Exercise A0.III.6.01

Preliminaries

Lemma (L3.1)

Let $R$ be a commutative ring. Then given any $R$-module $M$, the $R$-module morphism $\varphi : R \to M$ is uniquely determined by the image of $1$.

Proof

Since $\varphi$ is an $R$-module morphism, for any $r \in R$ we have $\varphi(r) = \varphi(r1) = r \varphi(1).$ $\blacksquare$

Lemma (L3.2)

Let $R$ be a commutative ring. Then given any $R$-module $M$ and an element $m \in M$ there exists a morphism $\varphi : R \to M$ such that $\varphi(1) = m.$

Proof

Construct a set function $\varphi(r) = rm.$ It is an $R$-module homomorphism, since

$$\varphi(r_1r_2) = r_1r_2 m = r_1(r_2 m) = r_1 \varphi(r_2),$$

and

$$\varphi(r_1 + r_2) = (r_1 + r_2)m = r_1m + r_2m = \varphi(r_1) + \varphi(r_2).$$

$\blacksquare$

Exercise

Proposition

Let $A$ be any set, let $R$ be a commutative ring, and let $F^R(A)$ be the free $R$-module on $A.$ Let $R^{\oplus A}$ be the direct sum of $R$ indexed by $A.$ Then $F^R(A) \cong R^{\oplus A}.$

Proof

Note that as a direct sum $R^{\oplus A}$ comes with an injection for each element of $A.$ We can think of it as a map $i : A \to R^{\oplus A}$ such that for any $a,b \in A$, $(i(a))_a = 1$ and if $a \neq b$ then $(i(a))_b = 0.$

Let $M$ be any $R$-module and $f : A \to M$ be any set-function. For each element $a \in A,$ construct using L3.1 a morphism $\varphi_a : R \to M$ such that $\varphi_a(1) = f(a).$ By L3.2 we know that there is one such morphism for each $a.$

Now, consider the map

$$\varphi : R^{\oplus A} \to M \quad \text{given by} \quad \varphi=\sum_{a\in A}(\varphi_a \circ \pi_a).$$

This function is well-defined, since any $v \in R^{\oplus A}$ is non-zero on finitely many indices and hence all but a finite subset of the summands vanish. Is it a homomorphism? For any $v \in R^{\oplus A}$ we have

$$\varphi(v) = \varphi_{m_1}(v_{m_1}) + \varphi_{m_2}(v_{m_2}) + \ldots + \varphi_{m_k}(v_{m_k}),$$

and so $\varphi$ clearly satisfies all requirements.

By construction, $\varphi(i(a)) = \varphi_a (\pi_a (i(a))) = \varphi_a(1) = f(a).$ Since $\varphi$ is a copairing of uniquely determined morphisms $\varphi_a$, it is itself uniquely determined.

Thus, we have constructed the morphism $\varphi$ required for $R^{\oplus A}$ to satisfy the universal property of $F^R(A).$ This forces unique isomorphism $F^R(A) \cong R^{\oplus A}.$

$\blacksquare$