Exercise A0.III.6.06

Proposition (P1)

Let $R$ be a commutative ring, and let $F = R^{\oplus n}$ be a finitely generated free $R$-module. Then $\Hom_{R-\mathrm{Mod}}(F,R) \cong F.$

Proof

Let $f$ be any map in $\Hom_{R-\mathrm{Mod}}(F,R)$. Take the set of “one-hot” $n$-tuples $j_1,\ldots,j_n \in F.$ Note that any element $v$ of $F$ can be written down as a sum

$$v = \sum_{k=1}^n r_kj_k$$

for some set of $r_1,\ldots,r_n \in R.$

We can find $f(v)$ by using the fact that it’s a homomorphism, since

$$\begin{align*}f(v) &= f\left(\sum_{k=1}^n r_kj_k\right) \\ &=\sum_{k=1}^n f(r_kj_k) \\&=\sum_{k=1}^n r_kf(j_k),\end{align*}$$

that is, $f$ is completely determined by the values it takes on $j_1,\ldots,j_n.$

Note also that every possible assignment of an element of $R$ to each $j_k$ gives rise to a unique homomorphism in $\Hom_{R-\mathrm{Mod}}(F,R).$ This follows from the universal property of a free object in $R\mathrm{-Mod}$.

Consider the map $\tau : \Hom_{R-\mathrm{Mod}}(F,R) \to F$ defined by

$$f \mapsto \sum_{k=1}^n f(j_k)j_k.$$

We have already established that $\tau (f)$ contains sufficient data to recover $f$ uniquely (injectivity) and that this mapping is surjective, so we only need to show that $\tau$ is an $R$-module homomorphism and this is straightforward. Let $g \in \Hom_{R-\mathrm{Mod}}(F,R)$ and $u \in R,$ then

$$\begin{align*} \tau (f + g) &= \sum_{k=1}^n (f + g)(j_k)j_k \\ &= \sum_{k=1}^n (f(j_k)j_k + g(j_k)j_k) \\ &= \sum_{k=1}^n f(j_k)j_k + \sum_{k=1}^n g(j_k)j_k \\ &= \tau(f) + \tau(g), \end{align*}$$

and

$$\begin{align*} \tau(uf) &= \sum_{k=1}^n (uf)(j_k)j_k \\ &= u \sum_{k=1}^n f(j_k)j_k \\ &= u \tau (f). \end{align*}$$

$\blacksquare$

NOTE

We need commutativity of $R$ to have an $R$-module structure on $\Hom_{R-\mathrm{Mod}}(F,R).$ Aluffi clarifies this towards the end of §5.2

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Proposition (P2)

There exists a ring $R$ and a non-zero $R$-module $M$ such that $\Hom_{R-\mathrm{Mod}}(M,R) = 0.$

Proof

Indeed, set $R=\mathbb{Z}$ and $M = \mathbb{Z}/2\mathbb{Z}.$ An abelian group homomorphism from $M$ to $R$ must send $1_M$ to an element of finite order in $R,$ but the only such element is $0_R,$ forcing every homomorphism $M \to R$ to be trivial.

$\blacksquare$