[proofplan] We prove the two implications separately. If $D$ is injective, we apply the extension property to the inclusion $n\mathbb{Z} \subset \mathbb{Z}$ and a homomorphism sending $n$ to a chosen element $d \in D$; the image of $1$ under the extension is an $n$-th divisor of $d$. Conversely, assuming divisibility, we verify the hypotheses of [Baer's criterion](/theorems/4203) for $\mathbb{Z}$ by extending every homomorphism from an ideal of $\mathbb{Z}$ to $D$ over all of $\mathbb{Z}$. [/proofplan]

[step:Use injectivity to solve the divisibility equation]Assume that $D$ is an injective $\mathbb{Z}$-module. Let $d \in D$ and let $n \geq 1$ be an integer. Define the inclusion homomorphism \begin{align*} \iota_n : n\mathbb{Z} &\to \mathbb{Z} \\ nk &\mapsto nk \end{align*} and define the $\mathbb{Z}$-[linear map](/page/Linear%20Map) \begin{align*} f_d : n\mathbb{Z} &\to D \\ nk &\mapsto kd. \end{align*} The expression $nk$ determines $k \in \mathbb{Z}$ uniquely because $n \neq 0$, so $f_d$ is well-defined. Since $D$ is injective and $\iota_n$ is an injective $\mathbb{Z}$-module homomorphism, there exists a $\mathbb{Z}$-linear map \begin{align*} F : \mathbb{Z} &\to D \end{align*} such that $F \circ \iota_n = f_d$. Set $x := F(1) \in D$. By $\mathbb{Z}$-linearity of $F$, \begin{align*} nx = nF(1) = F(n) = f_d(n) = d. \end{align*} Thus for every $d \in D$ and every integer $n \geq 1$, there exists $x \in D$ with $nx = d$. Hence $D$ is divisible.[/step]

[guided]Assume that $D$ is an injective $\mathbb{Z}$-module. To prove divisibility, we must solve the equation $nx = d$ for an arbitrary element $d \in D$ and an arbitrary integer $n \geq 1$. The injectivity of $D$ says that homomorphisms into $D$ extend across injective homomorphisms. We therefore encode the desired equation into a homomorphism defined on the submodule $n\mathbb{Z} \subset \mathbb{Z}$. Define \begin{align*} \iota_n : n\mathbb{Z} &\to \mathbb{Z} \\ nk &\mapsto nk \end{align*} to be the inclusion map, and define \begin{align*} f_d : n\mathbb{Z} &\to D \\ nk &\mapsto kd. \end{align*} This map is well-defined because $n \geq 1$: if $nk = n\ell$, then $k=\ell$. It is $\mathbb{Z}$-linear because for all $a,k,\ell \in \mathbb{Z}$, \begin{align*} f_d(a(nk)+n\ell) = f_d(n(ak+\ell)) = (ak+\ell)d = a(kd)+\ell d = a f_d(nk)+f_d(n\ell). \end{align*} Since $\iota_n$ is injective and $D$ is injective, the map $f_d$ extends to a $\mathbb{Z}$-linear map \begin{align*} F : \mathbb{Z} &\to D \end{align*} satisfying $F \circ \iota_n = f_d$. Now define $x := F(1)$. The value of $F$ at $1$ is the candidate divisor of $d$. Using $\mathbb{Z}$-linearity and the extension identity, \begin{align*} nx = nF(1) = F(n) = (F \circ \iota_n)(n) = f_d(n) = d. \end{align*} Thus $d$ has an $n$-th divisor in $D$. Since $d$ and $n$ were arbitrary, $D$ is divisible.[/guided]

[step:Extend homomorphisms from all ideals of $\mathbb{Z}$ using divisibility]Assume that $D$ is divisible. Let $I \trianglelefteq \mathbb{Z}$ be an ideal and let \begin{align*} f : I &\to D \end{align*} be a $\mathbb{Z}$-linear map. Since every ideal of $\mathbb{Z}$ has the form $n\mathbb{Z}$ for a unique integer $n \geq 0$, choose $n \geq 0$ such that $I = n\mathbb{Z}$. If $n=0$, then $I=\{0\}$ and the zero homomorphism \begin{align*} F : \mathbb{Z} &\to D \\ k &\mapsto 0 \end{align*} extends $f$. Now suppose $n \geq 1$. Set $d := f(n) \in D$. Since $D$ is divisible, choose $x \in D$ such that $nx=d$. Define \begin{align*} F : \mathbb{Z} &\to D \\ k &\mapsto kx. \end{align*} Then $F$ is $\mathbb{Z}$-linear. For every $k \in \mathbb{Z}$, \begin{align*} F(nk) = nkx = k(nx) = kd = kf(n) = f(kn), \end{align*} where the last equality uses $\mathbb{Z}$-linearity of $f$. Hence $F|_{n\mathbb{Z}} = f$.[/step]

[guided]Assume that $D$ is divisible. We want to prove injectivity, and Baer's criterion reduces this to extending homomorphisms from ideals of $\mathbb{Z}$ into $D$. Thus we fix an arbitrary ideal $I \trianglelefteq \mathbb{Z}$ and an arbitrary $\mathbb{Z}$-linear map \begin{align*} f : I &\to D. \end{align*} Every ideal of $\mathbb{Z}$ is principal, so there is a unique integer $n \geq 0$ with $I = n\mathbb{Z}$. There are two cases. If $n=0$, then $I=0\mathbb{Z}=\{0\}$. Since any $\mathbb{Z}$-linear map sends $0$ to $0$, the map $f$ is the zero map on $\{0\}$. Define \begin{align*} F : \mathbb{Z} &\to D \\ k &\mapsto 0. \end{align*} This map is $\mathbb{Z}$-linear and satisfies $F|_I=f$. Now suppose $n \geq 1$. The homomorphism $f$ is determined by the single value $f(n)$, because every element of $n\mathbb{Z}$ has the form $nk$ and $f(nk)=kf(n)$. Define $d := f(n) \in D$. Since $D$ is divisible, there exists $x \in D$ such that \begin{align*} nx=d. \end{align*} This element $x$ tells us what the extension should do to $1 \in \mathbb{Z}$. Define \begin{align*} F : \mathbb{Z} &\to D \\ k &\mapsto kx. \end{align*} The map $F$ is $\mathbb{Z}$-linear by the module axioms for $D$. For every $k \in \mathbb{Z}$, we compute \begin{align*} F(nk) = nkx = k(nx) = kd = kf(n) = f(kn). \end{align*} Since $kn=nk$, this shows $F(nk)=f(nk)$ for every element $nk \in n\mathbb{Z}$. Therefore $F|_{n\mathbb{Z}}=f$, so $f$ extends from $I$ to all of $\mathbb{Z}$.[/guided]

[step:Apply Baer's criterion to conclude injectivity] We have shown that for every ideal $I \trianglelefteq \mathbb{Z}$ and every $\mathbb{Z}$-linear map $f:I \to D$, there exists a $\mathbb{Z}$-linear map $F:\mathbb{Z}\to D$ such that $F|_I=f$. By Baer's criterion for injective modules over a ring (citing a result not yet in the wiki: Baer's criterion for injective modules), applied to the ring $\mathbb{Z}$ and the $\mathbb{Z}$-module $D$, this extension property implies that $D$ is injective as a $\mathbb{Z}$-module. Combining this implication with the first step proves that $D$ is injective if and only if $D$ is divisible. [/step]