# Finding specific isomorphism $\mathrm{Hom}(M, R)\otimes_R F \rightarrow \mathrm{Hom}(M, F)$

abstract-algebracommutative-algebramodulesring-theory

Let $$R$$ be a commutative ring with $$1_R$$. Let $$M$$ be a $$R$$-module and $$F$$ be a free $$R$$-module.

How do I find a specific isomorphism to show that $$\mathrm{Hom}(M, R)\otimes_R F \cong \mathrm{Hom}(M, F).$$

Due to the universal property I know that there exists a unique homomorphism $$\varphi: \mathrm{Hom}(M, R)\otimes_R F \rightarrow \mathrm{Hom}(M, F),$$
because $$\beta: \mathrm{Hom}(M, R) \times F \rightarrow \mathrm{Hom}(M, F)$$ with $$\beta: (\lambda, w) \mapsto \lambda(v)\cdot w$$ is bilinear.

How do I go about finding $$\varphi$$? Can I use $$\beta = \varphi \circ \tau$$, where $$\tau$$ is the natural inclusion or do I need to use the resemblance to dual spaces. How much does the case with $$R$$ a field, and $$M, F$$ vectorspaces differ from the one with rings?

In the case that $$R$$ is a field, $$M$$ and $$F$$ vector spaces, one typically shows injectivity and then uses a dimension counting argument to show that this map is an isomorphism.
One way to prove it in this setting would be to pick a basis $$\{e_i\}_{i\in I}$$ of $$F$$ (this can be done, as $$F$$ is free!). Then $$\text{Hom}_R(F,R)$$ is also free, with basis denoted by $$\{e^i\}_{i\in I}$$, where the duality pairing is defined by $$e^j(e_i) = \delta^j_i$$.
Then given a map $$f:M \to F$$, the inverse of the map $$\phi$$ can be written as $$\phi^{-1}(f)(m) = \sum_{i\in I} e^i(f(m))e_i,$$ for $$m\in M$$, where the sum only has finitely many nonzero terms.