Let $R$ be a Dedekind domain and $I$ be an ideal of $R$. Show that $I$ is a projective $R$-module.
My definition of a projective module is that it is a direct summand of a free module, i.e. there exists an $R$ - module $S$ such that $I \oplus S$ is free. Since $R$ is Dedekind I could take $I^{-1}$ to be the inverse of $I$ and guess that $I \oplus I^{-1}$ is free. I am pretty sure that $I \oplus I^{-1} \neq R$ in the general case, so maybe my guess of $I^{-1}$ beeing the direct summand is wrong.
Hint:
$I$ is finitely generated, and for every prime ideal $\mathfrak p\in\operatorname{Spec}R$, the localisation $I_\mathfrak{p}$ is a free $R_\mathfrak{p}$-module of rank 1$.