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For $a\ge 1$ and $p\nmid m$, the identity ${\Phi }_{p^{a}m}(X)={\Phi }_m(X^{p^a})/{\Phi }_m(X^{p^{a-1}})$ gives the displayed reduction, since taking a $p$th power equals substitution $X\mapsto X^p$ in characteristic $p$. The reduction of ${\Phi }_m$ is squarefree because it divides the separable polynomial $X^m-1$. The Kummer-Dedekind theorem therefore gives ramification index of a prime ideal $\phi (p^a)$ in the cyclotomic field. In particular, a conductor factor $2$ occurring only to the first power has index one and creates no ramification by itself.

If a prime ideal $\mathfrak{p}$ of $K$ lies over the prime number $p$ with ramification index of a prime ideal $e_{\mathfrak{p}}$, its normalized absolute value isIt extends the usual p-adic absolute value on $\mathbb{Q}$. Its local degree is $[K_{\mathfrak{p}}:{\mathbb{Q}}_p]=e_{\mathfrak{p}}{f}_{\mathfrak{p}}$, where $f_{\mathfrak{p}}$ is the residue-field degree.

Let $L=\mathbb{Q}(\alpha )$, where $\alpha$ is an algebraic integer with monic minimal polynomial $f\in \mathbb{Z}[X]$, and put $I=[{\mathcal{O}}_L:\mathbb{Z}[\alpha ]]$. The Kummer-Dedekind theorem applies to a prime number $p$ with $p\nmid I$. Write its reduction as a product of polynomial factors over ${\mathbb{F}}_p$:where the $g_j$ are distinct monic irreducible polynomials, choose monic integral lifts $G_j$. Then the distinct prime ideals above $p$ are ${\mathfrak{p}}_j=(p,G_j(\alpha ))$, andThus the multiplicity of a factor is its ramification index of a prime ideal, and its degree is the residue degree. The hypothesis on the order in a number field is essential: one cannot infer prime ideal factorization from an arbitrary defining polynomial at a prime dividing its index. There is also a relative version over a base number field $K$: if ${\mathcal{O}}_L$ and ${\mathcal{O}}_K[\alpha ]$ agree after localization at $\mathfrak{p}$, factor the minimal polynomial over ${\mathcal{O}}_K/\mathfrak{p}$ and use $(\mathfrak{p},G_j(\alpha ))$ with the same conclusions.

Let $L/K$ be an extension of number fields, let $\alpha \in {\mathcal{O}}_L$ be a primitive element of a field extension, and let $f\in {\mathcal{O}}_K[X]$ be its monic minimal polynomial of an algebraic element. Put $R={\mathcal{O}}_K[\alpha ]$ and $S={\mathcal{O}}_L$. The Kummer-Dedekind theorem applies at a nonzero prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_K$ when $R_{\mathfrak{p}}=S_{\mathfrak{p}}$: equivalently, $\mathfrak{p}$ is outside the support of the finite-index quotient $S/R$. Write the factorization over the residue field $k={\mathcal{O}}_K/\mathfrak{p}$ aswith distinct monic irreducible polynomials ${\overline{g}}_j$, and choose monic lifts $g_j\in {\mathcal{O}}_K[X]$. ThenHere $f({\mathfrak{P}}_j/\mathfrak{p})$ denotes the residue-field degree, and the ramification index of a prime ideal is $m_j$.

To prove this, the local index condition givesThe final isomorphism is the Chinese remainder theorem. Each factor is a local ring with unique maximal ideal generated by ${\overline{g}}_j$, so its inverse image in $S$ is exactly ${\mathfrak{P}}_j$, and its residue field is $k[X]/({\overline{g}}_j)$. This proves the list of primes and the residue-field degrees. For the exponents, localize at ${\mathfrak{P}}_j$. The ring of integers of a number field is a Dedekind domain, so $S_{{\mathfrak{P}}_j}$ is a discrete valuation ring. If $\mathfrak{p}{S}_{{\mathfrak{P}}_j}={\mathfrak{P}}_j^{e_j}{S}_{{\mathfrak{P}}_j}$, the quotient has $e_j$ successive layers isomorphic to $S/{\mathfrak{P}}_j$, and hence $k$-dimension $e_j\mathrm{deg}{\overline{g}}_j$. The corresponding polynomial factor has $k$-dimension $m_j\mathrm{deg}{\overline{g}}_j$. Thus $e_j=m_j$, completing the proof rather than merely identifying the radicals.

For the calculation, set $K=\mathbb{Q}(\sqrt{5})$ and $t=(\sqrt{5}-1)/2$. The ring of integers of a quadratic field gives ${\mathcal{O}}_K=\mathbb{Z}[t]$, with $t^2+t-1=0$. The algebraic norm of $4+\sqrt{5}$ is $11$, so its principal ideal has ideal norm $11$. Its quotient ring has prime order and is therefore the finite field ${\mathbb{F}}_{11}$; consequently $(4+\sqrt{5})$ is a prime ideal. In that residue field, $\sqrt{5}=-4=7$ and $t=3$.

Let $L=\mathbb{Q}({\zeta }_5)$. Since $t={\zeta }_5+{\zeta }_5^{-1}$ lies in $\mathbb{Z}[{\zeta }_5]$, the permitted identity ${\mathcal{O}}_L=\mathbb{Z}[{\zeta }_5]$ gives ${\mathcal{O}}_L={\mathcal{O}}_K[{\zeta }_5]$. Thus the local index condition holds at every prime. The relative minimal polynomial of an algebraic element is $X^2-tX+1$, whose reduction isThe roots are distinct. Applying the Kummer-Dedekind theorem givesFor the last equality, evaluation at $5$ or $9$ in ${\mathbb{F}}_{11}$ gives ${\zeta }_5+{\zeta }_5^{-1}=3$ and therefore $4+\sqrt{5}=5+2t=0$. Both descriptions are the same kernels of these evaluation maps. Each prime has ideal norm $11$, residue-field degree one and ramification index of a prime ideal one.

Write $e=e(\mathfrak{q}/\mathfrak{p})$, $f=f(\mathfrak{q}/\mathfrak{p})$ and $g$ for the number of primes above $\mathfrak{p}$. The Galois group acts transitively on those primes, with stabilizer $D_{\mathfrak{q}}$. The fundamental identity for prime decomposition and the orbit-stabilizer theorem therefore giveIt remains to establish that the residue action has image of order $f$, rather than assume this in the desired equality.

Use the decomposition group of a prime and local Galois group identification $D_{\mathfrak{q}}=\mathrm{Gal}(F_{\mathfrak{q}}/K_{\mathfrak{p}})$. A generator of the finite field extension $k_{\mathfrak{q}}/k_{\mathfrak{p}}$ has a separable monic minimal polynomial of an algebraic element. Lift its coefficients to ${\mathcal{O}}_{K_{\mathfrak{p}}}$. The Hensel lemma lifts each of its distinct residue roots to $F_{\mathfrak{q}}$. One such lift has field degree at most $f$, while its residue generates the degree-$f$ finite field extension, so its field degree is exactly $f$ and its ramification index of a prime ideal is one. It therefore generates an unramified extension of degree $f$, and sending it to each of the other lifts gives its $f$ embeddings. Because $F_{\mathfrak{q}}/K_{\mathfrak{p}}$ is a Galois extension, these embeddings extend to its automorphisms. Their reductions realize every automorphism of $k_{\mathfrak{q}}/k_{\mathfrak{p}}$. Thus the residue action is surjective, and

Now let $G$ be cyclic of order ${\ell }^2$, and let $H=\mathrm{Gal}(F/L)$ be its unique subgroup of order $\ell$. In a tower of Galois extensions, inertia maps onto inertia in the intermediate extension. This also follows from the just-proved order formula: its image has order $e(F/K)/e(F/L)=e(L/K)$, using multiplicativity of the ramification index of a prime ideal in a tower. If $\mathfrak{p}$ ramifies in $L/K$, this image in $G/H$ is nontrivial. Every proper subgroup of $G$ is contained in $H$, so inertia upstairs must be $G$. HenceThe fundamental identity for prime decomposition then also forces a single prime above $\mathfrak{p}$ and residue-field degree one.

Put $\alpha =\sqrt[n]{m}$. It is an algebraic integer because it satisfies the monic polynomial $X^n-m$. Choose a prime number $q$ dividing the square-free integer $m$, and choose a prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_L$ above $q$. Fromand prime ideal factorization, the $\mathfrak{p}$-adic exponents satisfybecause $q$ occurs to exponent one in $m$. Hence $n$ divides the ramification index of a prime ideal $e_{\mathfrak{p}}$. If $d=[L:\mathbb{Q}]$, then $e_{\mathfrak{p}}\le d$, while the equation ${\alpha }^n=m$ gives $d\le n$. Thereforeso

The prime ideal factorization of $pB$ iswhich defines the ramification index of a prime ideal $e_{q/p}=v_q(pB)$. The residue-field degree isLocalizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the residue field $R/p$. The Chinese remainder theorem and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the fundamental identity for prime decomposition gives

The polynomial of $\alpha$ satisfies the Eisenstein criterion at $p$. Let $\mathfrak{p}$ be any prime ideal of ${\mathcal{O}}_K$ above $p$, normalize its ideal valuation by $v_{\mathfrak{p}}(\mathfrak{p})=1$, and write $e=v_{\mathfrak{p}}(p)$ for its ramification index of a prime ideal. Put $t=v_{\mathfrak{p}}(\alpha )$. Inthe constant term has valuation $e$, while every other $a_j{\alpha }^j$ has valuation at least $e+jt$. The non-Archimedean valuation property then forcesIndeed, either strict inequality would leave a unique term of least valuation in the equation. Since $t$ is a positive integer and $e\le n$, it follows that $t=1$ and $e=n$.

The fundamental inequality ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}}{f}_{\mathfrak{p}}\le n$ now shows that $\mathfrak{p}$ is the unique prime above $p$ and has residue degree one. At this prime,while $(p,\alpha )$ has no factor away from $p$. Unique prime ideal factorization therefore givesFinally $v_P(\alpha )=1$, soThis is total ramification from an Eisenstein polynomial.

For an integral generator $\alpha$ of a number field, let $A=\mathbb{Z}[\alpha ]$ and suppose $p$ is coprime to $[{\mathcal{O}}_K:A]$. Localizing at $p$ makes $A$ and the ring of integers equal, so reduction gives the displayed quotient ring. If $\overline{m}=\prod g_i^{e_i}$ with distinct monic irreducible polynomials, the Chinese remainder theorem gives local factors ${\mathbb{F}}_p[X]/(g_i^{e_i})$. Their residue fields have degrees $\mathrm{deg}{g}_i$ and their maximal ideals have nilpotence indices of an ideal $e_i$. Localizing the Dedekind domain ${\mathcal{O}}_K$ at the corresponding prime ideals gives discrete valuation rings, where this same nilpotence index of an ideal is the ramification index of a prime ideal. Thus $(p)=\prod (p,g_i(\alpha ){)}^{e_i}$ and each residue degree is $\mathrm{deg}{g}_i$.