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Let $R$ be a (noncommutative) unital ring which is a domain and let $\mathcal{N}$ be a non-zero projective (right) module. Projectivity of course implies that $\mathcal{N}$ is flat, but does the fact that $R$ is also a domain imply that $\mathcal{N}$ is faithfully flat?

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    $\begingroup$ Is that really what you meant to ask? The zero module is projective and torsionfree, but not faithfully flat. $\endgroup$
    – Jeremy Rickard
    Commented Sep 28, 2021 at 10:47
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    $\begingroup$ There are still trivial counterexamples. For example, if $R=S\times\mathbb{Z}$ then $0\times\mathbb{Z}$ is projective and torsion-free, but not faithfully flat. $\endgroup$
    – Jeremy Rickard
    Commented Sep 28, 2021 at 10:56
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    $\begingroup$ @JeremyRickard I don't think your example is torsion-free: the element $(1,0)\in R$ annihilates the element $(0,1)\in \{0\}\times\mathbb Z$. $\endgroup$
    – Wojowu
    Commented Sep 28, 2021 at 15:02
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    $\begingroup$ @Wojowu For rings that are not domains, I think that “torsion free” usually means that no nonzero element is annihilated by a non zero divisor. Otherwise no nonzero module for a ring that is not a domain would be torsion free. Appeal to authority: en.m.wikipedia.org/wiki/Torsion-free_module $\endgroup$
    – Jeremy Rickard
    Commented Sep 28, 2021 at 15:10
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    $\begingroup$ @Wojowu Of course, you might argue that if one can’t even be bothered to be a domain, then one doesn’t deserve to have any torsion free modules. $\endgroup$
    – Jeremy Rickard
    Commented Sep 28, 2021 at 15:19

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