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:
- Flat modules are torsion free.
- Flatness is a local property.
- 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.