Local field

A local field is a nondiscrete locally compact topological field. A non-Archimedean local field is a finite extension of ${\mathbb{Q}}_p$ or of ${\mathbb{F}}_p((t))$.

A p-adic field is a finite extension of the P-adic number field ${\mathbb{Q}}_p$, equipped with its extended p-adic absolute value. Conversely every characteristic-zero, nontrivially valued, non-Archimedean locally compact field is such an extension. A finite quotient $\mathcal{O}/p\mathcal{O}$ supplies finitely many generators of $\mathcal{O}$ over ${\mathbb{Z}}_p$ by repeated reduction and completeness.

An integral lattice in a finite-dimensional ${\mathbb{Q}}_p$-vector space is a free ${\mathbb{Z}}_p$-submodule of full rank. Any two such lattices are commensurable: their intersection has finite index in each, because a sufficiently large power of $p$ multiplies either into the other. A finite group acting on the space admits invariant lattices by taking a sum of translates. This allows Herbrand quotient comparisons via finite quotient modules.

The degree of the residue field extension induced by a finite extension of complete discretely valued fields. Along with the ramification index, it satisfies $[L:K]=ef$. An unramified extension additionally has separable residue extension; a totally ramified extension has residue degree one.

The absolute residue degree is the degree of the residue field over its prime field. For a tower above $p$, degrees multiply: $[k_{\mathfrak{q}}:{\mathbb{F}}_p]=[k_{\mathfrak{q}}:k_{\mathfrak{p}}][k_{\mathfrak{p}}:{\mathbb{F}}_p]$. In a Galois extension over $\mathbb{Q}$, the decomposition group has a cyclic quotient group of this order by its inertia group. This can exclude residue degrees even when the whole extension has that degree: $S_3$ has no cyclic quotient group of order six.

A complete field with a nontrivial Non-Archimedean absolute value is locally compact exactly when its valuation is discrete and its residue field is finite. In that case its valuation ring is the inverse limit of the finite rings $\mathcal{O}/{\mathfrak{m}}^n$ and is compact. The nontrivial valuation in the local compactness criterion is essential.

Let a nontrivially valued local field have compact valuation ring $\mathcal{O}$, and choose any $t$ with $0<|t|<1$. The quotient $\mathcal{O}/t\mathcal{O}$ is finite. A fixed set of representatives containing zero gives every field element a unique convergent digit expansion, with finitely many nonzero negative-index terms. Recursive reduction modulo $t\mathcal{O}$ gives existence; comparing the first differing digit modulo this ideal gives uniqueness. The element $t$ need not be a uniformizer.

In a locally compact nontrivially valued field, the valuation ring is compact. A Cauchy sequence has a tail inside a translate of this compact set, hence a convergent subsequence. The Cauchy property forces the full sequence to converge.

For a locally compact field with a nontrivial Non-Archimedean absolute value, scaling a compact neighborhood makes its valuation ring compact. Its open maximal ideal gives a finite discrete residue field, so that ideal is also closed and compact. A nonzero element of maximal absolute value in the ideal generates it. Successive division by this uniformizer makes the value group discrete.

An infinite field with the trivial absolute value on a field is discrete, hence locally compact, but its residue field is the original infinite field. Thus local compactness forces a finite residue field and a discrete nontrivial value group only when the valuation is assumed nontrivial.

If a non-Archimedean local field $K$ has uniformizer $\pi$, valuation ring ${\mathcal{O}}_K$, maximal ideal ${\mathfrak{m}}_K$ and residue field ${\mathbb{F}}_q$, thenThe first factor records the discrete valuation, the Teichmuller representatives split reduction ${\mathcal{O}}_K^{\times }\to {\mathbb{F}}_q^{\times }$, and $1+{\mathfrak{m}}_K$ is the group of principal units.

The groups $U_m$ form a compact-open neighbourhood basis at one inside the units of a non-Archimedean local field. A smooth character trivial on $U_{i+1}$ but nontrivial on $U_i$ cannot have a nonzero fixed vector under $U_m$ for $m\le i$. This elementary conductor comparison controls admissibility and smooth duals of direct sums.

If $K/{\mathbb{Q}}_p$ is finite of degree $d$, then as a topological abelian groupwhere ${\mu }_{p^a}(K)$ is the finite group of all roots of unity in $K$ whose orders are prime powers dividing a power of $p$.

For a complete discrete valuation ring $R$, a simple root of a polynomial modulo the maximal ideal lifts uniquely to a root in $R$.

A monic polynomial over a complete discrete valuation ring factors with prescribed coprime monic residue factors. Given monic lifts $g_r,h_r$ correct modulo ${\pi }^r$, the degree-bounded map $(u,v)\mapsto u{\bar{h}}_r+v{\bar{g}}_r$ is a bijection over the residue field, by coprimality and equality of dimensions. Solve it for the error $(f-g_r{h}_r)/{\pi }^r$, and replace the lifts by $g_r+{\pi }^{r}u,h_r+{\pi }^{r}v$. Their coefficients converge by completeness and the limiting product is $f$. The completeness assumption cannot be replaced just by localization.

For a complete discrete valuation ring $R$, let $f(a_1)\in (\pi )$ and let $f^{\prime }(a_1)$ be a unit. If $f(a_n)\in ({\pi }^n)$, choose $c$ modulo $\pi$ solving $f(a_n)/{\pi }^n+c{f}^{\prime }(a_n)\equiv 0$ and put $a_{n+1}=a_n+c{\pi }^n$. Taylor expansion gives $f(a_{n+1})\in ({\pi }^{n+1})$. The sequence is Cauchy and its limit is a root. If two roots lift the same residue, factoring $f(x)-f(y)$ leaves $x-y$ times a unit, proving uniqueness.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension of an absolute value to every finite extension. Factor a monic $g\in {\mathcal{O}}_K[X]$ into monic irreducible factors over $K$. All factors have integral coefficients because their roots have absolute value at most one. By pure-power reduction of a monic irreducible polynomial, each factor reduces to a power of a single irreducible polynomial. If $\overline{g}={\overline{g}}_1{\overline{g}}_2$ is a coprime monic factorization, assign each whole irreducible factor of $g$, with its multiplicity, to the unique side containing its residue factor. The resulting monic $g_1,g_2\in {\mathcal{O}}_K[X]$ satisfy $g=g_1{g}_2$ and reduce to the prescribed factors. Completeness of $K$ is not required for this proof.

For a non-Archimedean local field $K$, an element of $K^{\times }$ is an $m$th power for infinitely many positive integers $m$ exactly when it belongs to ${\mathcal{O}}_K^{\times }$. Valuations prove necessity; for sufficiency, choose infinitely many $m$ coprime to the residue characteristic and to $|k^{\times }|$, for which exponentiation by $m$ is an automorphism of the unit group.

The polynomial $X^4+9X^2-2$ has three irreducible factors over ${\mathbb{Q}}_2$, of degrees one, one, and two. Its quadratic in $X^2$ has one unit root congruent to one modulo eight, hence a square, and one root of valuation one, whose associated quadratic factor is Eisenstein.

Let $K$ be complete for a discrete valuation $v$. If $f\in {\mathcal{O}}_K[X]$ andthen Newton iteration over a valued field converges to a root $\alpha \in {\mathcal{O}}_K$ with $v(\alpha -a)>v(f^{\prime }(a))$.

Starting from $a_0$, Newton iteration sets $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Under the strong form of Hensel lemma inequality, the correction valuations grow at least geometrically, so $(a_n)$ is a Cauchy sequence converging to a root.

Let $K$ be complete, let $\alpha$ be separable over $K$, and let $\beta$ be algebraic over $K$. If $\beta$ is closer to $\alpha$ than any other $K$-conjugate of $\alpha$, then

Let $L/K$ be a finite Galois extension of a field complete for a Non-Archimedean absolute value. Suppose $\alpha$ is uniquely closest to $\beta$ among the roots of a polynomial over $K$ lying in $L$. Every element of $\mathrm{Gal}(L/K(\beta ))$ preserves absolute values on a field and permutes these roots. It sends $\alpha$ to an equally close root, hence fixes $\alpha$. The Fundamental theorem of Galois theory gives $K(\alpha )\subseteq K(\beta )$. The converse inclusion need not hold: for roots $0,\sqrt{p},-\sqrt{p}$, the element $p\sqrt{p}$ is uniquely closest to zero but does not belong to ${\mathbb{Q}}_p$.

A p-adic field has only finitely many extensions of a fixed degree up to isomorphism. There are finitely many possible residue degrees and a unique unramified extension of each such degree. Over each maximal unramified subfield, a totally ramified extension is generated by a root of an Eisenstein polynomial. The coefficient space of these polynomials of fixed degree is compact. Hensel lemma and Krasner's lemma show that sufficiently close polynomials generate isomorphic extensions, giving a finite cover by neighborhoods of a constant extension type.

The roots of a separable polynomial over a complete non-Archimedean field vary continuously with its coefficients after the roots are suitably matched. In particular, a sufficiently close coefficientwise approximation has one root in each sufficiently small disjoint ball around an original root.

Every finite extension of ${\mathbb{Q}}_p$ is isomorphic to the completion $F_{\mathfrak{p}}$ of some number field $F$ at a prime $\mathfrak{p}$ above $p$. Approximate a primitive element's minimal polynomial by a polynomial over $\mathbb{Q}$ and use Krasner's lemma to preserve the generated local field.

The completion ${\mathbb{C}}_p$ of an algebraic closure of ${\mathbb{Q}}_p$ is an algebraically closed field. Approximate a polynomial over ${\mathbb{C}}_p$ by one over ${\overline{\mathbb{Q}}}_p$, use continuity of roots over a non-Archimedean field to obtain a nearby algebraic root, and then use Krasner's lemma to show that the original root already belongs to ${\mathbb{C}}_p$.

Over a complete local ring, a power series whose first unit coefficient has index $d$ is a unit times a unique monic distinguished polynomial of degree $d$. Consequently its sufficiently small zeros, counted with multiplicity, are the zeros of that polynomial.

A distinguished polynomial over ${\mathbb{Z}}_p$ is monic and has every non-leading coefficient divisible by $p$. The Weierstrass preparation theorem expresses any nonzero element of ${\mathbb{Z}}_p[[T]]$ as a power of $p$ times a unit times a distinguished polynomial.

The Laurent series field $k((t))$ consists of formal series ${\sum }_{n\ge N}{a}_n{t}^n$ with finitely many negative-degree terms. Its $t$-adic valuation is the least exponent with nonzero coefficient.

For arbitrary positive integers $e,f$, the displayed extension has degree $ef$, ramification index $e$, and residue degree $f$. If $|t|=p^{-1}$, its unique extended absolute value on a field is $|\sum a_n{u}^n|=p^{-\min \{n:a_n\ne 0\}/e}$. A basis consists of ${\gamma }_i{u}^j$, where the ${\gamma }_i$ form an basis of ${\mathbb{F}}_{p^f}$ and $0\le j<e$. Regrouping a Laurent series by its exponent modulo $e$ proves the basis assertion. When $p$ divides $e$ the extension can be inseparable; the valuation calculation still applies.

A finite extension of the complete discretely valued field $\mathbb{C}((t))$ has residue field $\mathbb{C}$, hence is totally ramified. If $\pi$ is its uniformizer, write $t=u{\pi }^n$. A root of the residue of $u$ lifts by Hensel lemma to an $n$th root $v$ of $u$, since the residue characteristic is zero. Then $s=v\pi$ satisfies $s^n=t$. Distinct valuations modulo $n$ make $1,s,\dots ,s^{n-1}$ linearly independent over the base, so $L=\mathbb{C}((t))(s)$. Expanding in $s$ with constant coefficients identifies it with $\mathbb{C}((s))$. Its automorphisms are $s\mapsto {\zeta }_{n}s$, giving a cyclic Galois group.

Inside an algebraic closure, choose compatible roots $t^{1/n}$. The fields $\mathbb{C}((t^{1/n}))$ exhaust all finite extensions by finite extensions of the complex Laurent series field. Restriction from a degree-$m$ field to a degree-$n$ field for $n\mid m$ sends an exponent modulo $m$ to its reduction modulo $n$. The absolute Galois group is their inverse limit, namely the profinite completion of $\mathbb{Z}$. This is a procyclic group, rather than the direct-limit torsion group of all complex roots of unity.

For an irreducible polynomial $h\in {\mathbb{F}}_p[t]$ of degree $d$, the $h$-adic completion of ${\mathbb{F}}_p(t)$ is isomorphic to ${\mathbb{F}}_{p^d}((T))$. Its residue field is ${\mathbb{F}}_p[t]/(h)$ and $h$ is a uniformizer.

The $p$-adic exponential $\exp (x)={\sum }_{n\ge 0}{x}^n/n!$ converges near zero.

On ${\mathbb{C}}_p$, the power series $\exp (x)={\sum }_{n\ge 0}{x}^n/n!$ has radius of convergenceIndeed, Legendre formula gives $v_p(n!)/n\to 1/(p-1)$, and the Cauchy-Hadamard theorem gives the stated radius.

The $p$-adic logarithm is the convergent series $\log (1+x)={\sum }_{n\ge 1}(-1{)}^{n+1}{x}^n/n$ on a sufficiently small group of principal units.
It is locally inverse to the p-adic exponential function.

For a mixed-characteristic local field with residue characteristic $\ell$, the p-adic logarithm and p-adic exponential function are mutually inverse continuous group homomorphisms on these domains. Bounds on the valuations of the denominators show that higher terms have greater valuation than the linear term. Dividing the logarithm by ${\pi }^r$ identifies the multiplicative higher principal-unit group with the additive integer ring; in particular it is torsion-free.

The Teichmuller character sends a nonzero residue class to its unique root-of-unity lift in the valuation ring.

The Teichmuller representative of a nonzero residue class $\bar{a}\in {\mathbb{F}}_q^{\ast }$ is the unique lift $\omega$ satisfying ${\omega }^{q-1}=1$ and $\omega \equiv a$ modulo the maximal ideal.

For a prime $p$ and $n\ge 1$, the unique residue $y$ modulo $p^n$ satisfying $y\equiv x(\mathrm{mod}p)$ and $y^p\equiv y(\mathrm{mod}{p}^n)$ is the displayed residue. Existence follows from Fermat's little theorem and prime-power contraction under pth powers. Uniqueness follows by raising any other such residue repeatedly to the $p$th power and applying the same contraction to its congruence with $x$. This is the finite-modulus version of a Teichmuller representative, including the zero class.

For $K/{\mathbb{Q}}_p$ finite, let $x_n^{p^n}=x$ in its finite residue field, and choose arbitrary lifts $y_n\in {\mathcal{O}}_K$. With normalized discrete valuation,Applying this binomial estimate repeatedly to $y_n$ and $[x_n]$ proves the displayed bound, and hence convergence to $[x]$.

For residue field ${\mathbb{F}}_q$, these two properties characterize the Teichmuller lift on every residue class, including zero. The Hensel lemma applies to $T^q-T$, whose derivative reduces to $-1$. Uniqueness implies $[ab]=[a][b]$, $[0]=0$, $[1]=1$, and $[a^p]=[a{]}^p$. In mixed characteristic this multiplicative section need not be additive.

If $K$ is a complete discretely valued field whose residue field $k$ is a perfect field, and $\pi$ is a uniformizer, the multiplicative Teichmuller lift $[\cdot ]:k\to {\mathcal{O}}_K$ gives each $x\in {\mathcal{O}}_K$ a unique expansion

If a complete discretely valued field $K$ has positive characteristic and perfect residue field $k$, its Teichmuller lift is the unique ring section $k\to {\mathcal{O}}_K$. A choice of uniformizer $t$ then gives

For a characteristic-p local field with residue field ${\mathbb{F}}_q$, the simple roots of $X^q-X$ lift uniquely by Hensel lemma. Their lifts form a copy of ${\mathbb{F}}_q$, since qth powers preserve addition and multiplication. Choose a uniformizer $\pi$. Repeatedly subtracting the lifted residue and dividing by $\pi$ produces a unique convergent expansion ${\sum }_{i\ge 0}{a}_i{\pi }^i$ for each integral element. Allowing finitely many negative powers gives a valued-field isomorphism $K\cong {\mathbb{F}}_q((T))$, $T\mapsto \pi$.

A finite extension of local fields is unramified when its ramification index is one and its residue-field degree equals the field degree. A local field has a unique unramified extension of each positive degree inside a fixed algebraic closure.

For non-Archimedean local fields, let $L/K$ be an unramified extension of degree $f$, and let $E/K$ be finite. The degree-$f$ unramified extension of $E$ has a residue field containing the degree-$f$ residue extension of $K$. Lift an integral residue generator of $L$ into that field by Hensel lemma. This embeds $L$ into an unramified extension of $E$, so $LE/E$ is an unramified subextension. Applying this argument to each completion of a number field at a prime ideal proves the finite-prime assertion for number fields, even when $E/K$ is ramified.

For an unramified extension $L/K$ of non-Archimedean local fields and any finite extension $M/K$, reduction is a bijection between $K$-field homomorphisms $L\to M$ and $k_K$-field homomorphisms $k_L\to k_M$. Choose an integral generator with separable irreducible reduction. Each specified residue image lifts uniquely to a root in $M$ by Hensel lemma. Conversely a $K$-embedding preserves the unique extended valuation and therefore induces the residue embedding. The same argument applies when $M$ is any non-Archimedean local extension with the given compatible embedding of $K$.

A finite extension of complete discretely valued fields is unramified exactly when it has an integral generator whose reduced minimal polynomial of an algebraic element is separable. In one direction, lift a residue-field primitive element. In the other, the factorization form of Hensel's lemma makes the separable reduction irreducible, forcing residue degree to equal the entire extension degree.

For an unramified extension of local fields $L/K$, an integral element whose residue generates $k_L/k_K$ generates $L/K$: its degree is at least the residue degree $f=[L:K]$. The reduced powers $1,\bar{\alpha },\dots ,{\bar{\alpha }}^{f-1}$ are a basis of the residue field extension. If $x=\sum c_j{\alpha }^j$, factor out a coefficient of least valuation. Reduction of the remaining sum is nonzero by that residue basis, so $v_L(x)={\min }_j{v}_K(c_j)$. Consequently $x$ is integral exactly when all $c_j$ are integral.

Let $L/K$ be a finite unramified extension of complete fields with a Non-Archimedean absolute value, with separable residue extension and $[L:K]=[k_L:k_K]=n$. If $x$ is integral and its residue generates $k_L/k_K$, then $1,x,\dots ,x^{n-1}$ form a $K$-basis. Divide any nonzero combination by a coefficient of largest absolute value. Its remaining coefficients are integral, at least one is a unit, and the residue-basis independence makes the reduction nonzero. The displayed norm formula follows. In particular, the combination is integral exactly when every coefficient is integral, proving ${\mathcal{O}}_L={\mathcal{O}}_K[x]$ without requiring a discrete valuation.

For a non-Archimedean local field $K$, every finite extension $\ell /k_K$ of its residue field is realized by an unramified extension of $K$, unique up to $K$-isomorphism. Lift the irreducible polynomial of a residue generator to a monic integral polynomial and adjoin a root. Its residue degree equals its field degree, forcing ramification index one. In any other such extension the Hensel lemma lifts the same simple residue root, giving an isomorphism.

For a finite residue field $k$ of cardinality $q$, arithmetic Frobenius is $x\mapsto x^q$. Its lift generates the Galois group of a finite unramified extension; geometric Frobenius is its inverse.

In a finite Galois extension $L/K$, choose a prime ideal $\mathfrak{P}$ above an unramified $\mathfrak{p}$. The Frobenius automorphism at $\mathfrak{P}$ is characterized on its residue field by $x\mapsto x^{N\mathfrak{p}}$. Changing $\mathfrak{P}$ conjugates this element, so its conjugacy class is well-defined. It is the identity class exactly at a completely split prime.

Let a monic integer polynomial have distinct roots in its splitting field. Outside the finitely many primes dividing its polynomial discriminant or ramifying in that field, its roots reduce to distinct residue roots. The Frobenius automorphism permutes them by $a\mapsto a^p$; a reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed. Distinctness of reduction identifies this with a fixed root of the Galois permutation itself. Combining this observation with a derangement in a transitive group action and the Chebotarev density theorem proves that an irreducible polynomial of degree at least two has rootless reduction at a positive Dirichlet density of primes. The degree bound is essential: every monic linear integer polynomial has a root modulo every prime.

The maximal unramified extension $K^{\mathrm{nr}}$ is the union of all finite unramified extensions of a non-Archimedean local field $K$. Its Galois group is canonically $\hat{\mathbb{Z}}$, generated topologically by Frobenius.

Every finite extension $L/K$ of non-Archimedean local fields has a unique intermediate field $K_0$ whose residue field is $k_L$ and for which $K_0/K$ is unramified. The extension $L/K_0$ is totally ramified.

For a finite Galois extension of local fields, the lower ramification groups measure how closely automorphisms fix the valuation ring. The inertia group is $G_0$, and $G_1$ is the wild inertia group.

For $\sigma \in G_i$, the residue of $(\sigma {\pi }_L-{\pi }_L)/{\pi }_L^{i+1}$ gives a homomorphism to the additive residue field with kernel $G_{i+1}$. On composing two automorphisms, their leading displacement coefficients add because each fixes residues and changes the uniformizer only in degree at least two. These successive quotients are elementary abelian groups in residue characteristic $p$. The filtration terminates, so the wild inertia group is a finite $p$-group.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

For $L={\mathbb{Q}}_p({\zeta }_{p^n})$, the uniformiser ${\zeta }_{p^n}-1$ and the uniformizer criterion for lower ramification groups give the displayed valuation for $a\ne 1$. Thus $G_0$ is the full unit group modulo $p^n$, and $G_i$ consists of units congruent to one modulo $p^j$ when $p^{j-1}\le i\le p^j-1$, $1\le j<n$. All groups are trivial for $i\ge p^{n-1}$. The formula includes $p=2$, where the first congruence subgroup is already the full group.

For all real $s\ge 0$, this definition uses the normalized integral valuation on $L$. Equivalently $G_s=G_{\lceil s\rceil }$. The group $G_0$ is the inertia group and $G_1$ is the wild inertia group.

The Herbrand function is a continuous strictly increasing piecewise linear function on the nonnegative reals. Its inverse is denoted ${\psi }_{L/K}$. It reparametrizes the lower ramification numbering to the upper ramification numbering.

The upper numbering is obtained by applying the inverse Herbrand function to the index in the lower ramification numbering. It is compatible with quotient Galois groups, whereas lower numbering is compatible with subgroups.

For a normal subgroup $H$ of a finite local Galois group, the upper ramification numbering on its quotient is obtained by projecting the upper groups. The Herbrand function is the change of variable that ensures this compatibility. In contrast, the lower ramification numbering restricts directly to subgroups.

If $L/K$ is a finite totally ramified Galois extension and ${\pi }_L$ is a uniformizer, thenfor every $i\ge 0$.

For $L={\mathbb{Q}}_3({\zeta }_3,\alpha )$ with ${\alpha }^3=2$, the shifted cubic for $\alpha +1$ and quadratic for ${\zeta }_3-1$ are Eisenstein. Their compositum is totally ramified of degree six. The element $\pi =({\zeta }_3-1)/(\alpha +1)$ has valuation one. Order-three automorphisms move it with valuation two and transpositions with valuation one. The lower breaks are zero and one, the upper breaks zero and one-half, and the different exponent from ramification groups is seven.

For ${\mathbb{Q}}_2({\zeta }_8)/{\mathbb{Q}}_2$, the shifted polynomial $(1+T{)}^4+1$ is Eisenstein and ${\zeta }_8-1$ is a uniformizer. The three nonidentity automorphisms have displacement valuations two, four and two. Thus the lower breaks are one and three, the upper breaks one and two, and the different exponent from ramification groups is eight.

Let $K$ be an equal-characteristic-p local field and let $c\in K$ have $v_K(c)=-m<0$, with $p\nmid m$. If ${\alpha }^p-\alpha =c$, the Artin–Schreier extension $L=K(\alpha )$ has degree $p$ and is totally ramified: a root in $K$ would force the pole order to be divisible by $p$, and $p{v}_L(\alpha )=-me(L/K)$ forces $e=p$. Choose $1\le s<p$ and an integer $r$ with $rp-sm=1$. For a base uniformizer $t$, the element ${\pi }_L=t^r{\alpha }^s$ has valuation one. Under ${\sigma }_a(\alpha )=\alpha +a$, $a\in {\mathbb{F}}_p^{\times }$, the leading term in ${\sigma }_a({\pi }_L)-{\pi }_L$ has valuation $rp-(s-1)m=m+1$. The uniformizer criterion for lower ramification groups therefore gives the single lower break $m$. The Herbrand function is the identity up to that break, so the upper break is also $m$.

For the splitting field $M/{\mathbb{Q}}_3$ of $T^3-3T+3$, the cubic is Eisenstein and its discriminant is $-135$. The quadratic subfield is ${\mathbb{Q}}_3(\sqrt{3})$, and the degree-six extension is totally ramified. If $\alpha$ is a root and $s=\sqrt{3}$, then $\pi ={\alpha }^2/s$ is a uniformizer. Transpositions have $v_M(\sigma \pi -\pi )=1$, while three-cycles have value $2$, giving the displayed ramification groups. Its different exponent from ramification groups is $7$.

For odd $p$, the splitting field $K={\mathbb{Q}}_p({\zeta }_p,\sqrt[p]{p})$ is totally ramified of degree $p(p-1)$. Withthe lower ramification groups are

To verify the filtration, write ${\alpha }^p=p$ and $\lambda ={\zeta }_p-1$. The Eisenstein polynomials for $\alpha$ and $\lambda$ give coprime degrees $p$ and $p-1$, so their compositum has degree $p(p-1)$. Its element $\varpi =\lambda /\alpha$ has $p$-adic valuation $1/[p(p-1)]$, forcing total ramification. For an automorphism ${\sigma }_{a,b}$ with $\alpha \mapsto {\zeta }_p^a\alpha$ and ${\zeta }_p\mapsto {\zeta }_p^b$,Its residue is $b$, so the normalized displacement valuation is one if $b\ne 1$. If $b=1$ and $a\ne 0$, that valuation is $1+v_K({\zeta }_p^{-a}-1)=p+1$. The uniformizer criterion for lower ramification groups now gives every group displayed above.

If $a^3=3$ and ${\zeta }_3$ is a primitive cube root of unity, then ${\pi }^3=1+2{\zeta }_3$ and ${\pi }^6=-3$. The Eisenstein polynomial $X^6+3$ therefore presents the entire degree-six splitting field, and $\pi$ is a uniformizer. Its normalised valuation gives $v_L(3)=6$, $v_L(a)=2$ and $v_L({\zeta }_3-1)=3$. The order-three automorphisms displace $\pi$ with valuation four, whereas transpositions displace it with valuation one.

For a finite totally ramified Galois extension, the mapinduces an injection $G_0/G_1\hookrightarrow k_L^{\times }$. In particular, the tame inertia quotient is cyclic and has order prime to the residue characteristic.

For a Finite Galois extension of $K(T)$ with algebraically closed characteristic-zero constants, the local decomposition and inertia groups coincide and are cyclic. A fixed compatible choice of primitive roots of unity selects a generator through its uniformizer character of inertia. Changing the prime above a base point conjugates this generator, and changing the uniformizer does not change its character. Without the roots-of-unity convention, only the inertia-subgroup conjugacy class is intrinsic. Over $\mathbb{C}$, positive local orientation gives the standard convention.

In characteristic zero with algebraically closed constants, a local unit has a square root. Thus $K(T)(\sqrt{a})$ near a point is equivalent to adjoining the square root of the local parameter to the power $v_p(a)$. Even order is unramified; odd order gives ramification index two. For a polynomial $a$, its order at infinity is minus its degree, so infinity branches exactly for odd degree.

For a finite Galois extension of non-Archimedean local fields, this is a homomorphism from the inertia group to $k_L^{\times }$ with kernel the wild inertia group. Inertia acts trivially on residue units, which proves multiplicativity and independence of the chosen uniformizer. Passing to the maximal unramified subextension reduces the kernel calculation to a totally ramified extension. Hence it embeds the tame inertia quotient into the cyclic multiplicative group of the residue field.

For a finite separable extension of local fields, the different is the inverse fractional ideal of the codifferent. If ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$ and $f$ is the minimal polynomial of $\alpha$, then ${\mathfrak{D}}_{L/K}=(f^{\prime }(\alpha ))$.

The different exponent at a prime ideal $\mathfrak{P}$ is its exponent in the prime ideal factorization of the different ideal. It equals the local different exponent of the corresponding extension of completions. The different exponent and tame ramification theorem detects unramified, tame, and wild behavior without assuming a Galois extension.

For $K_r={\mathbb{Q}}_p({\zeta }_{p^r})$, the uniformizer ${\zeta }_{p^r}-1$ has different exponent $p^{r-1}(r(p-1)-1)$. Differentiate the cyclotomic polynomial at ${\zeta }_{p^r}$: the numerator contributes $r{p}^{r-1}(p-1)$ to the valuation, while the denominator ${\zeta }_p-1$ contributes $p^{r-1}$.