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.
Best Answer
Following Baer's criterion, let $I\subset R$ be an ideal and $f:I\to M$ be a homomorphism. We want to extend $f$ to a homomorphism $\bar{f}:R\to M$. To define a homomorphism $\bar{f}(r)$, you just have to pick an element $m\in M$ to be $\bar{f}(1)$ and then define $\bar{f}(r)=rm$ for each $r\in R$. For $\bar{f}$ to be an extension of $f$, you need to have $im=f(i)$ for each $i\in I$.
So you want to find $m\in M$ such that $im=f(i)$ for all $i\in I$. The condition that $M$ is divisible suggests a way to find such an $m$: just pick some nonzero $i\in I$, and then divisiblity of $I$ guarantees there exists $m\in M$ such $im=f(i)$. (This step requires $I\neq 0$; I'll let you figure out how to handle the case $I=0$ separately.)
Now of course, you only know that $im=f(i)$ for the particular $i\in I$ you chose. You still need to check that $jm=f(j)$ for all other $j\in I$. You might think it's not reasonable to expect this to be true--maybe we picked the wrong $m$. However, since $M$ is torsion-free, there is only one possible $m$ such that $im=f(i)$ (since $im=f(i)=im'$ implies $m=m'$). So if there is any $m$ that works, it must be the one we chose!
So see if you can prove that $jm=f(j)$ for all $j\in I$, using the fact that $M$ is torsion-free. The details are hidden below.