[proofplan] We construct an injective target using the character group $\operatorname{Hom}_{\mathbb{Z}}(-,\mathbb{Q}/\mathbb{Z})$. First we prove that $\mathbb{Q}/\mathbb{Z}$ is an injective cogenerator in the category of abelian groups. Then we use the coinduced left $R$-module $\operatorname{Hom}_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$, whose injectivity follows from the Hom-adjunction and the abelian-group injectivity just proved. Finally, all characters of the underlying abelian group of $M$ define an $R$-[linear map](/page/Linear%20Map) from $M$ into a product of these coinduced modules, and the cogenerator property separates nonzero elements. [/proofplan]

[step:Prove that $\mathbb{Q}/\mathbb{Z}$ is an injective cogenerator of abelian groups] Let $D$ denote the abelian group $\mathbb{Q}/\mathbb{Z}$. [claim:$D$ is injective as a $\mathbb{Z}$-module] [/claim] [proof] Let $A$ be an abelian group, let $B \subset A$ be a subgroup, and let $\varphi: B \to D$ be a group homomorphism. We prove that $\varphi$ extends to a group homomorphism $\Phi: A \to D$. Let $\mathcal{P}$ be the set of pairs $(C,\psi)$ such that $C$ is a subgroup of $A$ with $B \subset C$ and $\psi: C \to D$ is a group homomorphism satisfying $\psi|_B=\varphi$. Order $\mathcal{P}$ by declaring $(C,\psi) \le (C',\psi')$ when $C \subset C'$ and $\psi'|_C=\psi$. Every chain in $\mathcal{P}$ has an upper bound obtained by taking the union of the subgroups in the chain and the uniquely induced homomorphism on that union. By [Zorn's lemma](/theorems/1226), choose a maximal element $(C,\psi)$ of $\mathcal{P}$. We show $C=A$. Suppose not, and choose $a \in A \setminus C$. Let $C' := C+\mathbb{Z}a$ be the subgroup generated by $C$ and $a$. Define \begin{align*} N := \{n \in \mathbb{N} : n a \in C\}. \end{align*} If $N=\varnothing$, then every element of $C'$ has a unique form $c+n a$ with $c \in C$ and $n \in \mathbb{Z}$. Choose any element $u \in D$, and define $\psi': C' \to D$ by \begin{align*} \psi'(c+n a) := \psi(c)+n u. \end{align*} The uniqueness of the representation makes $\psi'$ well-defined, and direct addition of representatives shows that $\psi'$ is a group homomorphism extending $\psi$, contradicting maximality. It remains to treat the case $N \ne \varnothing$. Let $m := \min N$. Then $m a \in C$. We claim that every relation $c+n a=0$ with $c \in C$ and $n \in \mathbb{Z}$ satisfies $m \mid n$. Indeed, if $n=qm+r$ with $q \in \mathbb{Z}$ and $0 \le r < m$, then \begin{align*} r a = n a - q m a \in C. \end{align*} By the minimality of $m$, this forces $r=0$. Since $D=\mathbb{Q}/\mathbb{Z}$ is divisible, there exists $u \in D$ such that \begin{align*} m u = \psi(m a). \end{align*} Define $\psi': C' \to D$ by \begin{align*} \psi'(c+n a) := \psi(c)+n u. \end{align*} To verify well-definedness, suppose $c+n a=c'+n'a$. Then $(c-c')+(n-n')a=0$, so $m \mid n-n'$. Write $n-n'=qm$. Since $(n-n')a=-(c-c')$, we have $qma=c'-c$. Therefore \begin{align*} \psi(c)-\psi(c')+(n-n')u &= \psi(c-c')+qm u \\ &= \psi(c-c')+q\psi(m a) \\ &= \psi(c-c'+qma) \\ &= \psi(0) \\ &=0. \end{align*} Thus $\psi(c)+nu=\psi(c')+n'u$, so $\psi'$ is well-defined. It is a group homomorphism by direct addition of representatives and extends $\psi$, again contradicting maximality. Hence $C=A$, and $D$ is injective as a $\mathbb{Z}$-module. [/proof] [claim:$D$ separates points of abelian groups] [/claim] [proof] Let $A$ be an abelian group and let $a \in A$ be nonzero. Let $C:=\mathbb{Z}a$ be the cyclic subgroup generated by $a$. If $a$ has infinite order, define $\theta: C \to D$ by \begin{align*} \theta(n a) := n\left(\frac{1}{2}+\mathbb{Z}\right). \end{align*} Then $\theta(a)=\frac{1}{2}+\mathbb{Z} \ne 0$. If $a$ has finite order $m \ge 2$, define $\theta: C \to D$ by \begin{align*} \theta(n a) := \frac{n}{m}+\mathbb{Z}. \end{align*} This is well-defined because $m a=0$ and $m(\frac{1}{m}+\mathbb{Z})=0$, and $\theta(a)=\frac{1}{m}+\mathbb{Z} \ne 0$. In either case, the injectivity of $D$ extends $\theta$ to a homomorphism $\Theta: A \to D$. Since $\Theta(a)=\theta(a)\ne 0$, the group $D$ separates points. [/proof] [/step]

[step:Build the coinduced injective left $R$-module] Define the abelian group \begin{align*} C := \operatorname{Hom}_{\mathbb{Z}}(R,D). \end{align*} Give $C$ the left $R$-module structure defined by \begin{align*} (r\cdot f)(s) := f(sr) \end{align*} for $r,s \in R$ and $f \in C$. The associativity and identity axioms follow from associativity and the identity element of $R$: \begin{align*} ((r_1r_2)\cdot f)(s)=f(s r_1 r_2)=(r_1\cdot(r_2\cdot f))(s), \qquad (1_R\cdot f)(s)=f(s). \end{align*} We prove that $C$ is injective as a left $R$-module. Let $N \subset P$ be a submodule inclusion of left $R$-modules, and let $F: N \to C$ be an $R$-linear homomorphism. Define the underlying abelian-group homomorphism \begin{align*} \alpha: N &\to D \\ n &\mapsto F(n)(1_R). \end{align*} Since $D$ is injective as a $\mathbb{Z}$-module, there exists a group homomorphism $\beta: P \to D$ such that $\beta|_N=\alpha$. Define \begin{align*} \widetilde{F}: P &\to C \\ p &\mapsto \bigl(r \mapsto \beta(rp)\bigr). \end{align*} For each $p \in P$, the map $r \mapsto \beta(rp)$ is a group homomorphism $R \to D$, so $\widetilde{F}(p)\in C$. The map $\widetilde{F}$ is additive because $\beta$ is additive. For $a \in R$, $p \in P$, and $r \in R$, \begin{align*} \widetilde{F}(a p)(r) = \beta(r a p) = \widetilde{F}(p)(r a) = (a\cdot \widetilde{F}(p))(r), \end{align*} so $\widetilde{F}$ is $R$-linear. Finally, if $n \in N$ and $r \in R$, then $rn \in N$, and the $R$-linearity of $F$ gives \begin{align*} \widetilde{F}(n)(r) = \beta(rn) = \alpha(rn) = F(rn)(1_R) = (r\cdot F(n))(1_R) = F(n)(r). \end{align*} Thus $\widetilde{F}|_N=F$. Therefore every $R$-linear map from a submodule into $C$ extends across the ambient module, so $C$ is injective. [/step]

[step:Show that products of injective left $R$-modules are injective] Let $(I_j)_{j \in J}$ be a family of injective left $R$-modules indexed by a set $J$, and define their product left $R$-module \begin{align*} I := \prod_{j \in J} I_j. \end{align*} Let $N \subset P$ be a submodule inclusion, and let $F: N \to I$ be an $R$-linear homomorphism. For each $j \in J$, let \begin{align*} \pi_j: I \to I_j \end{align*} be the $j$-th coordinate projection, and define $F_j:=\pi_j\circ F: N \to I_j$. Since $I_j$ is injective, there exists an $R$-linear homomorphism $\widetilde{F}_j: P \to I_j$ extending $F_j$. Define \begin{align*} \widetilde{F}: P &\to I \\ p &\mapsto \bigl(\widetilde{F}_j(p)\bigr)_{j \in J}. \end{align*} The map $\widetilde{F}$ is $R$-linear coordinatewise. For $n \in N$, each coordinate satisfies \begin{align*} \pi_j(\widetilde{F}(n))=\widetilde{F}_j(n)=F_j(n)=\pi_j(F(n)). \end{align*} Hence $\widetilde{F}|_N=F$. Thus $I$ is injective. [/step]

[step:Embed $M$ into a product of coinduced injective modules] Let \begin{align*} \Lambda := \operatorname{Hom}_{\mathbb{Z}}(M,D) \end{align*} be the set of all group homomorphisms from the underlying abelian group of $M$ to $D$. Define \begin{align*} I := \prod_{\lambda \in \Lambda} C, \end{align*} where each factor is the injective left $R$-module $C=\operatorname{Hom}_{\mathbb{Z}}(R,D)$ constructed above. By the previous step, $I$ is injective. Define \begin{align*} \iota: M &\to I \\ m &\mapsto \bigl(\iota_\lambda(m)\bigr)_{\lambda \in \Lambda}, \end{align*} where, for each $\lambda \in \Lambda$, \begin{align*} \iota_\lambda(m): R &\to D \\ r &\mapsto \lambda(rm). \end{align*} For fixed $m \in M$, the map $\iota_\lambda(m)$ is a group homomorphism $R \to D$, so $\iota_\lambda(m)\in C$. The map $\iota$ is additive because every $\lambda$ is additive. If $a \in R$, $m \in M$, $\lambda \in \Lambda$, and $r \in R$, then \begin{align*} \iota_\lambda(a m)(r) = \lambda(r a m) = \iota_\lambda(m)(r a) = (a\cdot \iota_\lambda(m))(r), \end{align*} so $\iota_\lambda(a m)=a\cdot \iota_\lambda(m)$ for every $\lambda$, and therefore $\iota$ is $R$-linear. It remains to prove that $\iota$ is injective. Let $m \in M$ with $m \ne 0$. By the point-separating property of $D$, applied to the underlying abelian group of $M$, there exists $\lambda \in \Lambda$ such that $\lambda(m)\ne 0$. Evaluating the $\lambda$-coordinate of $\iota(m)$ at $1_R$ gives \begin{align*} \iota_\lambda(m)(1_R)=\lambda(1_Rm)=\lambda(m)\ne 0. \end{align*} Hence $\iota(m)\ne 0$. Thus $\ker \iota=\{0\}$, so $\iota$ is an injective $R$-linear homomorphism from $M$ into the injective left $R$-module $I$. [/step]