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Let $R$ be a PID which is not a field. I would like to know if the following is true:

If $M$ is a torsion free $R$-module, then $M$ is faithfully flat iff $M$ is not divisible.

Since PID $\implies$ Dedekind domain $\implies$ Prüfer domain, any torsion free $\Bbb Z$-module is flat. Therefore, a torsion free $\Bbb Z$-module $M$ is faithfully flat iff $M \otimes_{\Bbb Z} N \neq 0$ for any $N \neq 0$. Moreover, I know that if $R$ is commutative ring, $M$ is a torsion $R$-module, $N$ is a divisible $R$-module, then $M \otimes_R N = 0$.

Therefore, if $M$ is divisible, then let $I$ be a non-zero proper ideal of $R$ (it exists since $R$ is not a field), so that $R/I$ is a non-zero (since $I \neq R$) torsion $R$-module, annihilated by any $x \in I \setminus \{0\} \neq \varnothing$. Thus, $R/I \otimes_R M = 0$ even if $R/I \neq 0$, which means that $M$ is not faithfully flat.

But I'm stuck for proving $\Longleftarrow$. If $M$ is flat but not faithfully flat, then there is $N \neq 0$ such that $M \otimes_R N = 0$. If $m \in M, r \in R \setminus \{0\}$, how to find $m' \in M$ such that $m=r \cdot m'$ ? I know that since $R$ PID, $M$ is divisible iff $M$ is injective.

Thank you!

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    Have you considered $\mathbb{Z}_p$ (localization at a prime $p$)? It is torsion free (and hence flat), not divisible, but also not faithfully flat over $\mathbb{Z}$. – 2017-01-28
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    For some reason, my comments were removed. I want to point out that my question comes from [this](https://math.stackexchange.com/revisions/1181418/1) answer which was previously wrong. More precisely, my question and Mohan's answer were helpful to find a little flaw in this answer. – 2017-01-30

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If we work over the ring of integers, it is clear that for a torsion free module (which is then flat), if faithfully flat, then it is not divisible. But, the converse is not true. There exists torsion free, non-divisible modules which are not faithfully flat. As an example, consider $\mathbb{Z}_p$, the localization at the multiplicatively closed set $\{1,p, p^2,\ldots\}$, where $p$ is a prime. It is torsion free,, not divisible and not faithfully flat.

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    Thank you! $\Bbb Z_p$ is not f.f. since the maximal ideal $p\Bbb Z$ extends to the whole $\Bbb Z_{p}$. We could also take $\Bbb Z_{(p)}$ as counter-example, right? It is torsion free, not divisble and for any prime $q \neq p$, the maximal ideal $q\Bbb Z$ extends to the whole $\Bbb Z_{(p)}$. – 2017-01-29