For $F$ a finitely generated free $R$-module, prove $F \cong \text{Hom}_R (F, R)$

Question

Aluffi III.6.6 suggests proving the following:

Let $R$ be a ring, and let $F = R^{\oplus n}$ be a finitely generated free $R$-module. Prove that $\text{Hom}_{R-\text{Mod}} (F, R) \cong F$.

I ended up with the following construction. Let $H = \text{Hom}_{R-\text{Mod}} (F, R)$, $A = \{ 1, \dots, n \}$ and consider set-function $f : A \rightarrow H$, $f : i \mapsto \alpha_i$ where $\alpha_i : (r_1, \dots, r_n) \mapsto r_i$. This set-function gives rise to an unique $\sigma : F \rightarrow H$ (since $F$ is free). Then we just need to build a homomorphism $\psi : H \rightarrow F$ and prove that $\sigma \psi = \text{id}_F$ and we're done (since $\psi \sigma = \text{id}_H$ follows from $F$ being free).

Now, let $\alpha \in H$. It feels natural to map $\psi : \alpha \mapsto (\alpha(1, 0, \dots, 0), \alpha(0, 1, 0, \dots, 0), \dots, \alpha(0, \dots, 0, 1))$. Unless I messed up with the (boring) proof boiling down to substituting the definitions for $\sigma$ and $\psi$, $\sigma \psi = \text{id}_F$ really holds, and indeed we're done.

Does this look correct?

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I also tried going straight from the definition of a free module and tried proving that $\text{Hom}_{R-\text{Mod}} (F, R)$ satisfies the corresponding universal property, but I wasn't particularly successful at that. So, is there a better (more algebraic or category-theoretic) proof of the claim, preferably accessible to me at this point in the book?

Answer 1

Note that, due to the universal property of $F$, for any $f\in F$ there is a unique homomorphism $h:F\rightarrow R$ making the corresponding diagram commutative. So, just send $f$ to $h$.

P.S. In fact, proving this really gives a homomorphism, I found out that the action of $R$ is preserved only if $R$ is commutative.

Answer 2

$\DeclareMathOperator{\Hom}{Hom}$A more "category-theoretic" approach is to use the fact that the contravariant hom functor $\Hom_{R-\mathsf{Mod}}({-},R) : R-\mathsf{Mod} \to R-\mathsf{Mod}$ "turns coproducts into products" (or "preserves products" if you prefer to think of it as a covariant functor $(R-\mathsf{Mod})^{\text{op}} \to R-\mathsf{Mod}$). This is just a restatement of the universal property of coproducts.

What this means precisely is that the natural morphism $$\Hom_{R-\mathsf{Mod}}(R^{\oplus n},R) \to \prod_n \Hom_{R-\mathsf{Mod}}(R,R)$$ is an isomorphism of $R$-modules. Now, since a finite direct product of $R$-modules is equal to the finite direct sum of those $R$-modules, we have $$\Hom_{R-\mathsf{Mod}}(R^{\oplus n},R) \cong \Hom_{R-\mathsf{Mod}}(R,R)^{\oplus n}.$$ Finally, you need to show that $\Hom_{R-\mathsf{Mod}}(R,R) \cong R$, which is a standard exercise.

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Edit: reorganized the proof to make it a little more elegant.

Edit 2: I did assume the ring is commutative here. If $R$ is allowed to be noncommutative, then $\Hom_{R-\mathsf{Mod}}({-},R)$ is no longer a contravariant endofunctor of $R-\mathsf{Mod}$. Instead, it is a contravariant functor from left $R$-modules to right $R$-modules. To fix this, we consider the variant $\Hom_{R-\mathsf{Mod}}({-},R) : ((R,R)-\mathsf{Bimod})^{\text{op}} \to (R,R)-\mathsf{Bimod}$, which makes everything work out exactly as in the commutative case.