Is it true that a flat module is torsion-free over an arbitrary ring? Does the reverse implication hold for finitely generated modules?

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?

Solution 1:

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 $M$.

Solution 2:

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 $\mathbf C$.