin{align} \text{Hom}_{R}(A,B \otimes_S C) \end{align}$ given by

$\Phi(f \otimes v)(a) : = f(a)\otimes v$, for all $a\in A, f\in \text{Hom}_{R},(A,B) $ and $v\in C$, is an isomorphism. Thank you very much!

Answer 1

For a simple counterexample, take $R=S=\mathbb{C}[t]$ a polynomial ring, $A=C=\mathbb{C}=R/(t)$, and $B=_R\!\!R_R$.

Then $\operatorname{Hom}_R(\mathbb{C},R)\otimes_R\mathbb{C}=0$ as there are no non-zero $R$-module homomorphisms $\mathbb{C}\to R$, but $\operatorname{Hom}_R(\mathbb{C},R\otimes_R\mathbb{C})\cong \operatorname{Hom}_R(\mathbb{C},\mathbb{C})$ is non-zero.

Here $B\otimes_SC=R\otimes_R\mathbb{C}\cong\mathbb{C}$ is finite dimensional, and there are even similar examples with everything finite dimensional, as there are finite dimensional algebras $\Lambda$ with a non-zero finite dimensional module $M$ for which $\operatorname{Hom}_\Lambda(M,\Lambda)=0$, so you can take $R=S=B=\Lambda$ and $A=C=M$.