Is a finitely generated submodule of a direct sum of a field of fractions of a Dedekind domain projective? This is a question out of Donald Passman's "A Course in Ring Theory". 

Let $R$ be a Dedekind domain with field of fractions $F$, and let $V$ be a finitely generated $R$ submodule of $F^n:=F\oplus\dots\oplus F$. Passman asks to show whether $V$ is projective. 

Now I have already proven that over a semihereditary ring any finitely generated submodule of a free module is projective. Obviously $R$ is semihereditary, being a Dedekind domain, but I am struggling to see whether $F$ is a free $R$ module. If it is then my proof is done, but if not I'll need another approach. So if anyone can provide hints on how to prove that $F$ is a free $R$ module it would be much appreciated, if it is not then a push in the right direction would be very useful.
EDIT: As proven here, there is a counterexample to $F$ being free, which means that my proposed strategy won't work. Unfortunately I cannot think of another way to proceed, so any prod in the right direction will be much appreciated.
References
Donald Passman. A Course in Ring Theory. Wadsworth and Brooks/Cole, 1st edition, 1991.

Even though $F$ itself is not free, your approach is still valid: 


*

*First, note that $F$ is the directed union of its finitely-generated, free submodules. Namely, if $x=\frac{a}{b}\in F$ then $x\in U_b := R\frac{1}{b}\subset F$ which is free since $F$ is torsion-free. Moreover, $U_b\subset U_{bb^{\prime}}$ for any two $b,b^{\prime}\in R\setminus\{0\}$, so the family of $\{U_b\}$ is directed.

*Analogously, $F^n$ is the directed union of its finitely-generated, free submodules.

*Now, given a finitely-generated $R$-submodule $M$ of $F^n$, step (2) shows that it is contained in a finitely generated and free submodule of $F^n$, too, and you deduce that it is projective. (In more detail, you would first pick generators $m_1,...,m_n\in M$, then choose finitely-generated and free $U_i\subset F^n$ with $m_i\in U_i$, and finally find a finitely-generated and free $U$ such that $U_i\subset U$. Then $M\subset U$.)
