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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?

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    $\begingroup$ 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. $\endgroup$ May 23, 2022 at 11:51

2 Answers 2

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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.

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  • $\begingroup$ I don't think your $\phi^{-1}(f)$ has the right type. The way it's written it looks like $\phi^{-1}(f)=f$! $\endgroup$ May 23, 2022 at 16:38
  • $\begingroup$ 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. $\endgroup$ May 23, 2022 at 16:42
  • $\begingroup$ Although I guess your assertion that $\mathrm{Hom}_R(F,R)$ is free also requires $F$ to be finitely generated. $\endgroup$ May 23, 2022 at 16:50
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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 $\rho^{-1}(e_i) = i$.

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