# Does torsion-free implies flat over a "locally principal" domain?

Let $$R$$ be a commutative ring. An $$R$$-module $$M$$ over $$R$$ is said to be torsion-free if for every $$r\in R$$ which is not zero divisor and for every $$0\neq m\in M$$, we have $$r\cdot m\neq 0$$.

I know that:

1. Flat modules are torsion-free.
2. Flatness is a local property.
3. Over a PID, torsion-free modules are flat.

For me, a Dedekind domain is a domain which is Noetherian and locally a PID for every prime ideal.

As a consequence of the above, torsion-free modules over a Dedekind domain are flat.

But do we need the ring to be a Noetherian domain to conclude that? Does it not suffice to assume that the (commutative) ring is locally a PID at every prime ideal to conclude that torsion-free modules are flat? I certainly can't see when Noetherianess is needed.

In fact, we do not even need the locally PID condition: it's enough to assume that $R$ is locally a valuation ring, that is, $R$ is a Prüfer domain. (In fact, an integral domain is Prüfer iff every torsion-free module over it is flat.)