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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:

$$\text{Flat module} \implies \text{Torsion-free}\ ?$$

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?

darij grinberg
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Sam Williams
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2 Answers2

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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 $M$.

darij grinberg
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Jeremy Rickard
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  • 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... – rschwieb Sep 24 '15 at 15:01
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    @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!) – Jeremy Rickard Sep 24 '15 at 16:14
  • You're right: I promptly forgot the first two sentences while focusing on the third and after. That makes my concern unfounded. Thanks – rschwieb Sep 25 '15 at 01:22
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For a commutative domain $R$, torsion-free $\Rightarrow$ flat if and only if $R$ 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 $\mathbf C$.

darij grinberg
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Bernard
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  • Hmm i see, thank you. What if I am not working over a commutative domain, though? – Sam Williams Jun 10 '15 at 13:26
  • Do you mean not commutative, or not a domain? – Bernard Jun 10 '15 at 13:44
  • Not commutative. – Sam Williams Jun 10 '15 at 14:40
  • I'm sorry, I know very few things about noncommutative rings. Not even sure noncommutative rings of fractions are flat. – Bernard Jun 10 '15 at 14:46
  • Yes, it seems to be a problem whenever we leave commutativity behind. thanks in any case. – Sam Williams Jun 10 '15 at 14:54
  • @Bernard it depends by what is meant by ring of fractions in the non-commutative case. If you mean an Ore localization, then yes it is flat. On the other hand, sometimes people will say that a division ring $D$ is a ring of fractions of a ring $R$ if $D$ contains $R$ as a subring, and there is no proper subdivision ring of $D$ containing $R$. In this case, $D$ need not be flat as an $R$-module. Examples exist where $R$ is the group algebra $\mathbb Q[F]$ of a free group $F$. – SFSH Aug 16 '25 at 20:13