TITLE: A torsion-free divisible module over a commutative integral domain is injective
QUESTION [5 upvotes]: Show that a torsion-free divisible module over a commutative integral domain is injective. ($M$ is torsion-free if $rx=0$, $0\not=x\in M \implies r=0$.)
I tried to use Baer's criterion for injectivity to prove that $M$ is injective but I don't see how $R$ being a commutative integral domain gives me additional information. The first two conditions, namely $M$ being torsion-free and divisible tell me about the injectivity of the maps $\varphi_x(r)=rx$ and $\varphi_r(x) = rx$ for $r\in R-\{0\}$ and also the surjectivity of $\varphi_r$ but I don't know how to proceed from this point.

REPLY [4 votes]: There is an extremely direct proof by contradiction. Let $E(M)$ be the injective envelope of $M$, and suppose $x\in E(M)\setminus M$.

There exists an $r\in R$ such that $xr\in M\setminus\{0\}$ (since $E(M)$ is an essential extension.)
There exists a $y\in M$ such that $yr=xr$ (since $M$ is divisible.)
$x=y$ (since $M$ is torsion-free.)
This is a contradiction since $x\notin M$. Therefore $M=E(M)$. QED