Torsion-Free and Divisible Modules over an Integral Domain

Definitions.

Definition (Torsion-Free and Divisible). Let $R$ be an integral domain and let $M$ be an $R$-module. The module $M$ is torsion-free if for every non-zero $r\in R$ and any $x\in M$, the relation $rx=0$ implies $x=0$. The module $M$ is divisible if for every non-zero $r\in R$ and every $x\in M$, there exists some $y\in M$ such that $ry=x$.

Main Result.

Theorem (Characterization via Torsion-Free and Divisibility). Let $R$ be an integral domain with fraction field $K=\mathrm{Frac}(R)$, and let $M$ be an $R$-module. Then $M$ can be given the structure of a $K$-vector space that extends its $R$-module structure if and only if $M$ is a torsion-free and divisible $R$-module.

Proof. ($\Rightarrow$) Suppose $M$ is a $K$-vector space with an action compatible with the $R$-action. Let $r\in R\setminus \{0\}$ and $x\in M$. If $rx=0$, then since $r$ is invertible in $K$, $x=(r^{-1}r)x=r^{-1}(rx)=r^{-1}(0)=0$. Thus, $M$ is torsion-free. For divisibility, let $r\in R\setminus \{0\}$ and $x\in M$. Let $y=r^{-1}x\in M$. Then $ry=r(r^{-1}x)=(r{r}^{-1})x=1x=x$. Thus, $M$ is divisible.

($\Leftarrow$) Assume $M$ is a torsion-free and divisible $R$-module. We define the action of $k=a/b\in K$ on $x\in M$ to be the unique element $y\in M$ such that $by=ax$. Existence of such a $y$ is guaranteed by divisibility, as $ax\in M$ and $b\ne 0$. Uniqueness is guaranteed by torsion-freeness, since if $b{y}_1=b{y}_2$, then $b(y_1-y_2)=0$, which implies $y_1=y_2$ as $b\ne 0$.

The action is well-defined. If $a/b=c/d$, then $ad=bc$. Let $y_1=(a/b)\cdot x$ and $y_2=(c/d)\cdot x$, so $b{y}_1=ax$ and $d{y}_2=cx$. Then

$$db(y_1-y_2)=d(b{y}_1)-b(d{y}_2)=d(ax)-b(cx)=(ad-bc)x=0.$$

Since $db\ne 0$ and $M$ is torsion-free, $y_1-y_2=0$.

The vector space axioms are verified by clearing denominators and applying torsion-freeness. For $k=a/b,k^{\prime }=c/d\in K$ and $x,z\in M$:

$$b(k\cdot (x+z)-k\cdot x-k\cdot z)=a(x+z)-ax-az=0.$$ $$bd((k+k^{\prime })\cdot x-k\cdot x-k^{\prime }\cdot x)=(ad+bc)x-d(ax)-b(cx)=0.$$ $$bd((k{k}^{\prime })\cdot x-k\cdot (k^{\prime }\cdot x))=acx-a(d(k^{\prime }\cdot x))=acx-a(cx)=0.$$

The identity axiom $1\cdot x=x$ follows from the definition with $1=1/1$. The action extends the $R$-action, since for $r\in R$, $(r/1)\cdot x$ is the unique $y$ with $1\cdot y=rx$, so $y=rx$. $\square$

Generalization.

Definition (S-Torsion-Free and S-Divisible Module). Let $R$ be a commutative ring and let $S\subseteq R$ be a multiplicative set consisting of non-zero-divisors. An $R$-module $M$ is $S$-torsion-free if for every $s\in S$ and $x\in M$, the relation $sx=0$ implies $x=0$. The module $M$ is $S$-divisible if for every $s\in S$ and every $x\in M$, there exists some $y\in M$ such that $sy=x$.

Theorem (Extension to the Ring of Fractions). Let $R$ be a commutative ring, $S\subseteq R$ a multiplicative set of non-zero-divisors, and $M$ an $R$-module. Let $Q=S^{-1}R$ be the ring of fractions. The $R$-module $M$ admits a unique $Q$-module structure that extends the $R$-action if and only if $M$ is $S$-torsion-free and $S$-divisible.

Proof. ($\Rightarrow$) Assume $M$ has a $Q$-module structure extending the $R$-action. For any $s\in S$, its image $s/1\in Q$ is a unit. If $sx=0$, then $x=(1/s)(sx)=0$, so $M$ is $S$-torsion-free. For any $x\in M$, let $y=(1/s)x$. Then $sy=x$, so $M$ is $S$-divisible.

($\Leftarrow$) Assume $M$ is $S$-torsion-free and $S$-divisible. Define the action of $a/s\in Q$ on $x\in M$ to be the unique $y\in M$ such that $sy=ax$. Existence follows from $S$-divisibility and uniqueness from $S$-torsion-freeness. For well-definedness, if $a/s=a^{\prime }/s^{\prime }$, there exists $t\in S$ with $t(a{s}^{\prime }-a^{\prime }s)=0$. Let $y=(a/s)\cdot x$ and $y^{\prime }=(a^{\prime }/s^{\prime })\cdot x$. Then

$$ts{s}^{\prime }(y-y^{\prime })=t{s}^{\prime }(sy)-ts(s^{\prime }{y}^{\prime })=t{s}^{\prime }(ax)-ts(a^{\prime }x)=t(a{s}^{\prime }-a^{\prime }s)x=0.$$

Since $ts{s}^{\prime }\in S$ and $M$ is $S$-torsion-free, $y=y^{\prime }$. The module axioms are verified by clearing denominators, analogously to the integral domain case. The uniqueness of the structure follows because any compatible structure $\ast$ must satisfy $sy=ax$ for $y=(a/s)\ast x$, which uniquely determines $y$. $\square$