[proofplan] We embed an arbitrary left $R$-module $M$ into its double character module with respect to $\mathbb{Q}/\mathbb{Z}$. The key algebraic facts are that $\mathbb{Q}/\mathbb{Z}$ is an injective cogenerator of abelian groups, and that for every right $R$-module $N$, the character module $\operatorname{Hom}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ is injective as a left $R$-module. Taking $N=\operatorname{Hom}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$ gives an injective left $R$-module, and the evaluation map embeds $M$ into it. [/proofplan]

[step:Show that $\mathbb{Q}/\mathbb{Z}$ is an injective cogenerator of abelian groups] Let $D:=\mathbb{Q}/\mathbb{Z}$, regarded as a $\mathbb{Z}$-module. The group $D$ is divisible: for every $d=q+\mathbb{Z}\in D$ and every integer $n\geq 1$, the element $q/n+\mathbb{Z}\in D$ satisfies \begin{align*} n(q/n+\mathbb{Z})=q+\mathbb{Z}=d. \end{align*} [claim:Every divisible abelian group is injective as a $\mathbb{Z}$-module] Let $D_0$ be a divisible abelian group. For every inclusion of abelian groups $A\subset B$ and every group homomorphism $f:A\to D_0$, there exists a group homomorphism $F:B\to D_0$ such that $F|_A=f$. [/claim] [proof] Consider the set $\mathcal{S}$ of pairs $(C,g)$ where $C$ is a subgroup of $B$ with $A\subset C$, and $g:C\to D_0$ is a group homomorphism satisfying $g|_A=f$. Order $\mathcal{S}$ by extension: $(C,g)\leq (C',g')$ if $C\subset C'$ and $g'|_C=g$. Every chain has an upper bound obtained by taking the union of the subgroups and the compatible union of the homomorphisms, so [Zorn's lemma](/theorems/1226) gives a maximal pair $(C_{\max},g_{\max})$. We prove $C_{\max}=B$. Suppose not, and choose $b\in B\setminus C_{\max}$. Define \begin{align*} I_b:=\{n\in \mathbb{Z}: nb\in C_{\max}\}. \end{align*} Then $I_b$ is an ideal of $\mathbb{Z}$, hence either $I_b=\{0\}$ or $I_b=m\mathbb{Z}$ for a unique integer $m\geq 1$. If $I_b=\{0\}$, define $C':=C_{\max}+\mathbb{Z}b$ and define \begin{align*} g':C'&\to D_0\\ c+kb&\mapsto g_{\max}(c) \end{align*} for $c\in C_{\max}$ and $k\in\mathbb{Z}$. The condition $I_b=\{0\}$ implies that this representation is unique, so $g'$ is well-defined and extends $g_{\max}$, contradicting maximality. If $I_b=m\mathbb{Z}$ with $m\geq 1$, then $mb\in C_{\max}$. Since $D_0$ is divisible, choose $d\in D_0$ such that \begin{align*} md=g_{\max}(mb). \end{align*} Define $C':=C_{\max}+\mathbb{Z}b$ and \begin{align*} g':C'&\to D_0\\ c+kb&\mapsto g_{\max}(c)+kd. \end{align*} To check well-definedness, suppose $c+kb=0$. Then $kb=-c\in C_{\max}$, so $k\in I_b=m\mathbb{Z}$. Write $k=mt$ with $t\in\mathbb{Z}$. Since $c=-mtb$, we get \begin{align*} g_{\max}(c)+kd &=g_{\max}(-tmb)+mtd\\ &=-t g_{\max}(mb)+tmd\\ &=0. \end{align*} Thus $g'$ is a well-defined homomorphism extending $g_{\max}$, again contradicting maximality. Therefore $C_{\max}=B$, and $D_0$ is injective. [/proof] Applying the claim to $D=\mathbb{Q}/\mathbb{Z}$, we conclude that $D$ is injective as a $\mathbb{Z}$-module. It remains to show that $D$ is a cogenerator. Let $A$ be a nonzero abelian group and choose $a\in A$ with $a\neq 0$. If $a$ has infinite order, define a homomorphism \begin{align*} h_0:\mathbb{Z}a&\to D\\ ka&\mapsto k(1/2+\mathbb{Z}). \end{align*} If $a$ has finite order $m\geq 2$, define \begin{align*} h_0:\mathbb{Z}a&\to D\\ ka&\mapsto k(1/m+\mathbb{Z}). \end{align*} In both cases $h_0(a)\neq 0$. Since $D$ is injective, $h_0$ extends to a homomorphism $h:A\to D$. Hence every nonzero abelian group admits a nonzero homomorphism into $D$. [/step]

[step:Build an injective left $R$-module by coinduction from abelian groups] Define the abelian group \begin{align*} E_0:=\operatorname{Hom}_{\mathbb{Z}}(R,D), \end{align*} and make it a left $R$-module by \begin{align*} (r\varphi)(s):=\varphi(sr) \end{align*} for $r\in R$, $\varphi\in E_0$, and $s\in R$. The associativity and unit axioms follow from multiplication in $R$: \begin{align*} ((rs)\varphi)(t) &=\varphi(t rs) =(s\varphi)(t r) =(r(s\varphi))(t),\\ (1_R\varphi)(t) &=\varphi(t1_R) =\varphi(t) \end{align*} for every $t\in R$. We prove that $E_0$ is injective as a left $R$-module. Let $A$ and $B$ be left $R$-modules, let $\alpha:A\to B$ be an injective left $R$-module homomorphism, and let $f:A\to E_0$ be a left $R$-module homomorphism. Define the abelian group homomorphism \begin{align*} f_1:A&\to D\\ a&\mapsto f(a)(1_R). \end{align*} Since $D$ is injective as a $\mathbb{Z}$-module and $\alpha:A\to B$ is an injective homomorphism of abelian groups after forgetting the $R$-module structure, there exists an abelian group homomorphism $g_1:B\to D$ such that $g_1\circ \alpha=f_1$. Define \begin{align*} g:B&\to E_0\\ b&\mapsto g(b), \end{align*} where \begin{align*} g(b):R&\to D\\ s&\mapsto g_1(sb). \end{align*} For each $b\in B$, the map $g(b):R\to D$ is a homomorphism of abelian groups because the scalar multiplication map $R\to B$, $s\mapsto sb$, is additive and $g_1$ is additive. The map $g$ is additive for the same reason. It is left $R$-linear: for $r\in R$, $b\in B$, and $s\in R$, \begin{align*} g(rb)(s) &=g_1(srb)\\ &=g(b)(sr)\\ &=(rg(b))(s). \end{align*} Finally, for $a\in A$ and $s\in R$, the $R$-linearity of $\alpha$ and $f$ gives \begin{align*} g(\alpha(a))(s) &=g_1(s\alpha(a))\\ &=g_1(\alpha(sa))\\ &=f_1(sa)\\ &=f(sa)(1_R)\\ &=(sf(a))(1_R)\\ &=f(a)(1_Rs)\\ &=f(a)(s). \end{align*} Thus $g\circ \alpha=f$. Hence every left $R$-module homomorphism $A\to E_0$ extends across every monomorphism $A\hookrightarrow B$, so $E_0$ is injective as a left $R$-module. [/step]

[step:Embed an arbitrary left $R$-module into a product of coinduced injectives] Let $M$ be a left $R$-module. Define the abelian group \begin{align*} M^\vee:=\operatorname{Hom}_{\mathbb{Z}}(M,D). \end{align*} Let $E$ be the product of copies of $E_0=\operatorname{Hom}_{\mathbb{Z}}(R,D)$ indexed by the set $M^\vee$: \begin{align*} E:=\prod_{\lambda\in M^\vee}E_0. \end{align*} We give $E$ the product left $R$-module structure. The product $E$ is injective. Indeed, if $A\hookrightarrow B$ is a monomorphism of left $R$-modules, then applying $\operatorname{Hom}_R(-,E)$ gives \begin{align*} \operatorname{Hom}_R(B,E) &\cong \prod_{\lambda\in M^\vee}\operatorname{Hom}_R(B,E_0),\\ \operatorname{Hom}_R(A,E) &\cong \prod_{\lambda\in M^\vee}\operatorname{Hom}_R(A,E_0). \end{align*} Since each $E_0$ is injective by the preceding step, every component map $\operatorname{Hom}_R(B,E_0)\to \operatorname{Hom}_R(A,E_0)$ is surjective; hence their product is surjective. Thus every left $R$-module homomorphism $A\to E$ extends across $A\hookrightarrow B$, so $E$ is injective. Define \begin{align*} \iota:M&\to E\\ m&\mapsto (\iota_\lambda(m))_{\lambda\in M^\vee}, \end{align*} where, for each $\lambda\in M^\vee$, \begin{align*} \iota_\lambda(m):R&\to D\\ s&\mapsto \lambda(sm). \end{align*} For each $m\in M$ and $\lambda\in M^\vee$, the map $\iota_\lambda(m):R\to D$ is an abelian group homomorphism, so $\iota_\lambda(m)\in E_0$. The map $\iota$ is left $R$-linear because for $r\in R$, $m\in M$, $\lambda\in M^\vee$, and $s\in R$, \begin{align*} \iota_\lambda(rm)(s) &=\lambda(srm)\\ &=\iota_\lambda(m)(sr)\\ &=(r\iota_\lambda(m))(s). \end{align*} We show that $\iota$ is injective. Let $m\in M$ with $m\neq 0$. Since $D$ is a cogenerator of abelian groups, there exists an abelian group homomorphism $\lambda:M\to D$ such that $\lambda(m)\neq 0$. This $\lambda$ is an element of $M^\vee$, and the $\lambda$-component of $\iota(m)$ satisfies \begin{align*} \iota_\lambda(m)(1_R)=\lambda(1_Rm)=\lambda(m)\neq 0. \end{align*} Hence $\iota(m)\neq 0$, so $\ker \iota=0$. Thus $\iota:M\to E$ is an injective $R$-[linear map](/page/Linear%20Map) into an injective left $R$-module. [/step]

[step:Conclude that $R\operatorname{-Mod}$ has enough injectives] For the arbitrary left $R$-module $M$, we have constructed an injective left $R$-module \begin{align*} E=\prod_{\lambda\in \operatorname{Hom}_{\mathbb{Z}}(M,D)}\operatorname{Hom}_{\mathbb{Z}}(R,D) \end{align*} and an injective $R$-linear map $\iota:M\to E$. Since $M$ was arbitrary, every object of $R\operatorname{-Mod}$ embeds into an injective object. Therefore $R\operatorname{-Mod}$ has enough injectives. [/step]