Module categories have enough injectives

Regard $M$ as an abelian group. By [L1], choose an embedding $j:M\to D$ into a divisible abelian group $D$. Under AC, [L5] makes $D$ injective as an abelian group.

Now assume $R$ commutative and put $D_0=\mathbb{Q}\mathrm{/}\mathbb{Z}$. This group is divisible, hence injective by [L5]. For every nonzero $m\in M$, define a nonzero map on the cyclic subgroup $\mathbb{Z}m$ by sending $m$ to $1\mathrm{/}n+\mathbb{Z}$ when $m$ has finite order $n$, or to $1\mathrm{/}2+\mathbb{Z}$ when $m$ has infinite order; injectivity extends it to $M$. Therefore evaluation $M\to M^{\vee \vee }$ is injective.

Define $\eta :M\to {\mathrm{Hom}}_{\mathbb{Z}}(R,D)$ by $\eta (m)(r)=j(rm)$. It is $R$-linear for the coinduced action, and evaluation at $1$ gives $\eta (m)(1)=j(m)$, so $\eta$ is injective.

Let $\epsilon :R^{(M^{\vee })}\to M^{\vee }$ be the canonical free cover given by [L4]. Precomposition with its surjection embeds $M^{\vee \vee }$ into $J(M)=(R^{(M^{\vee })}{)}^{\vee }$.

The target in step 2.1 is injective by [L2], proving enough injectives for arbitrary unital rings.

A homomorphism from a direct sum to $D_0$ is the same as a family of homomorphisms from its summands, so $J(M)\cong {\prod }_{\varphi \in M^{\vee }}{R}^{\vee }$. Each $R^{\vee }$ is injective by [L2], and the product is injective by [L3].

A module map $M\to N$ induces $N^{\vee }\to M^{\vee }$, then a map between the canonical free modules, and dualizing reverses direction again; these maps commute with evaluation and the free covers, so $M\mapsto J(M)$ and the embeddings are functorial.

Steps 1.1, 2.1, and 3.1 prove enough injectives over every unital ring. Steps 1.2, 2.2, 3.2, and 4.1 give the stated functorial commutative-ring construction. Full AC enters through injectivity of divisible groups and arbitrary products, not through the divisible-hull embedding itself.