For an integral domain $R$ let $\mathrm{Frac}(R)$ denote its field of fractions. Then $R$ is embedded in $\mathrm{Frac}(R)$ and we can consider $\mathrm{Frac}(R)$ as an $R$-module. 


Can we characterize all non-field integral domains $R$ such that every proper non-zero submodule of the $R$-module $\mathrm{Frac}(R)$ is projective ? 


If $R$ satisfies my condition, then since every $R$-submodule of $K$ is projective hence so are in particular the fractional ideals of $R$, and hence every non-zero fractional ideal of $R$ is invertible, thus $R$ is a Dedekind domain. But $R=\mathbb Z$ is a Dedekind domain which does not satisfy my condition . So the family of integral domains, that satisfies the condition I stated , should be Dedekind domain + something more ; I can't figure out what this something more should be . 



On integral domains over which special kind of modules are projective
An integral domain $R$ such that every proper non-zero $R$-submodule of $\text{Frac}(R)$ is projective is a local principal ideal ring. (The converse is David Handelman's comment above).


Indeed, we have 



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> > **Claim.** Let $R$ be a [bounded factorization domain](https://en.wikipedia.org/wiki/Atomic_domain), e.g., a Noetherian domain, with at least two maximal ideals. Then $\text{Frac}(R)$ contains a non-zero proper $R$-submodule which is not projective.
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> > *Proof.* Let $\mathfrak{p}$ and $\mathfrak{q}$ be to distinct maximal ideals of $R$. Let $p \in \mathfrak{p} \setminus \mathfrak{q}$ and let $q \in \mathfrak{q} \setminus \{ 0 \}$. Then $1/q \notin R[1/p]$, since otherwise we would get $p \in \mathfrak{q}$, a contradiction. Thus $R[1/p] \subsetneq \text{Frac}(R)$. But $R[1/p]$ is not projective over $R$ because it doesn't embed into a free $R$-module. Indeed, $1 \in R[1/p]$ is a non-zero element which is divisible by any power of $p$ and no such element exists in a free $R$-module. 
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