Injectivity of Divisible Modules over a PID

Definition.

Let $R$ be an integral domain. An $R$-module $M$ is divisible if for every $m\in M$ and every non-zero $r\in R$, there exists an element $x\in M$ such that $rx=m$.

Theorem.

Let $R$ be a Principal Ideal Domain (PID). An $R$-module $D$ is injective if and only if it is divisible.

Proof. Assume $D$ is an injective $R$-module. Let $d\in D$ and $0\ne r\in R$. Consider the ideal $I=(r)$. Define an $R$-module homomorphism $f:(r)\to D$ by $f(sr)=sd$. This map is well-defined as $R$ is an integral domain. By the injectivity of $D$, there exists an extension $\tilde{f}:R\to D$ such that $\tilde{f}{|}_{(r)}=f$. Let $x=\tilde{f}(1)\in D$. Then $rx=r\tilde{f}(1)=\tilde{f}(r)=f(r)=d$. Thus, $D$ is divisible.

Conversely, assume $D$ is a divisible $R$-module. By Baer’s Criterion, it suffices to show that any homomorphism $f:I\to D$ from an ideal $I$ of $R$ extends to a homomorphism $\tilde{f}:R\to D$. Since $R$ is a PID, $I=(r)$ for some $r\in R$. If $r=0$, the extension is trivial. Assume $r\ne 0$. Let $f(r)=d\in D$. As $D$ is divisible, there exists an element $x\in D$ such that $rx=d$. Define $\tilde{f}:R\to D$ by $\tilde{f}(s)=sx$. This is an $R$-module homomorphism. For any element $sr\in (r)$, we have $\tilde{f}(sr)=(sr)x=s(rx)=sd$. We also have $f(sr)=sf(r)=sd$. Thus, $\tilde{f}$ extends $f$. By Baer’s Criterion, $D$ is an injective $R$-module. $square$

Theorem.

The $\mathbb{Z}$-module $\mathbb{Q}/\mathbb{Z}$ is injective.

Proof. The ring of integers $\mathbb{Z}$ is a PID. We show that $\mathbb{Q}/\mathbb{Z}$ is a divisible $\mathbb{Z}$-module. Let $q+\mathbb{Z}\in \mathbb{Q}/\mathbb{Z}$ for some $q\in \mathbb{Q}$, and let $n\in \mathbb{Z}$ be a non-zero integer. We seek an element $x+\mathbb{Z}\in \mathbb{Q}/\mathbb{Z}$ such that $n(x+\mathbb{Z})=q+\mathbb{Z}$. Let $x=q/n$. Since $q\in \mathbb{Q}$ and $n\in \mathbb{Z}$ is non-zero, $x\in \mathbb{Q}$. Then $n(x+\mathbb{Z})=nx+\mathbb{Z}=n(q/n)+\mathbb{Z}=q+\mathbb{Z}$. Thus, $\mathbb{Q}/\mathbb{Z}$ is divisible. Since $\mathbb{Z}$ is a PID and $\mathbb{Q}/\mathbb{Z}$ is a divisible $\mathbb{Z}$-module, it is injective by the preceding theorem. $\square$