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

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?

• The universal property only guarantees that $\phi$ is a homomorphism, not necessarily an iso. Such $\phi$ exists whether or not $F$ is free. Further argument is needed to show that $\phi$ is an iso when $F$ is free. Commented May 23, 2022 at 11:51

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.

• I don't think your $\phi^{-1}(f)$ has the right type. The way it's written it looks like $\phi^{-1}(f)=f$! Commented May 23, 2022 at 16:38
• I think you meant $\phi^{-1}(f) = \sum_{i\in I} (e^i\circ f)\otimes e_i$. But note that this only makes sense when $F$ is a finitely generated free module. Commented May 23, 2022 at 16:42
• Although I guess your assertion that $\mathrm{Hom}_R(F,R)$ is free also requires $F$ to be finitely generated. Commented May 23, 2022 at 16:50

A free module $$F$$ on a set of generators $$I$$ is isomorphic to the $$I$$-indexed direct sum: $$F \cong \oplus_{i\in I} R$$. Recall that tensor products distribute over direct sums. Thus we have: \begin{align*} \mathrm{Hom}(M,R)\otimes_R F &\cong \mathrm{Hom}(M,R)\otimes_R (\oplus_{i\in I} R)\\ &\cong \oplus_{i\in I} (\mathrm{Hom}(M,R)\otimes_R R)\\ &\cong \oplus_{i\in I} \mathrm{Hom}(M,R). \end{align*}

Now when $$I$$ is finite, or when $$M$$ is finitely generated, we have $$\oplus_{i\in I} \mathrm{Hom}(M,R)\cong \mathrm{Hom}(M,\oplus_{i\in I} R)$$. So in either of these cases: \begin{align*} \oplus_{i\in I} \mathrm{Hom}(M,R) &\cong \mathrm{Hom}(M,\oplus_{i\in I} R)\\ &\cong \mathrm{Hom}(M,F). \end{align*}

If $$I$$ is not finite and $$M$$ is not finitely generated, these modules are not isomorphic in general. One can establish this with some crude dimension-counting in the case of vector spaces. Let $$R = \mathbb{Q}$$, let $$M$$ be a vector space of dimension $$\aleph_0$$, and let $$F$$ be a vector space of dimension $$\kappa$$, where $$\kappa>2^{\aleph_0}$$ and $$\kappa$$ has cofinality $$\aleph_0$$. Then the dimension of $$\mathrm{Hom}(M,\mathbb{Q})\otimes_{\mathbb{Q}}F$$ is $$\aleph_0^{\aleph_0}\cdot \kappa = \kappa$$, while the dimension of $$\mathrm{Hom}(M,F)$$ is $$\kappa^{\aleph_0} > \kappa$$.

To extract an explicit description of the isomorphism (in the cases when it is an isomorphism), we need some notation. Remember that $$I$$ is our generating set for $$F$$. An arbitrary element $$x\in F$$ can be written as $$x = \sum_{i\in I} r_i i$$, and corresponds to the element $$(r_i)_{i\in I}\in \oplus_{i\in I} R$$ under the isomorphism $$\rho\colon F\cong \oplus_{i\in I} R$$. Let's write $$\lambda_i\colon F\to R$$ for the map $$\lambda_i(x) = r_i$$. This is $$\pi_i\circ \rho$$.

In the left-to-right direction, we just need to check what happens to a basic tensor: \begin{align*} f\otimes x &= f\otimes (\sum_{i\in I} r_ii)\in \mathrm{Hom}(M,R) \otimes_R F \\ &\mapsto f\otimes (r_i)_{i\in I} \in \mathrm{Hom}(M,R) \otimes_R (\oplus_{i\in I} R)\\ &\mapsto (f\otimes r_i)_{i\in I} \in \oplus_{i\in I} (\mathrm{Hom}(M,R) \otimes_R R)\\ &\mapsto (r_if)_{i\in I} \in \oplus_{i\in I} \mathrm{Hom}(M,R)\\ &\mapsto (m\mapsto (r_if(m))_{i\in I}) \in \mathrm{Hom}(M,\oplus_{i\in I} R)\\ &\mapsto (m\mapsto \sum_{i\in I} r_if(m)i)\in \mathrm{Hom}(M,F)\\ &= (m\mapsto f(m) x)\in \mathrm{Hom}(M,F). \end{align*} So this is exactly the map you identified in the question using the universal property of the tensor product.

In the reverse direction: \begin{align*} g\in \mathrm{Hom}(M,F) & \mapsto \rho\circ g\in \mathrm{Hom}(M,\oplus_{i\in I} R)\\ &\mapsto (\lambda_i\circ g)_{i\in I}\in \oplus_{i\in I} \mathrm{Hom}(M,R)\\ &\mapsto (\lambda_i\circ g \otimes 1_R)_{i\in I} \in \oplus_{i\in I} (\mathrm{Hom}(M,R)\otimes_R R)\\ &\mapsto \sum_{i\in I} ((\lambda_i\circ g)\otimes e_i)\in \mathrm{Hom}(M,R)\otimes_R (\oplus_{i\in I} R)\\ &\mapsto \sum_{i\in I} ((\lambda_i\circ g)\otimes i))\in \mathrm{Hom}(M,R)\otimes_R F. \end{align*} Here $$e_i$$ is the element of $$\oplus_{i\in I} R$$ which is $$1_R$$ in component $$i$$ and $$0_R$$ in all other components. We have $$\rho^{-1}(e_i) = i$$.