Suppose $R$ is a PID (Principal Ideal Domain), and $M$ a torsion-free $R$-module of rank $1$. Is it true that $M$ is either cyclic or injective?
I suppose it is true because I have read (without proof) that this holds when $R$ is the ring of $p$-adic integers (and it is clear that the same holds for $R=\Bbb{Z}$).
So I'm looking for a down to earth proof of this fact (or a reference).
This holds if $R$ is a DVR. Let $K$ be the field of fractions of $R$. Since $M$ is torsion-free, the map $M\to K\otimes_RM$ is injective, and since $M$ has rank one we know $K\otimes_RM\cong K$. Thus we can regard $M$ as an $R$-submodule of $K$.
Take a generator $t$ for the maximal ideal of $R$ (a uniformising element), and consider the set of $r$ such that $t^{-r}\in M$. If this set has a maximal element $n$, then $M=t^{-n}R$ is cyclic (and free); otherwise $M=K$ is injective.
Edit (first attempt was wrong):
By the structure theorem for PIDs, every finitely-generated torsion-free module is free, so if it has rank one, then it is cyclic. The question is thus about non-finitely generated submodules of the field of fractions. If the PID has at least two primes $p,q$, then the submodule generated by all $1/p^r$ is neither cyclic nor divisible.