Exercise A0.III.6.08

Proposition

Let $R$ be a commutative ring and $A$ any set. Then $\Hom_{R\mathrm{-Mod}}(R^{\oplus A},R)$ satisfies the universal property of the product $\prod_{a \in A} R$.

Proof

We will use $H$ to denote $\Hom_{R\mathrm{-Mod}}(R^{\oplus A},R).$

Define $\pi_a : H \to R$ as $f \mapsto f(g_a)$ where $g_a : A \to R$ is a finite support map given by

$$g_a(a')={\begin{cases}0_R&{\text{if }}a \neq a',\\1_R &{\text{if }}a=a'.\end{cases}}$$

Let $X$ be some $R$-module and let $\{f_a\}_{a \in A}$ be a family of morphisms $f_a : X \to R.$ We want to show that there exists a unique morphism $\sigma : X \to H$ such that for every $a \in A$ we have $\pi_a \circ \sigma = f_a.$

Let $x \in X$ be arbitrary and fix $a \in A.$ Now, we know that $\pi_a(\sigma(x)) = \sigma(x)(g_a)$ and to make this equal to $f_a(x)$ we must have

$$\sigma(x)(h) = \sum_{a \in A} h(a)f_{a}(x).$$

This map is well-defined. Even though the sum is over an infinite set, $h$ has finite support and hence the whole sum has only finitely many non-zero terms.

It is easy to check that $\sigma(x)(-)$ is an $R$-module homomorphism: Let $r \in R$, $h, h' \in R^{\oplus A}$, then

$$\begin{align*} \sigma(x)(rh) &= \sum_{a \in A} (rh)(a)f_{a}(x) \\ &= \sum_{a \in A} rh(a)f_{a}(x) \\ &= r\sum_{a \in A} h(a)f_{a}(x) \\ &= r\sigma(x)(h) , \end{align*}$$

and

$$\begin{align*} \sigma(x)(h+h') &= \sum_{a \in A} (h+h')(a)f_{a}(x) \\ &= \sum_{a \in A} (h(a)+h'(a))f_{a}(x) \\ &= \sum_{a \in A} (h(a)f_{a}(x)+h'(a)f_{a}(x)) \\ &= \sum_{a \in A} h(a)f_{a}(x) + \sum_{a \in A}h'(a)f_{a}(x) \\ &= \sigma(x)(h)+\sigma(x)(h'). \end{align*}$$

We can also check that $\sigma(-)$ is an $R$-module homomorphism: Let $x,x' \in X$, then

$$\begin{align*} \sigma(x + x')(h) & = \sum_{a \in A} h(a)f_{a}(x + x') \\ & = \sum_{a \in A} h(a)(f_{a}(x) + f_{a}(x')) \\ & = \sum_{a \in A} (h(a)f_{a}(x) + h(a)f_{a}(x')) \\ & = \sum_{a \in A} h(a)f_{a}(x) + \sum_{a \in A}h(a)f_{a}(x') \\ & = \sigma(x)(h) + \sigma(x')(h), \end{align*}$$

and

$$\begin{align*} \sigma(rx)(h) & = \sum_{a \in A} h(a)f_{a}(rx) \\ & = \sum_{a \in A} h(a)rf_{a}(x) \\ & = r\sum_{a \in A} h(a)f_{a}(x) \\ & = r\sigma(x)(h). \end{align*}$$

Suppose another homomorphism $\sigma' : X \to H$ satisfies $\pi_a \circ \sigma' = \pi_a \circ \sigma$ for all $a \in A.$ Then we can look at $\pi_a(\sigma'(x)) = \sigma'(x)(g_a)$ for all $x \in X$ and observe that $\sigma'(x)(g_a) = \sigma(x)(g_a).$

But since $\sigma'(x)(-)$ is an $R$-module homomorphism, it is completely determined by the values it takes on $\{g_a\}_{a \in A}.$ More explicitly, take any $h \in R^{\oplus A}$ and write it as a sum of the form

$$h = \sum_{a \in A}h(a)g_a.$$

This sum has finitely many non-zero terms, and so poses no issue to us. Then

$$\begin{align*} \sigma'(x)(h) &= \sum_{a \in A} \sigma'(x)(h(a)g_a) \\ &= \sum_{a \in A} h(a)\sigma'(x)(g_a) \\ &= \sum_{a \in A} h(a)\sigma(x)(g_a) \\ &= \sigma(x)(h).\\ \end{align*}$$

So $\sigma : X \to H$ is a unique morphism such that for every $a \in A$ we have $\pi_a \circ \sigma = f_a.$ Via $\sigma$, $H$ satisfies the universal property of the product $\prod_{a \in A} R.$

$\blacksquare$

NOTE

This proof demonstrates that $\Hom_{R\mathrm{-Mod}}(R^{\oplus A},R) \cong R^{A}.$ It subsumes 6.6 as a special case where $A$ is finite. Since we have established that $R^A \not\cong R^{\oplus A}$ in general (6.7), we now can conclude that in general

$$\Hom_{R\mathrm{-Mod}}(R^{\oplus A},R) \not\cong R^{\oplus A}.$$