Injectivity of Divisible Modules over PIDs and of Torsion-Free Divisible Modules over Domains

Injectivity of Divisible Modules over PIDs and of Torsion-Free Divisible Modules over Domains.

Definition (Integral Domain). A commutative ring $R$ with $1\ne 0$ is an integral domain if for all $a,b\in R$, $ab=0$ implies $a=0$ or $b=0$.

Definition (Principal Ideal Domain). A commutative ring $R$ with $1\ne 0$ is a principal ideal domain if $R$ is an integral domain and every ideal $I\subseteq R$ has the form $I=(a)$ for some $a\in R$.

Definition (Torsion-Free Module). Let $R$ be a domain. An $R$-module $M$ is torsion-free if for all $0\ne r\in R$ and $m\in M$, $rm=0$ implies $m=0$.

Definition (Divisible Module). Let $R$ be a domain. An $R$-module $M$ is divisible if for every $0\ne r\in R$ and every $m\in M$ there exists $x\in M$ with $rx=m$.

Definition (Injective Module). Let $R$ be a commutative ring with $1\ne 0$. An $R$-module $M$ is injective if for every monomorphism of $R$-modules $A\hookrightarrow B$, every $R$-linear map $A\to M$ extends to an $R$-linear map $B\to M$.

Theorem (Baer’s Criterion). Let $R$ be a commutative ring with $1\ne 0$ and let $M$ be an $R$-module. The module $M$ is injective if and only if for every ideal $I\subseteq R$, every $R$-linear map $f:I\to M$ extends to an $R$-linear map $F:R\to M$ with $F{|}_I=f$.

Theorem (Divisible over PID $\Rightarrow$ Injective). Let $R$ be a principal ideal domain and let $M$ be a divisible $R$-module. Then $M$ is injective.

Proof. By Baer’s criterion it suffices to extend $f:I\to M$ for an ideal $I\subseteq R$. If $I=(0)$ take $F=0$. If $I=(a)$ with $a\ne 0$, set $m:=f(a)$ and choose $x\in M$ with $ax=m$ by divisibility. Define $F:R\to M$ by $F(r):=rx$. Then for all $r\in R$,

$$\begin{array}{rl}F(ar)& =arx=rax=rf(a)=f(ar),\end{array}$$

so $F{|}_{(a)}=f$. Hence $M$ is injective. $\square$

Theorem (Torsion-Free Divisible over Domain $\Rightarrow$ Injective). Let $R$ be an integral domain and let $M$ be an $R$-module that is torsion-free and divisible. Then $M$ is injective.

Proof. By Baer’s criterion it is enough to extend $f:I\to M$ for an ideal $I\subseteq R$. If $I=(0)$ take $F=0$. Otherwise choose $0\ne a\in I$ and set $m_b:=f(b)$ for $b\in I$. By divisibility choose $x\in M$ with $ax=m_a$. For any $b\in I$,

$$\begin{array}{rl}a(bx-m_b)& =abx-a{m}_b=b(ax)-a{m}_b=b{m}_a-a{m}_b\\ & =f(ba)-f(ab)=0.\end{array}$$

Since $a\ne 0$ and $M$ is torsion-free, $bx-m_b=0$. Thus $bx=f(b)$ for all $b\in I$. Define $F:R\to M$ by $F(r):=rx$. Then $F$ is $R$-linear and $F{|}_I=f$. Hence $M$ is injective. $\square$