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So when you work over a commutative ring, this result is quite well known. I am wondering if the same holds true for an arbitrary ring; that is, if $R$ is some (possibly noncommutative) ring, does the following implication hold:

$$\textit{Flat module} \implies \textit{Torsion free}\ ?$$

In particular, I am considering a ring which has no zero divisors (i.e. a domain).

If I add in the condition that the module is finitely generated, can I also claim the reverse implication?

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  • $\begingroup$ @sam What is your definition of torsion free for an arbitrary ring? $\endgroup$ – rschwieb Sep 24 '15 at 14:58
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Pretty much the same proof as in the commutative case shows that flat implies torsion free for modules over a noncommutative ring $R$:

Let $a\in R$, not a right zero divisor. Then the map $\mu_a:R\to R$ given by $r\mapsto ra$ is an injective left $R$-module homomorphism. If $M$ is a flat right $R$-module, then $\operatorname{id}_M\otimes\mu_a:M\otimes_RR\to M\otimes_RR$ must also be injective. But this is isomorphic to the map $M\to M$ sending $m\in M$ to $ma$. So $a$ does not annihilate any non-zero element of $M$.

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  • $\begingroup$ Over a Von Neumann regular ring, all modules are flat, but there are certainly modules with elements annihilated by nonzero elements of R, so something doesn't seem quite right... $\endgroup$ – rschwieb Sep 24 '15 at 15:01
  • $\begingroup$ @rschwieb The definition of "torsion-free" I was assuming was that no (non-zero) elements are annihilated by non-zero-divisors. I think this is the most commonly used definition for rings with zero divisors (if you require that no non-zero element of $R$ annihilates any non-zero element of $M$, then no non-zero module for a ring with zero divisors is torsion-free!) $\endgroup$ – Jeremy Rickard Sep 24 '15 at 16:14
  • $\begingroup$ You're right: I promptly forgot the first two sentences while focusing on the third and after. That makes my concern unfounded. Thanks $\endgroup$ – rschwieb Sep 25 '15 at 1:22
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For a commutative domain $R$, torsion-free $\Rightarrow$ flat if and only $R$ is a Prüfer domain, i.e. all localisations of $R$ are valuation domains. If in addition, $R$ is noetherian, this means $R$ is a Dedekind domain.

To have a counter-example, take a noetherian integral domain of Krull dimension $1$ that is not integrally closed. It is not a Dedekind domain, hence it has ideals which are not flat, albeit torsion-free, else all its ideals would finitely generated and flat, hence projective, which implies $R$ is Dedekind.

Concretely, you can take the ring $\mathbf Z[\sqrt{-3}]$. Its integral closure is $\mathbf Z[\omega]$, where $\omega$ is one of the complex cubic roots of unity in $\mathbf C$.

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  • $\begingroup$ Hmm i see, thank you. What if I am not working over a commutative domain, though? $\endgroup$ – Sam Williams Jun 10 '15 at 13:26
  • $\begingroup$ Do you mean not commutative, or not a domain? $\endgroup$ – Bernard Jun 10 '15 at 13:44
  • $\begingroup$ Not commutative. $\endgroup$ – Sam Williams Jun 10 '15 at 14:40
  • $\begingroup$ I'm sorry, I know very few things about noncommutative rings. Not even sure noncommutative rings of fractions are flat. $\endgroup$ – Bernard Jun 10 '15 at 14:46
  • $\begingroup$ Yes, it seems to be a problem whenever we leave commutativity behind. thanks in any case. $\endgroup$ – Sam Williams Jun 10 '15 at 14:54

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