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A prime ideal of a number field splits completely in a finite extension of degree $n$ when it factors into $n$ distinct prime ideals, each of residue-field degree one. In a Galois extension, an unramified prime ideal splits completely exactly when its Frobenius conjugacy class is the identity class. The Chebotarev density theorem then gives these primes Dirichlet density $1/n$.

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.

The relative discriminant of a finite extension of number fields is the relative norm of a fractional ideal applied to its different ideal. Consequently, at $\mathfrak{p}$,where $f_{\mathfrak{P}/\mathfrak{p}}$ is the residue-field degree and $d_{\mathfrak{P}}$ is the different exponent. Over $\mathbb{Q}$, $N({\mathfrak{D}}_{E/\mathbb{Q}})=|D_E|$.

Set $L=K_{\mathfrak{P}}$ and $F=k_{\mathfrak{p}}$, with valuation rings $S,R$. Let $e$ be the ramification index, $f$ the residue-field degree, and $p$ the residue characteristic. A base uniformizer $\pi$ generates ${\mathfrak{P}}^e$ in $S$, up to a unit. The free $R$-module $S$ has rank $ef$. For an integral $a\in S$, its field trace is the matrix trace of multiplication by $a$ on this module. Reduction modulo $\pi$ preserves the matrix trace.

The quotient $S/\pi S$ has a filtrationEach of the $e$ successive quotients is a one-dimensional vector space over ${\kappa }_L=S/\mathfrak{P}$. On each, multiplication by $a$ acts as multiplication by its residue $\bar{a}$, whose trace over ${\kappa }_F=R/\pi R$ is ${\mathrm{Tr}}_{{\kappa }_L/{\kappa }_F}(\bar{a})$. A basis adapted to the filtration makes the multiplication matrix block triangular, so its trace is the sum of these quotient traces. This proves trace reduction in a ramified local extension:Since the hypothesis says $p\mid e$, the right side is zero in ${\kappa }_F$. ThusNo normality hypothesis is needed. The original PDF only requires divisibility of $e$, not divisibility exactly once; the converted TeX has corrupted that condition and several local subscripts.

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.

For a finite extension $E/K$ of number fields, the inverse different is the fractional idealThe different ideal is its inverse. The trace pairing is nondegenerate because extensions of number fields are separable, so this definition gives a full fractional ideal. The inclusion ${\mathcal{O}}_E\subseteq {\mathfrak{D}}_{E/K}^{-1}$ shows that ${\mathfrak{D}}_{E/K}$ is integral.

Write its prime ideal factorization as ${\mathfrak{D}}_{E/K}={\prod }_{\mathfrak{P}}{\mathfrak{P}}^{d_{\mathfrak{P}}}$. For $\mathfrak{P}$ above $\mathfrak{p}$, with ramification index $e_{\mathfrak{P}}$ and residue characteristic $\ell$, the different exponent and tame ramification theorem saysThe equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak{P}}=0$ precisely at unramified prime ideals; in the wild case $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}$. This theorem does not require a Galois extension. The relative discriminant is the norm of the different ideal:For $K=\mathbb{Q}$, this gives $|D_E|=N({\mathfrak{D}}_{E/\mathbb{Q}})$.

Now take ${\alpha }^3=m$ and $K=\mathbb{Q}(\alpha )$. For any prime number $q\mid m$, the square-free integer hypothesis makes $X^3-m$ an Eisenstein polynomial at $q$. It is therefore irreducible, and $K$ is a pure cubic number field of degree $3$. Since $\alpha$ is an algebraic integer, $A=\mathbb{Z}[\alpha ]$ is an order in a number field. The discriminant of elements of a number field for its basis $1,\alpha ,{\alpha }^2$ isFor example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the polynomial discriminant is negative. If $I=[{\mathcal{O}}_K:A]$, the discriminant-index formula for an integral lattice givesOnly prime numbers dividing $3m$ can therefore divide $I$.

For $q\mid m$, the Eisenstein polynomial gives a totally ramified extension of ${\mathbb{Q}}_q$ of degree $3$. Thus $K$ has a unique prime ideal $\mathfrak{P}$ above $q$, with ramification index $3$ and residue-field degree $1$. Since $q\ne 3$, this is tame ramification, and the different exponent and tame ramification theorem gives $d_{\mathfrak{P}}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the discriminant-index formula for an integral lattice yields $v_q(I)=0$.

At $3$, use the shifted Eisenstein polynomial of $\beta =\alpha -m$:Because $3\nmid m$, the constant term is divisible by $3$. Moreover,when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an Eisenstein polynomial at $3$. The resulting completion is a totally ramified extension of degree $3$, with residue-field degree $1$, and its ramification index is divisible by the residue characteristic. Thus its different exponent is at least $3$.

It follows that $v_3(D_K)\ge 3$. But $v_3(-27m^2)=3$, soforces $v_3(I)=0$ and $v_3(D_K)=3$. No prime number divides $I$. We have proved the integral basis of a nonexceptional pure cubic field:Using local Eisenstein polynomials here only establishes the local ramification indices; it does not assume that $\mathbb{Z}[\alpha ]$ is already the full ring of integers of a number field.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.

For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also $p-1$ and, with the stated real-index convention,As a consistency check, ${\pi }_L$ satisfies the Eisenstein polynomial $X^p-t^{p-1}X-t$. Its derivative in characteristic $p$ is $-t^{p-1}$, giving different exponent $p(p-1)$, equal to ${\sum }_{i\ge 0}(|G_i|-1)=p(p-1)$ from the different exponent from ramification groups.

The completion of a number field at a prime ideal does not change either invariant. More precisely, ${\hat{B}}_q$ is the valuation ring of $L_q$, its maximal ideal is $q{\hat{B}}_q$, andThus the completed extension has the same ramification index and the same residue-field extension, hence the same residue-field degree $f_{q/p}$. This is completion preserves ramification and residue data.

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

A finite extension $L/K$ of non-Archimedean local fields is an unramified extension when its ramification index is one and its residue-field degree equals $[L:K]$. Equivalently, its maximal ideal is generated by a uniformizer of $K$ and its residue-field extension is separable.

A finite extension of number fields is an everywhere unramified extension of number fields when every nonzero prime ideal of $K$ is unramified in $L$. Under the convention that includes infinite places, one also requires every real embedding of $K$ to remain real.

The cyclotomic extension of a p-adic fieldhas degree $p-1$, is Galois with group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, and is totally ramified. Indeed, ${\zeta }_p-1$ is a root of the Eisenstein polynomial

Put $\alpha =\sqrt[p-1]{-p}$. Its polynomial $X^{p-1}+p$ is Eisenstein, so $L_2={\mathbb{Q}}_p(\alpha )$ is totally ramified of degree $p-1$. Every $(p-1)$st root of unity lies in ${\mathbb{Q}}_p$ by the Teichmuller lifts, so every root $\omega \alpha$ of this polynomial lies in $L_2$. Hence $L_2/{\mathbb{Q}}_p$ is also Galois.

For either $L_i/{\mathbb{Q}}_p$, the norm has every possible valuation because the residue-field degree is one. The norm units in a tamely totally ramified extension lie in the principal units $1+p{\mathbb{Z}}_p$: reduction of a unit norm is the $(p-1)$st power of its residue, hence is $1$. Part (b) says that this unit norm subgroup has index $p-1$, exactly the index of $1+p{\mathbb{Z}}_p$ in ${\mathbb{Z}}_p^{\times }$. ConsequentlyThe uniqueness clause in the existence theorem of local class field theory now gives

For a finite extension $L/K$ of local fields, let $e=e(L/K)$ be its ramification index and $f=f(L/K)=[k_L:k_K]$ its residue-field degree. An unramified extension has $e=1$. A totally ramified extension has $e=[L:K]$, equivalently $f=1$. A tamely ramified extension has separable residue extension and ramification index coprime to the residue characteristic.

Let $\bar{\theta }$ generate the finite extension $k_L/k_K$, and let $\bar{g}$ be its minimal polynomial. Lift $\bar{g}$ to a monic $g\in {\mathcal{O}}_K[X]$ and choose any lift $t\in {\mathcal{O}}_L$ of $\bar{\theta }$. Since finite fields are perfect fields, ${\bar{g}}^{\prime }(\bar{\theta })\ne 0$. The simple-root form of Hensel lemma, applied inside $L$, gives $\theta \in {\mathcal{O}}_L$ withSet $K_0=K(\theta )$. Its residue field contains $k_K(\bar{\theta })=k_L$, soThe equation $g(\theta )=0$ gives the reverse inequality. Thus $[K_0:K]=f$, its residue-field degree is $f$, and $e(K_0/K)=1$; hence $K_0/K$ is unramified. Since $k_{K_0}=k_L$, the extension $L/K_0$ has residue-field degree one and is totally ramified. This constructs the maximal unramified subextension of a local field extension.

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

A local homomorphism of local rings $(R,\mathfrak{m})\to (S,\mathfrak{n})$ induces the residue-field extension $R/\mathfrak{m}\to S/\mathfrak{n}$. For an extension of non-Archimedean local fields, its degree is the residue-field degree.

A finite extension of local fields is totally ramified when its residue-field degree is one, equivalently when its ramification index equals its field degree.