0
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ 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 $\mathbb{Z}$. $\endgroup$ – Mohan Jan 28 '17 at 22:42
  • $\begingroup$ For some reason, my comments were removed. I want to point out that my question comes from this answer which was previously wrong. More precisely, my question and Mohan's answer were helpful to find a little flaw in this answer. $\endgroup$ – Watson Jan 30 '17 at 12:30
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\Bbb Z_{(p)}$. $\endgroup$ – Watson Jan 29 '17 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.