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Let $G=\mathrm{Gal}(K/k)$. Each field automorphism preserves the ring of integers and the base prime ideal, so it permutes the prime ideals above $\mathfrak{p}$. Suppose there were more than one orbit, and fix one orbit $\mathcal{C}$. By the Chinese remainder theorem for ideals, choose $a\in {\mathcal{O}}_K$ such thatThe field norm $N_{K/k}(a)={\prod }_{\sigma \in G}\sigma (a)$ lies in ${\mathcal{O}}_k$. It lies in every prime ideal of $\mathcal{C}$, since for a fixed member each conjugate of $a$ is zero modulo that member. Thus its contraction lies in $\mathfrak{p}$. On the other hand, for $\mathfrak{Q}$ outside $\mathcal{C}$, every ${\sigma }^{-1}(\mathfrak{Q})$ is also outside it, so every $\sigma (a)$ equals one modulo $\mathfrak{Q}$. The product is consequently one modulo $\mathfrak{Q}$. This contradicts its membership in $\mathfrak{p}{\mathcal{O}}_K\subseteq \mathfrak{Q}$. Therefore the Galois action on the primes above $\mathfrak{p}$ is transitive. The normal extension of number fields is separable, so its conjugates are exactly the elements of $G$ used in the product.

For pairwise comaximal ideals $I_1,\dots ,I_r$ of $R$, reduction gives the Chinese remainder theorem for ideals:Its kernel is ${\bigcap }_i{I}_i$. For two comaximal ideals, write $1=u+v$ with $u\in I$, $v\in J$. If $x\in I\cap J$, then $x=xu+xv\in IJ$, showing $I\cap J=IJ$. Induction gives the product formula: a product of ideals individually comaximal with $J$ is still comaximal with $J$, by multiplying identities modulo $J$.

Put $J_i={\prod }_{j\ne i}{I}_j$. Since $I_i+J_i=R$, choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$. The element ${\sum }_i{a}_i{e}_i$ has any prescribed residues $a_i(\mathrm{mod}{I}_i)$, proving surjectivity. Its residue class modulo the product is unique because we already identified the kernel. In a Dedekind domain, powers of distinct nonzero prime ideals are pairwise comaximal, giving the prime-power form. The theorem requires comaximality; it is not a statement about arbitrary lists of ideals.

Let $R={\mathcal{O}}_k$, $S={\mathcal{O}}_K$ and $n=[K:k]$. For each prime ideal $\mathfrak{P}$ of $S$ above $\mathfrak{p}$ of $R$, set $f_{\mathfrak{P}}=[S/\mathfrak{P}:R/\mathfrak{p}]$. The relative norm of a fractional ideal can be specified on the unique prime ideal factorization byHere is why this definition is the intrinsic relative ideal norm from norms of elements, and why the residue weights occur. Localize at $\mathfrak{p}$. The finite torsion-free $R_{\mathfrak{p}}$-module $S_{\mathfrak{p}}$ is free of rank $n$, because $R_{\mathfrak{p}}$ is a discrete valuation ring. Multiplication by a nonzero integral $a$ has determinant $N_{K/k}(a)$. The Smith normal form givesDecompose the quotient at the prime ideals above $\mathfrak{p}$. In $S_{\mathfrak{P}}$ each quotient ${\mathfrak{P}}^j/{\mathfrak{P}}^{j+1}$ is one-dimensional over $S/\mathfrak{P}$, so has length $f_{\mathfrak{P}}$ over $R_{\mathfrak{p}}$. ThereforeFor an integral ideal $I$, all norms of elements of $I$ have valuations at least $\sum f_{\mathfrak{P}}{v}_{\mathfrak{P}}(I)$. These minima can be attained simultaneously at the finitely many prime ideals above a fixed $\mathfrak{p}$. In fact the Chinese remainder theorem for ideals, applied inside the finitely generated module $I$, chooses an $a\in I$ with nonzero image in each $I/\mathfrak{P}I$. Equivalently $v_{\mathfrak{P}}(a)=v_{\mathfrak{P}}(I)$ at all these primes. The generated ideal of the element norms consequently has exactly the displayed valuations. Clearing a denominator extends this description to fractional ideals.

Multiplication of fractional ideals adds their prime valuations. The displayed formula therefore provesIt also gives $N_{K/k}((a))=(N_{K/k}(a))$.

Finally, reduction of the free rank-$n$ module $S_{\mathfrak{p}}$ modulo $\mathfrak{p}$ has dimension $n$ over $R/\mathfrak{p}$. The factorization $\mathfrak{p}S={\prod }_i{\mathfrak{P}}_i^{e_i}$ and the Chinese remainder theorem for ideals identify this reduction with the product of the quotients by ${\mathfrak{P}}_i^{e_i}$. Their filtrations have $e_i$ successive quotients of dimension $f_i$ over $R/\mathfrak{p}$. Adding dimensions proves the fundamental identity for prime decomposition:

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

For a nonzero fractional ideal of a number field, the displayed generated ideal equals the relative norm of a fractional ideal. Locally, multiplication by $x$ on the integral lattice has determinant $N_{L/K}(x)$. A Smith normal form calculation and the residue filtrations give $v_{\mathfrak{p}}(N(x))={\sum }_{\mathfrak{P}\mid \mathfrak{p}}{f}_{\mathfrak{P}/\mathfrak{p}}{v}_{\mathfrak{P}}(x)$. The Chinese remainder theorem for ideals chooses $x\in I$ simultaneously attaining the minimum valuation of $I$ at each prime above a fixed $\mathfrak{p}$. Thus the generated ideal has valuation $\sum f_{\mathfrak{P}/\mathfrak{p}}{v}_{\mathfrak{P}}(I)$. Adding these valuations proves multiplicativity, including fractional ideals after clearing denominators.