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For a complete discretely valued field with perfect residue field $k$, the Witt residue sequence is the split exact sequenceThe first map identifies the unramified Brauer classes. The residue map is canonical; a splitting is chosen by a uniformizer $\pi$ and sends an unramified character $\chi$ to the cyclic Brauer class $(\chi ,\pi )$. Changing $\pi$ may change this splitting by an unramified class. This is the version of Witt's theorem used here.

Now suppose $k={\mathbb{F}}_q$. Fix arithmetic Frobenius $F_k:a\mapsto a^q$. Define the local Brauer invariant byHere is an explicit cohomological construction. Set $E=K^{\mathrm{nr}}$ and $G=\mathrm{Gal}(E/K)\cong \hat{\mathbb{Z}}$. The permitted vanishing $\mathrm{Br}(E)=0$, together with Hilbert 90 and inflation, identifies $\mathrm{Br}(K)$ with $H^2(G,E^{\times })$. Normalize the valuation $v(\pi )=1$. Its induced map sends a multiplicative two-cocycle $c$ to the integer-valued two-cocycle $v(c)$, givingThe middle isomorphism is the connecting map for $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}\to 0$, since positive-degree continuous cohomology of the trivial module $\mathbb{Q}$ vanishes.

To see that the valuation map is an isomorphism, its module sequence splits using $\pi$, and the unit part has zero second cohomology. For each finite unramified cyclic extension $E_m/K$, cyclic cohomology identifies that unit contribution with ${\mathcal{O}}_K^{\times }/N_{E_m/K}{\mathcal{O}}_{E_m}^{\times }$. The norm on residue-field units is surjective. On the principal-unit quotient of level $r\ge 1$, the norm is the residue trace, through $N(1+{\pi }^{r}b)\equiv 1+{\pi }^r\mathrm{Tr}(\overline{b})(\mathrm{mod}{\pi }^{r+1})$. The trace of a finite separable field extension is surjective; successive corrections and completeness therefore make the norm surjective on all units. Passing to the direct limit gives the asserted vanishing. Finally, a continuous character of $\hat{\mathbb{Z}}$ is uniquely specified by its value on arithmetic Frobenius, and that value can be any element of $\mathbb{Q}/\mathbb{Z}$. The invariant map is an isomorphism.

In concrete terms, if $E_m/K$ is the unramified extension of degree $m$ and $\sigma$ lifts arithmetic Frobenius, the cyclic algebra with $ub=\sigma (b)u$ and $u^m=a$ hasIndeed its standard two-cocycle has value $a$ when the powers of $\sigma$ wrap past $m$ and value one otherwise. Valuation turns this into the connecting cocycle for the character with Frobenius value $v_K(a)/m$. This also fixes the sign of the normalization.

For restriction and corestriction of local Brauer invariants, let $l$ be the residue field of $L$, $e$ the ramification index and $f=[l:k]$. A finite extension of local fields has $[L:K]=ef$, $v_L{|}_K=e{v}_K$ and $v_K(N_{L/K}b)=f{v}_L(b)$. Every class of $\mathrm{Br}(K)$ is $(\chi ,{\pi }_K)$, with $\chi$ unramified. After restriction to $L$, its character has Frobenius value $f\chi (F_k)$ and its parameter has valuation $e$. ThusFor corestriction, write a class of $\mathrm{Br}(L)$ as $({\chi }_l,{\pi }_L)$. Choose a character ${\chi }_k$ of the finite-field absolute group such that ${\mathrm{Res}}_{l/k}{\chi }_k={\chi }_l$. This is possible because restriction is multiplication by $f$ on $\mathbb{Q}/\mathbb{Z}$, which is divisible. Inflate both characters to the local fields. The norm projection formula for cyclic Brauer pairings givesThis projection identity follows from the cohomological transfer and cup-product identity; on multiplicative degree-zero coefficients transfer is the field norm, and the identity in higher degree follows by the cochain definition or dimension shifting. The norm parameter has $K$-valuation $f$, so its invariant is $f{\chi }_k(F_k)={\chi }_l(F_l)$. ThereforeThis argument uses the character-and-norm description directly and does not cancel multiplication by $[L:K]$ in $\mathbb{Q}/\mathbb{Z}$, where such cancellation would be invalid. For purely inseparable steps in equal characteristic, absolute Galois groups identify, restriction uses inclusion of multiplicative coefficient modules, and Brauer corestriction uses the field norm on them. The same projection calculation applies. Factoring a general finite extension into separable and purely inseparable steps covers all finite extensions in the question.

Lastly, a class with invariant $a/b$ in lowest terms has Brauer period $b$. The unramified extension of degree $b$ splits it, so its division-algebra degree divides $b$. Conversely a maximal subfield of the division algebra splits it and has degree equal to that division-algebra degree; the restriction formula forces $b$ to divide that degree. These standard splitting-field properties are among the permitted inputs. Thus the index of a central simple algebra equals its period over a local field. The central division algebras of degree $n$ are exactly those with invariants $a/n$ and $\gcd (a,n)=1$. The Brauer class determines its division representative uniquely, so there areisomorphism classes, where $\phi$ is the Euler totient function. This includes the single degree-one algebra $K$.

For this non-Archimedean local field, the Schwartz-Bruhat space $\mathcal{S}(F)$ consists of locally constant, compactly supported complex-valued functions. Choose a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ of modulus one and an additive Haar measure $dx$. We use the plus-sign convention for the Fourier transform over a local field:Compact support makes the integral absolutely convergent. The additive character and the scale of the Haar measure are part of the definition; without them there is no canonical numerical transform.

Let $q=|{\mathcal{O}}_F/{\pi }_F{\mathcal{O}}_F|$, and write $V=\mathrm{vol}({\mathcal{O}}_F)$. Define the integer $c$ by the annihilator of the valuation ring ${\mathcal{O}}_F^{\perp }={\pi }_F^c{\mathcal{O}}_F$. Equivalently, $\psi$ is trivial on ${\pi }_F^c{\mathcal{O}}_F$ and not on ${\pi }_F^{c-1}{\mathcal{O}}_F$. Such a conductor exists: continuity puts the image of some additive ball inside an arc containing no nontrivial circle subgroup, so the character is trivial on that ball; nontriviality bounds the possible ball exponents below. Then $H_n={\pi }_F^n{\mathcal{O}}_F$ has volume $V{q}^{-n}$ and character annihilator $H_n^{\perp }={\pi }_F^{c-n}{\mathcal{O}}_F$.

Translation givesThe integral is the volume if $y\in H_n^{\perp }$. Otherwise translate it by $t_0\in H_n$ with $\psi (t_{0}y)\ne 1$; Haar measure invariance multiplies the same integral by a nonidentity scalar, so it must vanish. ThusEvery Schwartz-Bruhat function is a finite linear combination of such coset indicators: compactness of its support supplies a common sufficiently small translation subgroup on which it is constant, and finitely many of its cosets cover the support. The transform of each indicator has compact support and is locally constant, since $\psi$ has open kernel. Consequently $\hat{f}\in \mathcal{S}(F)$ for every $f\in \mathcal{S}(F)$.

Apply the transform again to an indicator. The same character-orthogonality calculation, together with $(H_n^{\perp }{)}^{\perp }=H_n$, givesChoose the self-dual Haar measure, characterized here by $V=q^{c/2}$. By linearity the desired Fourier inversion isIn particular, one may rescale any nontrivial additive character to make $c=0$, and then take $V=1$. For a concrete construction, start with ${\psi }_0(x)={\psi }_{{\mathbb{Q}}_p}({\mathrm{Tr}}_{F/{\mathbb{Q}}_p}x)$, where the standard rational additive character has kernel ${\mathbb{Z}}_p$. Its annihilator is the inverse different ${\mathfrak{D}}_{F/{\mathbb{Q}}_p}^{-1}$. If that ideal is ${\pi }_F^{-d}{\mathcal{O}}_F$, the rescaled character $\psi (x)={\psi }_0({\pi }_F^{-d}x)$ has ${\mathcal{O}}_F^{\perp }={\mathcal{O}}_F$. This supplies the stated normalization and also explains how the different ideal enters local Fourier analysis.

The valuation ring and local compactness. The opening, unlettered requests useThe ultrametric inequality makes ${\mathcal{O}}_K$ a valuation ring; its elements outside $\mathfrak{m}$ are exactly the units, so $\mathfrak{m}$ is its unique maximal ideal, and $k$ is the residue field.

Here the customary nontrivial-valuation hypothesis for a local field is necessary. With the trivial absolute value, any infinite field is discrete and locally compact, but its residue field is infinite and there is no nonzero uniformizer. We use the nontrivial valuation in the conclusions below.

Local compactness first gives a compact neighborhood of zero. Choose a sufficiently small nonzero $t$ so that the closed set $t{\mathcal{O}}_K$ is contained in that compact neighborhood. Then $t{\mathcal{O}}_K$, and hence ${\mathcal{O}}_K$, is compact. Since $\mathfrak{m}$ is open, its cosets give a discrete compact quotient $k$, which must be finite. Write $|k|=q$. The subgroup $\mathfrak{m}$ now has finite index and is closed in ${\mathcal{O}}_K$, hence compact. The continuous function $|\cdot |$ attains on $\mathfrak{m}$ a maximum $c$ with $0<c<1$. Choose $\pi$ with $|\pi |=c$. Every $x\in \mathfrak{m}$ has $|x/\pi |\le 1$, soSuccessive division by $\pi$ shows that every nonzero element of ${\mathcal{O}}_K$ is a unit times a power of $\pi$: division must terminate because $|\pi {|}^j\to 0$. Thus the value group is discrete. A Cauchy sequence eventually lies in a translate of the compact ring ${\mathcal{O}}_K$, has a convergent subsequence, and consequently converges itself. These arguments prove the discrete valuation from nontrivial local compactness and the completeness of locally compact nontrivially valued fields.

The iterated-power limit. Normalize the discrete valuation by $v(\pi )=1$, and write $q={\ell }^f$, where $\ell$ is the residue characteristic. For $u\in 1+{\pi }^s{\mathcal{O}}_K$ with $s\ge 1$, the binomial theorem givesThe linear term gains a factor $q\in \mathfrak{m}$ (or is zero in positive characteristic), while all terms of degree at least two have valuation at least $2s\ge s+1$.

For $a\in {\mathcal{O}}_K^{\times }$, reduction gives $a^{q-1}\in 1+\mathfrak{m}$, because $k^{\times }$ has order $q-1$. Iterating the inequality yieldsThe sequence is Cauchy and converges to a unit $\omega (a)$. Since the reduction of $a^{q^j}$ is always $\bar{a}$, its limit has the same residue. Taking limits after shifting the sequence gives $\omega (a{)}^q=\omega (a)$, whenceThis is the Teichmuller representative. The Hensel lemma applied to $T^{q-1}-1$, whose derivative is a unit at every nonzero residue, shows uniqueness for each residue. Consequently the prime-to-$\ell$ roots give a canonical splitting

The filtration and its meaning. Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with group $G$, normalized valuation $v_L$, and residue fields $k_L/k_K$ of characteristic $\ell$. The lower ramification numbering isThe higher ramification groups are normal, decrease with $j$, and eventually become trivial: for each nonidentity automorphism some integral element is moved by a nonzero amount. The first group $G_0$ is the inertia group, withIts fixed field is the maximal unramified extension inside $L$. The wild inertia group is $G_1$. For a uniformizer ${\pi }_L$, there are injective homomorphismsThe kernels are the indicated next groups; the additive homomorphism assertion follows by expanding a product of automorphisms modulo ${\pi }_L^{j+2}$. These maps show that $G_0/G_1$ is cyclic of order prime to $\ell$, while $G_1$ is an $\ell$-group. Thus tame ramification is exactly the case $G_1=1$.

To calculate the groups, the uniformizer criterion for lower ramification groups saysIndeed, ${\mathcal{O}}_L$ is generated by ${\pi }_L$ over the integers of the maximal unramified subextension, which $G_0$ fixes. Differences of powers of ${\pi }_L$ are divisible by $\sigma {\pi }_L-{\pi }_L$, so the one test implies all the defining inequalities.

Eisenstein facts used in the calculations. For a complete discrete valuation ring $A$ with uniformizer $\varpi$, a monic polynomialis an Eisenstein polynomial if $\varpi \mid a_j$ for every $j<d$ and ${\varpi }^2\nmid a_0$. The Eisenstein criterion makes it irreducible. If $\theta$ is a root, its valuation relative to the base normalization is $1/d$, the extension has degree and ramification index $d$, its residue field is unchanged, andThe root valuation follows by comparing the terms of its equation. The ramification index must then be at least $d$, hence exactly $d$. In the basis $1,\theta ,\dots ,{\theta }^{d-1}$, the valuations of $b_j{\theta }^j$ are distinct modulo $d$, so an integral linear combination has every $b_j\in A$. This proves the integer-ring assertion. Conversely, a uniformizer in a totally ramified extension generates the field, and its minimal polynomial is Eisenstein; this follows from its value $1/d$, equal valuations of its conjugates, and the valuation one of its norm.

For a monogenic separable integer ring $A[\theta ]$, the different ideal is generated by $f^{\prime }(\theta )$. For a Galois extension, the different exponent from ramification groups isIn the totally ramified case these agree directly: $f^{\prime }({\pi }_L)={\prod }_{\sigma \ne 1}({\pi }_L-\sigma {\pi }_L)$, and an automorphism with displacement valuation $h$ contributes once to each of $G_0,\dots ,G_{h-1}$.

Upper numbering. Extend the lower indexing to real $s\ge 0$ by $G_s=G_{\lceil s\rceil }$, and setOn $[-1,0]$ take $\phi (s)=s$. This Herbrand function slows the indexing as the groups shrink. Lower numbering is compatible with subgroups, while upper ramification numbering is compatible with quotients: for normal $H$, $(G/H{)}^u=G^{u}H/H$. This is the fact that upper ramification groups commute with quotients. Thus upper breaks are particularly useful when comparing intermediate Galois extensions.

The eighth-root cyclotomic extension. Put $\zeta ={\zeta }_8$ and $\pi =\zeta -1$. The shifted cyclotomic polynomialis Eisenstein at two. Hence $L={\mathbb{Q}}_2(\zeta )$ is totally ramified of degree four, $v_L(2)=4$, and $\pi$ is a uniformizer. Its Galois group consists of ${\sigma }_a(\zeta )={\zeta }^a$ for $a=1,3,5,7$, and is $C_2\times C_2$.

Since $\zeta$ is a unit,For $a=3,7$, this is the valuation of $i-1$ or $-i-1$, namely two, because their squares are units times two. For $a=5$, it is $v_L(-2)=4$. The ramification groups of the eighth-root cyclotomic extension of the 2-adic field are thereforeThe lower breaks are one and three, andThus the upper breaks are one and two:The different exponent from ramification groups is $3+3+1+1=8$, agreeing with $v_L(4{\zeta }^3)=8$ from the derivative of $T^4+1$.

The cubic splitting field at three. Let ${\alpha }^3=2$, $\zeta ={\zeta }_3$, and putThe polynomialsare Eisenstein at three. They give totally ramified subextensions of degrees three and two. Their intersection is the base field, so $L={\mathbb{Q}}_3(\alpha ,\zeta )$ has degree six. Its ramification index is divisible by both three and two, hence equals six; therefore $L/{\mathbb{Q}}_3$ is totally ramified. It is the splitting field of $T^3-2$, with Galois group $S_3$.

Normalize $v_L$ so that $v_L(3)=6$. Then $v_L(\beta )=2$ and $v_L(\lambda )=3$, makinga uniformizer. Let $\sigma (\alpha )=\zeta \alpha ,\sigma (\zeta )=\zeta$ and $\tau (\alpha )=\alpha ,\tau (\zeta )={\zeta }^2$. They generate $S_3$. Since $\tau (\lambda )=(\zeta +1)\lambda$,For the order-three automorphism,Here $\alpha$ is a unit and $v_L(\zeta \alpha +1)=v_L(\sigma \beta )=2$. The same calculation gives valuation two for ${\sigma }^2$. The three transpositions are conjugate and the defining filtration is normal, so they all have displacement valuation one. The ramification groups of the splitting field of T3 minus 2 over Q3 areThe lower breaks are zero and one. Since $[G_0:G_s]=2$ for $0<s\le 1$,and the upper groups areThe different exponent from ramification groups is $5+2=7$. As an independent check, the cubic subfield has different exponent $v_{{\mathbb{Q}}_3(\alpha )}(3{\alpha }^2)=3$, and the quadratic extension above it is tame with different exponent one. Transitivity of the different ideal gives $1+2\cdot 3=7$, as required.

Normalize the discrete valuation on $L$ by $v_L(L^{\times })=\mathbb{Z}$, set $v_L(0)=+\infty$, and write ${\mathcal{O}}_L$ for its valuation ring. For a finite Galois extension of local fields, define $G_{-1}=G=\mathrm{Gal}(L/K)$ and, for integers $i\ge 0$,Equivalently, $G_i$ is the kernel of the action on ${\mathcal{O}}_L/{\mathfrak{m}}_L^{i+1}$. These are the lower ramification numbering of the ramification groups. In particular, $G_0$ is the inertia group, the kernel of the action on the residue field, and $G_1$ is the wild inertia group. They form a decreasing sequence of normal subgroups, eventually trivial.

For a real index $t\ge -1$, extend by $G_t=G_{\lceil t\rceil }$. Define the Herbrand function byIt is continuous, strictly increasing and piecewise linear; denote its inverse by ${\psi }_{L/K}$. The upper ramification numbering isThe definition includes the placement of the groups at break endpoints: the group at a break is the group before the drop. Lower numbering is compatible with subgroups; upper numbering is the one compatible with quotients. In a totally ramified extension, the uniformizer criterion for lower ramification groups permits testing the defining valuation inequality on one uniformizer instead of on every integral element.

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.

Fix a non-Archimedean local field $K$, its valuation ring $R={\mathcal{O}}_K$, a uniformizer $\pi$, and a residue field with cardinality $q$. The construction below applies both in mixed and equal characteristic. A one-dimensional commutative formal group law is $F(X,Y)\in R[[X,Y]]$ with identity, associativity and commutativity, and $F(X,Y)=X+Y+$ higher-degree terms. It has a unique formal inverse. Its morphisms are series $h(T)\in TR[[T]]$ with $h(F(X,Y))=G(h(X),h(Y))$; a unit linear coefficient makes such a morphism invertible by recursive coefficient comparison. Evaluating its integral series on the maximal ideal of a finite extension of $K$ is legitimate because the series converge there.

Choose a Lubin–Tate series $f(T)$ satisfyingA useful choice is $f(T)=\pi T+T^q$. The Lubin–Tate functional equation lemma constructs a unique series with any prescribed linear form $\ell$ and satisfyingHere is its coefficient argument. If terms below degree $d$ have been found, the unknown homogeneous degree-$d$ term appears with multiplier $\pi -{\pi }^d$. The already known error has every coefficient divisible by $\pi$: after reduction, the equation becomes $H(X{)}^q=H(X_1^q,\dots ,X_r^q)$, valid because the residue coefficients lie in ${\mathbb{F}}_q$. Since $\pi -{\pi }^d$ is $\pi$ times a unit for $d\ge 2$, the next term is uniquely determined and integral. Recursion proves existence and uniqueness.

Use $\ell =X+Y$ to construct $F_f$, and $\ell =aT$ to construct $[a{]}_f$ for every $a\in R$. Uniqueness gives $F_f(X,0)=X$ and $F_f(0,Y)=Y$. Comparing $[a{]}_f(F_f(X,Y))$ with $F_f([a{]}_f(X),[a{]}_f(Y))$, which satisfy the same functional equation and have the same linear form, proves that $[a{]}_f$ is an endomorphism of the law. The same uniqueness applied to three variables proves associativity: $F_f(F_f(X,Y),Z)$ and $F_f(X,F_f(Y,Z))$ have the same linear form and functional equation. It also proves commutativity and the identitiesThus this Lubin–Tate formal group is equipped with an action of the whole ring $R$, not only its integers. For a unit $a$, the inverse of $[a{]}_f$ is $[a^{-1}{]}_f$. The same recursion, with two series for the fixed $\pi$, gives a unique strict Lubin–Tate change of series. Its integral series and inverse converge on torsion points, so it identifies their torsion fields. We may compute with $f(T)=\pi T+T^q$ without losing the general construction for that uniformizer.

Write $f_0(T)=T$ and $f_n=f^{\circ n}$. The Lubin–Tate torsion of level $n$ isFor $n\ge 1$, its primitive points are the roots ofThis is a monic polynomial of degree $D_n=(q-1)q^{n-1}$. Modulo $\pi$ it is $T^{D_n}$, and its constant coefficient is exactly $\pi$. Thus the Eisenstein layers of Lubin–Tate torsion are Eisenstein polynomials. A root ${\lambda }_n$ generates a totally ramified extension of degree $D_n$, and has $v_K({\lambda }_n)=1/D_n$ in the uniquely extended valuation. Also $f_{n-1}({\lambda }_n{)}^{q-1}=-\pi \ne 0$, so it has exact level $n$.

The mapis well-defined and injective. If $a={\pi }^{r}u$ with $r<n$ and $u$ a unit, then $[a]{\lambda }_n=[u]f_r({\lambda }_n)\ne 0$, because $[u]$ is invertible and ${\lambda }_n$ is primitive. Conversely every multiple of ${\pi }^n$ kills it. The image therefore contains $q^n$ distinct roots of the degree-$q^n$ polynomial $f_n$, so it is all of ${\Lambda }_n$. This proves, without merely assuming the torsion cardinality,Its primitive generators are exactly $[a]{\lambda }_n$ with $a\in (R/{\pi }^{n}R{)}^{\times }$.

Define the Lubin–Tate extension $K_{\pi ,n}=K({\Lambda }_n)$. Every integral endomorphism series evaluated at ${\lambda }_n$ converges in the complete field $K({\lambda }_n)$, so all torsion points already lie there. HenceIt is the splitting field of $f_n$, which has $q^n$ distinct roots by the module calculation. Thus it is Galois. Every K-automorphism preserves the uniquely extended absolute value on a field, hence is continuous and commutes with limits of the integral endomorphism series. Its automorphisms therefore commute with the endomorphism action and send a primitive generator to a primitive generator. There is a unique unit class $a_{\sigma }$ such that $\sigma ({\lambda }_n)=[a_{\sigma }]{\lambda }_n$. The Lubin–Tate Galois action on primitive torsion gives an injective homomorphismBoth sides have order $D_n=(q-1)q^{n-1}$, so it is an isomorphism. In particular every finite layer is abelian and totally ramified.

For an example, take $K={\mathbb{Q}}_p$, $\pi =p$, and the equally admissible series $f(T)=(1+T{)}^p-1$. Its formal group law is the formal multiplicative group $F(X,Y)=X+Y+XY$, with $[a](T)=(1+T{)}^a-1$ for $a\in {\mathbb{Z}}_p$. The binomial polynomials make its coefficients integral. Its level-$n$ torsion points are $\zeta -1$ with ${\zeta }^{p^n}=1$. Thus the Lubin–Tate cyclotomic example gives $K_{p,n}={\mathbb{Q}}_p({\mu }_{p^n})$ and Galois action $\zeta \mapsto {\zeta }^a$. The order formula includes $p=2,n=1$, when the first field and its Galois group are trivial over ${\mathbb{Q}}_2$.

Finally choose primitive points compatibly, $[\pi ]{\lambda }_{n+1}={\lambda }_n$. They exist by taking a root of $f(T)-{\lambda }_n$ in the algebraic closure; its roots have positive valuation and exact level $n+1$. Then $K_{\pi ,n}\subseteq K_{\pi ,n+1}$, and restriction of Galois automorphisms corresponds to reduction of unit classes. The Lubin–Tate tower $K_{\pi }={\bigcup }_n{K}_{\pi ,n}$ consequently has the topological Galois groupThe final isomorphism uses completeness of the discrete valuation ring $R$. This computes the finite-layer and infinite-tower Galois groups directly from the formal action and the primitive Eisenstein polynomials.

Use the simple-root form of the Hensel lemma: for $F\in {\mathcal{O}}_K[T]$ and $\bar{a}\in k_K$ satisfying $\overline{F}(\bar{a})=0$ and $\overline{F^{\prime }}(\bar{a})\ne 0$, there is a unique $a\in {\mathcal{O}}_K$ with $\bar{a}$ as its reduction and $F(a)=0$.

Let $d=[\ell :k_K]$. The residue field $k_K$ is finite, so $\ell /k_K$ is separable, and the primitive element theorem gives $\ell =k_K(\bar{\alpha })$. Let $\bar{F}$ be the monic irreducible polynomial of $\bar{\alpha }$, of degree $d$, and choose a monic coefficient lift $F\in {\mathcal{O}}_K[T]$. Irreducible reduction implies that $F$ is irreducible over $K$: any monic factorization over $K$ has integral coefficients, since all its roots are integral over the valuation ring, and would reduce to a factorization of $\bar{F}$.

Adjoin a root $\alpha$ and put $L=K(\alpha )$. Then $[L:K]=d$ and $\alpha \in {\mathcal{O}}_L$. Its reduction satisfies $\bar{F}$, so $k_L$ contains a copy of $\ell$ over $k_K$. For finite extensions of non-Archimedean local fields, the fundamental degree relation isThe residue degree is at least $d$, while the left side equals $d$, henceThis constructs the required unramified extension.

For uniqueness, let $L^{\prime }/K$ be another unramified extension with a chosen identification $k_{L^{\prime }}\cong \ell$. The element corresponding to $\bar{\alpha }$ is a simple residue root of $F$, because finite fields are perfect. The Hensel lemma in $L^{\prime }$ gives a root ${\alpha }^{\prime }\in {\mathcal{O}}_{L^{\prime }}$ with this reduction. Since $F$ is irreducible of degree $d$, $[K({\alpha }^{\prime }):K]=d=[L^{\prime }:K]$, so $L^{\prime }=K({\alpha }^{\prime })$. Sending $\alpha$ to ${\alpha }^{\prime }$ yields a $K$-isomorphism $L\to L^{\prime }$, compatible with the prescribed residue identification. ThusThis proves the existence and uniqueness of unramified local extensions using the stated version of Hensel's lemma, without assuming the classification in advance.

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 Schwartz-Bruhat space $\mathcal{S}(F)$ of a non-Archimedean local field $F$ is the vector space of locally constant, compactly supported complex-valued functions on $F$. Fix a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ and a Haar measure $dx$. With the sign convention required in the question, the Fourier transform over a local field is

Letbe the annihilator of the valuation ring. Translation invariance givesIndeed, the integral is the volume when the character is trivial; otherwise translation by an element on which the character is nontrivial multiplies the integral by a scalar different from one, forcing it to vanish. With the usual character of conductor ${\mathcal{O}}_F$ and the normalization $\mathrm{vol}({\mathcal{O}}_F)=1$, this becomesFor the canonical character induced from ${\mathbb{Q}}_p$, the annihilator is instead the inverse different and the displayed general formula applies.

If $g(x)=f(ax+b)$ with $a\ne 0$, the substitution $u=ax+b$ and the scaling rule $dx=|a{|}^{-1}du$ give

Every locally constant compactly supported function is a finite linear combination of characteristic functions of cosets $b+a{\mathcal{O}}_F$: compactness extracts finitely many cosets on which the function is constant. The formula just proved, together with the transform of $1_{{\mathcal{O}}_F}$, shows that the transform of each such characteristic function is again locally constant and compactly supported. Therefore the Fourier transform over a local field maps $\mathcal{S}(F)$ to itself.

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.