Prime-ideal valuations of a fractional ideal have finite support and add under products

Choose $0\mathrm{\ne }x\in I$ and $0\mathrm{\ne }d\in R$ with $dI\subseteq R$, and put $a:=dx\in R$. If a nonzero prime ideal $\mathfrak{p}$ contains neither $a$ nor $d$, then both $a\mathrm{/}1$ and $d\mathrm{/}1$ are units in $R_{\mathfrak{p}}$, so $x=(a\mathrm{/}1)(d\mathrm{/}1{)}^{-1}$ is a unit of $R_{\mathfrak{p}}$. Hence $R_{\mathfrak{p}}=x{R}_{\mathfrak{p}}\subseteq I_{\mathfrak{p}}\subseteq d^{-1}{R}_{\mathfrak{p}}=R_{\mathfrak{p}}$, so $I_{\mathfrak{p}}=R_{\mathfrak{p}}$ and therefore $v_{\mathfrak{p}}(I)=0$.

Put $b:=ad$. If $b$ is a unit, then no nonzero prime ideal contains $b$. Otherwise every prime ideal containing $(b)$ is nonzero, hence maximal because [F2] gives $\dim R=1$. By [L2], every prime ideal of $R\mathrm{/}(b)$ is therefore maximal. The quotient $R\mathrm{/}(b)$ is Noetherian by [L3], so [L4] makes it Artinian, and then [L5] gives only finitely many maximal ideals. Translating back through [L2], only finitely many nonzero prime ideals of $R$ contain $b$.

Combining steps 1.1 and 2.1, only finitely many nonzero prime ideals can satisfy $v_{\mathfrak{p}}(I)\mathrm{\ne }0$.

Fix a nonzero prime ideal $\mathfrak{p}$. By [F1], write $I_{\mathfrak{p}}={\mathfrak{p}}^m{R}_{\mathfrak{p}}$ and $J_{\mathfrak{p}}={\mathfrak{p}}^n{R}_{\mathfrak{p}}$. Localizing the defining finite sums for a product gives $(IJ{)}_{\mathfrak{p}}=I_{\mathfrak{p}}{J}_{\mathfrak{p}}={\mathfrak{p}}^{m+n}{R}_{\mathfrak{p}}$, so $v_{\mathfrak{p}}(IJ)=m+n=v_{\mathfrak{p}}(I)+v_{\mathfrak{p}}(J)$.