I have a little problem with the definition of a fractional ideal. The definition I've been given is a set $f\subseteq Q=\text{Frac}(R)$ such that $\exists b\in R\backslash \{0\}$ such that $b.f\subseteq R$.
Apparently it's equivalent to $f$ being a finitely generated $R$-module. I don't understand why. I can't seem to find a reason for $f$ to be a group, or to be stable by multiplication by an element of $R$.
I saw at several places that we require both statements in the definition of a fractional ideal. That is $f$ must be a finitely generated $R$-module, and we require the existence of the $b$. Is it the case ? Is my definition incomplete ? And if not why does it imply that it's a finitely generated $R$-module ?
You may consider the isomorphism $f\rightarrow b\cdot f,x\mapsto bx$. As $b\cdot f$ is an integral ideal, it is finitely generated as $R$-module (recall that a Dedekind domain is Noetherian), so $f$, which is isomorphic to $b\cdot f$, is clearly a finitely generated $R$-module.