Unique factorization of nonzero fractional ideals into prime powers

By [L1], only finitely many integers $v_{\mathfrak{p}}(I)$ are nonzero, so $J:={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{v_{\mathfrak{p}}(I)}$ is a well-defined nonzero fractional ideal. Fix a nonzero prime ideal $\mathfrak{q}$. If $\mathfrak{p}\mathrm{\ne }\mathfrak{q}$, choose $x\in \mathfrak{p}\setminus \mathfrak{q}$; then $x\mathrm{/}1$ is a unit in $R_{\mathfrak{q}}$, so $({\mathfrak{p}}^{v_{\mathfrak{p}}(I)}{)}_{\mathfrak{q}}=R_{\mathfrak{q}}$. Hence only the $\mathfrak{q}$-factor survives after localizing, and $J_{\mathfrak{q}}={\mathfrak{q}}^{v_{\mathfrak{q}}(I)}{R}_{\mathfrak{q}}=I_{\mathfrak{q}}$.

Let $M:=(I+J)\mathrm{/}J$. Localizing the short exact sequence $0\to J\to I+J\to M\to 0$ and using [L2], we get $M_{\mathfrak{m}}\cong (I_{\mathfrak{m}}+J_{\mathfrak{m}})\mathrm{/}{J}_{\mathfrak{m}}$ for every maximal ideal $\mathfrak{m}$. Step 1.1 gives $I_{\mathfrak{m}}=J_{\mathfrak{m}}$, so $M_{\mathfrak{m}}=0$ for all $\mathfrak{m}$. Therefore [L3] gives $M=0$, hence $I\subseteq J$. The same argument with $(I+J)\mathrm{/}I$ gives $J\subseteq I$, so $I=J$. This proves existence of the factorization.

If also $I={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{n_{\mathfrak{p}}}$, localising at a fixed nonzero prime $\mathfrak{q}$ gives ${\mathfrak{q}}^{v_{\mathfrak{q}}(I)}{R}_{\mathfrak{q}}=I_{\mathfrak{q}}={\mathfrak{q}}^{n_{\mathfrak{q}}}{R}_{\mathfrak{q}}$. Uniqueness of powers in the DVR $R_{\mathfrak{q}}$ forces $n_{\mathfrak{q}}=v_{\mathfrak{q}}(I)$, so the factorization is unique.

If $I\subseteq R$, then $I_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$ for every nonzero prime ideal $\mathfrak{p}$, so the exponent in the DVR equality $I_{\mathfrak{p}}={\mathfrak{p}}^{v_{\mathfrak{p}}(I)}{R}_{\mathfrak{p}}$ must be nonnegative. Hence integral ideals have only nonnegative exponents.