 Solution

A Dedekind domain is a Noetherian integrally closed domain in which every nonzero prime ideal is maximal. The equivalent local characterization is that its localizations at nonzero prime ideals are discrete valuation rings. A field is harmless as the dimension-zero case.

Write $R$ for the domain, $F$ for its fraction field and $I$ for the nonzero fractional ideal. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $r{I}^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a fractional ideal; the Noetherian hypothesis also makes it finitely generated.

For a nonzero maximal prime ideal $\mathfrak{p}$, the discrete valuation ring $R_{\mathfrak{p}}$ makes $I_{\mathfrak{p}}$ principal, say $I_{\mathfrak{p}}=a_{\mathfrak{p}}{R}_{\mathfrak{p}}$. We need to justify that inversion commutes with this localization. One inclusion is immediate. Conversely, if $x{I}_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$, finite generation of $I$ supplies a common $s\notin \mathfrak{p}$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, provingIt follows that $(I{I}^{-1}{)}_{\mathfrak{p}}=R_{\mathfrak{p}}$ at every such prime ideal. Globally $I{I}^{-1}\subseteq R$. If this were a proper ideal, it would be contained in a maximal ideal and would remain proper after localizing there, a contradiction. In the field case every nonzero fractional ideal is the whole field. Hence in all casesThis proves that every nonzero fractional ideal of a Dedekind domain is an invertible fractional ideal, rather than presupposing its inverse exists.