4
$\begingroup$

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.

$\endgroup$

1 Answer 1

6
$\begingroup$

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.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .