Every nonzero ideal in a Dedekind domain is generated by two elements

Form the integral ideal $J:=(a)I^{-1}$. By [L1], only finitely many prime ideals $\mathfrak{p}$ satisfy $v_{\mathfrak{p}}(a)>v_{\mathfrak{p}}(I)$; call this finite set $S$. For each $\mathfrak{p}\in S$, choose $b_{\mathfrak{p}}\in I$ with $v_{\mathfrak{p}}(b_{\mathfrak{p}})=v_{\mathfrak{p}}(I)$. By [L3], choose elements $u_{\mathfrak{p}}\in R$ such that $u_{\mathfrak{p}}\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{p})$ and $u_{\mathfrak{p}}\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{q})$ for every distinct $\mathfrak{q}\in S$. Put $b:={\sum }_{\mathfrak{p}\in S}{u}_{\mathfrak{p}}{b}_{\mathfrak{p}}$, and if $S=\mathrm{\varnothing }$ put $b=0$. Then $b\in I$.

For $\mathfrak{p}\in S$, the term $u_{\mathfrak{p}}{b}_{\mathfrak{p}}$ is a unit multiple of $b_{\mathfrak{p}}$ in $R_{\mathfrak{p}}$, while every other summand lies in $\mathfrak{p}{I}_{\mathfrak{p}}$. Hence $v_{\mathfrak{p}}(b)=v_{\mathfrak{p}}(I)$. For $\mathfrak{q}\mathrm{\notin }S$, one has $v_{\mathfrak{q}}(a)=v_{\mathfrak{q}}(I)$ by definition of $S$. Therefore $(a,b)$ and $I$ have the same prime valuations.

By [L1], ideals with the same prime valuations are equal. Hence $(a,b)=I$.