Rings over which every torsion-free module is projective

Question

Let $R$ be an integral domain with the property that any finitely generated torsion-free $R$-module is projective. Certainly $R$ could be a Dedekind domain. But is it necessary that $R$ must be Dedekind domain?

Answer 1

Prüfer domains constitute a very important class of rings which are central to a few branches of commutative ring theory, in particular multiplicative ideal theory. They are the non-noetherian analogue of Dedekind domains.

Prüfer domains are most commonly characterized as domains $D$ satisfying either of the following equivalent properties:

(1) every f.g. ideal $I$ of $D$ is invertible (i.e. there exists a fractional ideal $J$ such that $IJ = D$
(2) $D_P$ is locally a valuation domain (i.e. a domain in which the ideals are totally ordered by inclusion)

An easy to find reference for this equivalence is theorems 58-64 of Kaplansky's Commutative Rings.

Note that in a domain, f.g. ideals are invertible iff they are projective, which hints at a module-theoretic characterization. One such characterization is that $D$ is Prüfer iff

(3) f.g. submodules of projective modules are projective.

You can read more about that in T.Y. Lam's Lectures on Modules and Rings (section 2E is all about (semi)-hereditary rings, which generalize Prüfer and Dedekind domains to rings with zero-divisors).

As Lam notes in the end of that section, this easily leads to the characterization of Prüfer domains in terms f.g. torsionfree modules being projective, because every f.g. torsionfree module in a domain can be embedded in a (finite-rank) free module.