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A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A profinite group has either of the following equivalent descriptions: a compact Hausdorff totally disconnected topological group, or an inverse limit of finite discrete groups. Totally disconnected means that every connected component is a singleton.

For clarity, an inverse system over a directed set $I$ consists of groups $G_i$ and maps ${\varphi }_{ij}:G_j\to G_i$ for $i\le j$, with ${\varphi }_{ii}=1$ and ${\varphi }_{ik}={\varphi }_{ij}{\varphi }_{jk}$. Its inverse limit iswith coordinatewise multiplication and the subspace topology. Equivalently, it is a group with compatible continuous projections such that every other compatible family of maps into the $G_i$ factors uniquely through it.

The displayed limit is a closed subgroup of the compact Hausdorff product of the finite groups, and is totally disconnected, since two different tuples are separated by a clopen coordinate condition. Conversely a compact Hausdorff totally disconnected group has a neighbourhood basis of open normal subgroups. One way to see the group part of this standard clopen-basis fact is to choose a clopen identity neighbourhood $U$ inside a prescribed neighbourhood. Compactness of $U$ and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$. The stabilizer $H=\{g:gU=U\}$ contains $V$, is open, and is contained in $U$ because $1\in U$. Its index is finite by part (ii); its normal core is the intersection of finitely many conjugates and is open and normal. The canonical mapis injective because these $N$ separate points, and surjective because compatible cosets have the finite intersection property and $G$ is compact. It is a homeomorphism by compactness and the Hausdorff property. This explains the equivalence of the two definitions.

The profinite Frattini subgroup is $\Phi (G)=\bigcap M$, where $M$ ranges over all maximal proper open subgroups; for the trivial group the empty intersection is $G$. For a pro-p group,Here powers and group commutators mean generated subgroups and the bar supplies closure. Every maximal open subgroup in a pro-p group is normal of index $p$, as follows by passing to the finite p-group quotient by its normal core. Such a subgroup therefore contains all powers and group commutators. Conversely the quotient by $D=\overline{G^p[G,G]}$ is an abelian pro-p group of exponent $p$. If an element of this quotient is nontrivial, some finite elementary abelian quotient detects it; a linear functional to ${\mathbb{F}}_p$ then detects it as well. Its group homomorphism kernel pulls back to a maximal open subgroup of $G$. Thus the intersection of these kernels is exactly $D$.

We also use the topological generation criterion: a subset generates $G$ topologically if and only if its image generates $G/\Phi (G)$ topologically. To prove the nontrivial direction, a proper closed generated subgroup $K$ lies in a proper open subgroup $KN$ for a suitably small open normal $N$, and hence in a maximal open subgroup obtained from the finite quotient $G/N$. Since that maximal subgroup contains $\Phi (G)$, the image of $K$ cannot generate the Frattini quotient.

Now suppose $G$ is a nontrivial procyclic pro-p group. If it had distinct maximal open subgroups $M_1,M_2$, the homomorphism to $C_p\times C_p$ induced by the two quotient maps would be onto: its projections are onto, and a proper such subgroup would be a one-dimensional graph, forcing the two kernels to coincide. Hence $G/(M_1\cap M_2)\cong C_p^2$, contradicting procyclicity. Thus there is exactly one maximal open subgroup $M$, and $\Phi (G)=M$. Any $g\notin M$ generates $G/\Phi (G)\cong C_p$, so the criterion givesFor the trivial group use its identity as generator. The additive group ${\mathbb{Z}}_p$ is an infinite example, topologically generated by $1$: its finite continuous quotients are cyclic $\mathbb{Z}/p^n\mathbb{Z}$.

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.

Equivalently $\mathcal{N}$ is the group of continuous $k$-automorphisms of $k((t))$ whose action on $(t)/(t^2)$ is the identity. Such an automorphism is determined by the image of $t$, which is $t+O(t^2)$. To identify this with ordinary series composition without reversing multiplication, send $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$. Its composite with the automorphism for $g$ is the one for $f\circ g$.

The depth filtration isEach ${\mathcal{N}}_m$ is open and normal, and the coefficient of $t^{m+1}$ identifies ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}$ with $(k,+)$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$, and $\mathcal{N}$ is the inverse limit of its finite p-group truncations.

Here is a local-field proof that every finite p-group embeds in $\mathcal{N}$. We first establish the realization step rather than assume the Nottingham embedding itself. Put $F=k((t))$ and $\Gamma =\mathrm{Gal}(F^{\mathrm{sep}}/F)$, and let $I⊴\Gamma$ be the inertia subgroup, the group homomorphism kernel of the action on the algebraic closure of the residue field. We use the standard additive normal basis theorem consequence $H^j(\Gamma ,F^{\mathrm{sep}})=0$ for $j>0$. It follows by taking direct limits of the finite-Galois normal-basis modules, whose additive cohomology vanishes. The Artin-Schreier sequencethen givesThe classes $t^{-m}$ for positive $m$ prime to $p$ are linearly independent: a nonzero finite linear combination has a most negative valuation prime to $p$, whereas a negative valuation of $z^p-z$ is divisible by $p$. Thus there are infinitely many independent continuous characters $\Gamma \to C_p$.

Induct on the order of a finite p-group $E$. Choose a central subgroup $C\cong C_p$ and put $Q=E/C$. By induction there is an onto map $\varphi :\Gamma \to Q$ with $\varphi (I)=Q$, corresponding to a totally ramified $Q$-extension. The central extension $1\to C\to E\to Q\to 1$ is specified by a two-cocycle. Its pullback along $\varphi$ is a coboundary because $H^2(\Gamma ,{\mathbb{F}}_p)=0$. Correcting a set-theoretic section by that coboundary gives a continuous homomorphism ${\rho }_0:\Gamma \to E$ lifting $\varphi$.

We must make the lift onto on inertia, not merely onto on the whole Galois group. Let $A=I\cap \ker \varphi$. On $A$, the lift ${\rho }_0$ takes values in $C$. Characters vanishing on $A$ form a finite-dimensional space: $\Gamma /A$ has a finite subgroup $I/A\cong Q$ and procyclic quotient $\Gamma /I\cong \hat{\mathbb{Z}}$, so a character is determined by finitely many values on generators of these two pieces. The infinite-dimensional character space therefore has infinitely many distinct restrictions to $A$. Choose a character $\chi :\Gamma \to C$ whose restriction to $A$ does not cancel ${\rho }_0{|}_A$, and set $\rho (\gamma )=\chi (\gamma ){\rho }_0(\gamma )$. Centrality of $C$ makes this another homomorphism lifting $\varphi$. Its image on $A$ is nontrivial and hence all of $C$, while its image on $I$ maps onto $Q$. Therefore $\rho (I)=E$. The fixed field of $\ker \rho$ is the required totally ramified finite Galois extension $K/F$ with group $E$. This proves the realization step for every finite p-group.

Since the residue field of $K$ is still $k$, choosing a uniformizer $u$ identifies $K$ with $k((u))$. Indeed successive subtraction of its residue coefficient and division by $u$ gives each element of its valuation ring a unique convergent expansion ${\sum }_{i\ge 0}{b}_i{u}^i$ with $b_i\in k$, and fractions give the Laurent expansions. Each element of $E$ preserves the valuation and fixes $k$, so sends $u$ to $a_{1}u+a_2{u}^2+\cdots$. The linear coefficient is a homomorphism $E\to k^{\times }$. Since $E$ is a $p$-group and $|k^{\times }|=p-1$, that homomorphism is trivial. Thus the faithful Galois action lies in $\mathcal{N}(k)$, completing the embedding proof.

An infinite profinite group is hereditarily just infinite when every open subgroup is infinite and every nontrivial closed normal subgroup of each such open subgroup is open. We sketch the normal-subgroup argument, giving the depth calculation in odd characteristic explicitly. For series of depths $r,s$,where $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. The coefficient follows by comparing $f\circ g$ and $g\circ f$ in degree $r+s+1$: their cross terms are $ab(r+1)$ and $ab(s+1)$.

Let $H$ be open and $1\ne K⊴H$ closed. Choose $m$ with ${\mathcal{N}}_m\le H$ and choose $f\in K$ of depth $r$. For every $j\ge m$ with $j\not\equiv r(\mathrm{mod}p)$, commutation with $t+b{t}^{j+1}$ supplies an element of $K$ of depth $r+j$, with any prescribed nonzero leading coefficient by varying $b$. Thus every sufficiently high layer except possibly depths congruent to $2r$ modulo $p$ occurs in $K$.

When $p$ is odd, choose one available depth $u\ge r+m$ with $u\not\equiv 2r,r(\mathrm{mod}p)$. Such a residue exists even for $p=3$. An element $h\in K$ of that depth can be commuted with $t+b{t}^{k-u+1}$ for every sufficiently large missing depth $k\equiv 2r(\mathrm{mod}p)$. The new leading coefficient contains $2u-k\equiv 2(u-r)\ne 0$, so those missing layers occur too. Hence $K$ contains an element with any chosen coefficient in every depth $k\ge M$, for some $M$.

Successive coefficient correction now proves ${\mathcal{N}}_M\le K$: match the coefficient of an arbitrary target in depth $M$ by an element of $K$, correct the error in depth $M+1$, and continue. The partial products converge $t$-adically to the target and remain in $K$, which is closed. Thus $K$ is open. This proves hereditary just infiniteness for odd $p$.

For characteristic two, the leading coefficient alone misses even depths. The standard characteristic-two normal-closure collection lemma supplies the needed extra input: for every $m\ge 1$ and every nonidentity $f\in \mathcal{N}({\mathbb{F}}_2)$, the closed subgroup generated by the conjugates $f^g$ with $g\in {\mathcal{N}}_m$ contains ${\mathcal{N}}_M$ for some $M$. Its proof retains the next coefficients in pairs of group commutators to fill the parity gaps, then uses the same successive coefficient correction. Apply this lemma to $f\in K$ and ${\mathcal{N}}_m\le H$: every such conjugate lies in $K$, so $K$ contains the tail and is open. This states the additional collection result used in the characteristic-two sketch; the odd-prime leading-term argument is not being applied there. Since every open subgroup contains a tail and is infinite, the Nottingham group is hereditarily just infinite for every prime $p$.

We construct an infinite residually finite torsion group using the Golod-Shafarevich infinitude criterion for pro-p groups. Its useful form says that a pro-$p$ presentation on $d$ generators with relators $r_i$ gives an infinite pro-p group wheneverHere $D$ is the Magnus degree of a free pro-p group element: map each generator to $1+X_j$ in a noncommutative power-series algebra over ${\mathbb{F}}_p$, and take the lowest degree of $w-1$. The criterion permits infinitely many relators with a convergent weight sum. It is a consequence of the Golod-Shafarevich inequality for the completed group algebra; see Ershov's survey, Theorem 3.1.

Take $p=2$, $d=2$, and $\tau =3/4$. Let $F$ be the free pro-p group on $a,b$, and enumerate all ordinary finite words $w_1,w_2,\dots$ in these generators and their inverses. Choose integers $e_i$ sufficiently large thatForm the quotient pro-p groupwhere the bar means the closed normal closure. If the Magnus image of $w_i$ is $1+v_i$, characteristic two givesfor nonidentity $w_i$; identity relators contribute zero weight. It follows thatThus the Golod-Shafarevich infinitude criterion for pro-p groups proves that $P$ is infinite.

Let $G=⟨\bar{a},\bar{b}⟩$ be the abstract subgroup generated by the two images inside $P$. This subgroup is dense because the two images topologically generate the quotient, and it is infinite: a finite subgroup of a Hausdorff profinite group is closed, so could not be dense in infinite $P$. Every element of $G$ is the image of some ordinary word $w_i$, and the imposed relation makes its order divide $2^{e_i}$. Therefore $G$ is a two-generated torsion group. Finally $P$ is an inverse limit of finite 2-groups. Any nonidentity element of $G$ is nonidentity in $P$ and hence has a nonidentity image in one of those finite groups. Restricting that map to $G$ proves that it is a residually finite group. We have obtainedThe abstract subgroup $G$ is essential here: the argument establishes finite order for ordinary words, not for every topological limit in the ambient pro-p group.

There is no surjective homomorphism from this group to an infinite soluble group. In fact finitely generated soluble torsion groups are finite. To prove this, let $S$ be finitely generated, soluble and torsion, and induct on its derived length. In the abelian case, finitely many generators of finite orders generate a finite group. In general the abelianization $S/S^{\prime }$ is a finitely generated abelian torsion group, hence finite. Its kernel $S^{\prime }$ therefore has finite index and is finitely generated by Schreier's lemma. Explicitly, for a finite coset transversal $T$ containing $1$ and a finite symmetric generating set $X$, the finitely many elements $tx{\overline{tx}}^{-1}$ generate the finite-index subgroup, where the bar denotes the chosen representative of the coset. Now $S^{\prime }$ is torsion and has smaller derived length, so induction makes it finite. Hence $S$ is finite as well. Every quotient of $G$ is finitely generated and torsion, so any soluble quotient must be finite.

Put $K=\mathbb{C}((t))$. This is a complete discretely valued field, with valuation ring $\mathbb{C}[[t]]$ and residue field $\mathbb{C}$; its infinite residue field means it is not a locally compact local field. The finite-extension facts needed here are uniqueness of the extended valuation, completeness of the extension, and $[L:K]=ef$ for complete discretely valued fields with perfect residue field. They apply in this characteristic-zero setting.

Let $[L:K]=n$. The finite residue extension of $\mathbb{C}$ must be $\mathbb{C}$ itself, so $f=1$ and $e=n$. Normalize $v_L$ integrally and choose a uniformizer $\pi$. Since $v_L(t)=n$, write $t=u{\pi }^n$ with $u\in {\mathcal{O}}_L^{\times }$.

The residue of $u$ has an $n$th root $b_0\in {\mathbb{C}}^{\times }$. Apply Hensel lemma to $X^n-u$: its derivative at $b_0$ is the unit $n{b}_0^{n-1}$ because the residue characteristic is zero. Hence there is $v\in {\mathcal{O}}_L^{\times }$ with $v^n=u$. Setting $s=v\pi$ givesThe elements $1,s,\dots ,s^{n-1}$ are linearly independent over $K$. In a nontrivial relation ${\sum }_{i=0}^{n-1}{a}_i{s}^i=0$, the nonzero summands have valuations $n{v}_K(a_i)+i$, distinct modulo $n$, so exactly one has least valuation and cannot cancel. Since $L$ has dimension $n$, this proves $L=K(s)$.

The constants already embed $\mathbb{C}$ into ${\mathcal{O}}_L$ with residue field $\mathbb{C}$. Subtracting the residue coefficient and dividing by $s$ repeatedly, exactly as in Question 1(c), expresses every integral element uniquely as ${\sum }_{j\ge 0}{c}_j{s}^j$ with $c_j\in \mathbb{C}$. Completeness and the first-nonzero-coefficient valuation show that allowing finite negative powers givesThis proof of finite extensions of the complex Laurent series field did not need a Galois hypothesis. All $n$ distinct $n$th roots of unity lie in the constant field, and the substitutions $s\mapsto \zeta s$ preserve $s^n=t$. They give all $n$ automorphisms, so every such finite extension is cyclic Galois:

Choose compatible roots $s_n=t^{1/n}$ in an algebraic closure, first choosing them successively along the factorial-degree tower and then taking powers to define the other roots, with $s_n=s_m^{m/n}$ whenever $n\mid m$. The fields $K_n=K(s_n)$ are nested under divisibility. Every finite extension inside the closure equals one of them: its constructed root $s$ of $X^n-t$ differs from $s_n$ by a constant root of unity. Therefore their union is the entire algebraic closure. Explicitly, every algebraic element generates a finite extension, so lies in the union; any polynomial over the union has its coefficients in one finite extension, and its roots are still algebraic over $K$, hence also lie there.

For $n\mid m$, restriction of automorphisms sends ${\zeta }_m\in {\mu }_m$ to ${\zeta }_m^{m/n}\in {\mu }_n$. Using the compatible primitive roots $\exp (2\pi i/n)$, these are the reduction maps on exponent groups. Thus the absolute Galois group of the complex Laurent series field isThe inverse limit carries its profinite topology; choosing compatible roots gives the displayed identification with the profinite completion of the integers.

Fix a non-Archimedean local field $K$, write $Q=|k_K|$, and use the arithmetic Frobenius normalization throughout: a uniformizer maps to the automorphism inducing $a\mapsto a^Q$ on the maximal unramified extension. The Local Artin map, or local reciprocity map, is the canonical continuous group homomorphismwhere $K^{\mathrm{ab}}$ is the maximal abelian extension inside a fixed algebraic closure with its separable part understood. Its finite projection to every finite abelian extension $L/K$ is onto and has kernel exactly the norm subgroup of a local field extension:In particular, the quotient has order $[L:K]$. This is how local class field theory turns abelian extension problems into computations inside a multiplicative group.

One precise definition of the finite reciprocity map uses the cup-product construction of local reciprocity. For a finite Galois extension with group $G$, the local fundamental class $u_{L/K}\in H^2(G,L^{\times })$ is specified by local Brauer invariant $1/[L:K]$. Cup product gives an isomorphismThe finite Local Artin map is its inverse, composed with the quotient map from $K^{\times }$. For abelian $G$ the target is $G$ itself. Taking the compatible inverse limit over finite abelian extensions defines ${\mathrm{Art}}_K$.

The essential inputs to this cohomological construction have concrete arithmetic content. The local Brauer invariant identifies $\mathrm{Br}(K)$ with $\mathbb{Q}/\mathbb{Z}$; restriction to $L$ multiplies the invariant by $[L:K]$, so its kernel is cyclic of that order and has the distinguished generator $u_{L/K}$. For every subgroup $H\subseteq G$, Hilbert theorem 90 gives $H^1(H,L^{\times })=0$, while the same invariant calculation over the fixed field gives $H^2(H,L^{\times })$ cyclic of order $|H|$. These are the hypotheses of the algebraic cup-product isomorphism in Tate cohomology of a finite group. It is obtained by dimension shifting in group cohomology with a projective resolution of the trivial module and the extension representing the fundamental class. Thus the norm-kernel statement comes from the degree-zero definition of Tate cohomology of a finite group, rather than from an arbitrary assignment of an automorphism to each element. In an unramified extension of degree $d$, the fundamental class is the cyclic algebra with arithmetic Frobenius generator and parameter a uniformizer; the isomorphism sends that Frobenius to the class of the uniformizer. This explains the chosen sign of reciprocity.

Several properties follow immediately from this construction. If $L\supseteq E\supseteq K$ are finite abelian extensions, restricting ${\mathrm{Art}}_{L/K}(a)$ to $E$ gives ${\mathrm{Art}}_{E/K}(a)$. The Local Artin map is therefore compatible with the finite projections defining its infinite target. For any finite separable extension $E/K$ and $a\in E^{\times }$, its norm compatibility isHere $K^{\mathrm{ab}}E/E$ is abelian, so the displayed restriction makes sense. An outline of the compatibility proof is to use the corestriction map in group cohomology: on multiplicative coefficients its degree-zero map is the norm. The cup-product projection formula, together with restriction and corestriction of the local fundamental class, gives exactly the commuting reciprocity diagram. The compatibility is an arithmetic instance of the functoriality of the cohomological definition.

The Local Artin map has dense image, because every finite abelian quotient receives a surjective map. Its kernel is trivial, and it extends to an isomorphismwhere this profinite completion uses the open finite-index subgroups of the topological multiplicative group. The original map is generally not onto: on the unramified quotient it sends the integer valuation into $\hat{\mathbb{Z}}$, rather than producing every profinite integer. Its units map onto the abelian inertia group. For a finite abelian extension, the corresponding unit image is its inertia group, while the valuation accounts for its residue extension. Indeed $v_K(Nz)=f(L/K)v_L(z)$, so the valuation quotient has order $f$; the full norm quotient has order $ef$, leaving unit image of order $e$. Since unit images act trivially on the residue field, they give precisely the inertia group. More finely, the image of the higher principal-unit group $1+{\mathfrak{m}}_K^r$ in a finite abelian extension is its rth upper ramification group, for each integer $r\ge 1$. The use of upper ramification numbering is essential for compatibility with quotients.

The existence theorem of local class field theory states that every open finite-index subgroup $H\subseteq K^{\times }$ is $N_{L/K}(L^{\times })$ for a unique finite abelian extension $L/K$. Inclusions are reversed: larger extensions have smaller norm subgroups. In the established reciprocity isomorphism, the proof is simply to take the fixed field of the open subgroup corresponding to $H$. Conversely a finite extension gives an open kernel of a finite continuous projection. Uniqueness follows because that kernel specifies its fixed field. The substantive existence problem is therefore to prove that the reciprocity isomorphism realizes all open finite-index subgroups; it is not just this final fixed-field argument.

The Lubin–Tate description of local reciprocity provides an explicit route to that existence problem. Choose a uniformizer $\pi$ and a Lubin–Tate series $f(T)=\pi T+T^Q$, with its associated Lubin–Tate formal group $F$. Solving $f(F(X,Y))=F(f(X),f(Y))$ successively by total degree constructs the law with linear part $X+Y$; the analogous equations construct the endomorphisms $[a]$ with linear coefficient $a$. The congruence $f(T)\equiv T^Q(\mathrm{mod}\pi )$ supplies the divisibility needed at each step. These endomorphisms satisfy $[a][b]=[ab]$ and make the torsion an ${\mathcal{O}}_K$-module.

If $f_r$ denotes the rfold iterate, the primitive level-r Lubin–Tate torsion points are roots ofReduction modulo $\pi$ is $T^{Q^{r-1}(Q-1)}$, and the constant term is $\pi$, so this is an Eisenstein polynomial. Its roots therefore generate a totally ramified extension $K_{\pi ,r}$ of degree $(Q-1)Q^{r-1}$. A primitive point generates the entire torsion module, and all points are its images under the integral convergent endomorphisms $[a]$. The torsion field is therefore a Galois extension of $K$. The action map into $({\mathcal{O}}_K/{\pi }^r{)}^{\times }$ is injective, and the computed degree equals the order of this unit group, proving it is an isomorphism. Passing to the inverse limit gives $\mathrm{Gal}(K_{\pi }/K)\simeq {\mathcal{O}}_K^{\times }$ for the Lubin–Tate tower $K_{\pi }$.

The unramified tower and $K_{\pi }$ are linearly disjoint field extensions: a finite common subextension would be both unramified and totally ramified, hence trivial. The Lubin–Tate description of local reciprocity on their compositum isThe inverse on the unit action goes with the arithmetic Frobenius convention; changing to geometric Frobenius inverts the whole reciprocity map. The reciprocity theorem identifies this explicit action with the finite cup-product maps, independently of the chosen uniformizer. Its proof compares the fundamental class with the norm relations between consecutive torsion fields. The construction makes the possible finite kernels concrete: on the compositum of the degree-d unramified field with $K_{\pi ,r}$ the kernel isThese subgroups are cofinal among open finite-index subgroups. Indeed any such $H$ contains $1+{\pi }^r{\mathcal{O}}_K$ for sufficiently large $r$, and contains ${\pi }^d$ for some positive $d$, since the class of $\pi$ in $K^{\times }/H$ has finite order. Taking the appropriate fixed subgroup in this explicit finite compositum realizes $H$. This proves the open-subgroup part of existence once the norm compatibility of the explicit and cohomological constructions is established. Together with the finite reciprocity maps it also shows that $K^{\mathrm{ab}}=K^{\mathrm{nr}}{K}_{\pi }$.

As a direct check, on a finite unramified extension the construction reduces toThe earlier elementary unit-norm proof shows its kernel is exactly ${\pi }^{d\mathbb{Z}}{\mathcal{O}}_K^{\times }$. At $K={\mathbb{Q}}_p$, the Lubin–Tate cyclotomic example takes $[p](T)=(1+T{)}^p-1$ and $F(X,Y)=X+Y+XY$. Its torsion is $\zeta -1$, so a unit $u$ acts by $\zeta \mapsto {\zeta }^{u^{-1}}$. These examples exhibit both the valuation and unit parts of local reciprocity with consistent signs.

A topological group is a group equipped with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous maps. The product uses the product topology. Translation and inversion are therefore homeomorphisms. Hausdorffness is sometimes included in the definition; the two open-subgroup arguments below do not require it.

For the inverse system, take a directed set $I$, topological spaces $X_i$, and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=\mathrm{id}$ and $f_{ik}=f_{ij}{f}_{jk}$. An inverse limit is a space $X$ with compatible continuous maps ${\pi }_i:X\to X_i$ such that every compatible family $q_i:Y\to X_i$ factors uniquely as $q_i={\pi }_{i}q$ through a continuous map $q:Y\to X$. Existence follows from the concrete constructionwith the subspace topology of the product topology. The only possible map $q$ is $q(y)=(q_i(y){)}_i$. Compatibility puts its image in $X$, and coordinatewise continuity makes it continuous. This proves the universal property of an inverse limit, even when $X$ is empty. Given two inverse limits, their universal properties supply maps each way; the composites have the same coordinates as the identity, so uniqueness makes them identities. The inverse limit exists and is unique up to the unique homeomorphism respecting its projections.

A profinite group can be defined as an inverse limit of finite discrete groups, or as a compact Hausdorff space that is a totally disconnected space, equipped with continuous group operations. For the first implication, use the Tychonoff theorem. The compatibility equalities define a closed subgroup of the compact Hausdorff product. Two distinct points are distinguished by a finite discrete coordinate, so no connected subset contains both. Consequently the inverse limit is a compact Hausdorff totally disconnected topological group.

For the converse, use the topological result that a compact Hausdorff totally disconnected space has a basis of clopen sets. Choose a clopen set $U$ containing $1$ inside any specified identity neighbourhood. For each $u\in U$, continuity gives an identity neighbourhood $V_u$ and an open neighbourhood $W_u$ of $u$ with $V_u{W}_u\subseteq U$. Finitely many $W_u$ cover the compact set $U$. Intersect their $V_u$ and its inverse to obtain a symmetric identity neighbourhood $V$ with $VU\subseteq U$. Symmetry also gives $V^{-1}U\subseteq U$, and hence $vU=U$ for every $v\in V$. The setis a subgroup containing $V$, so it is open. It is contained in $U$, since $1\in U$ and $h=h\cdot 1\in hU=U$. By part (ii), $H$ has finite index. Its normal core of a subgroup is a finite intersection of conjugates of $H$, hence an open normal subgroup $N$ contained in $U$. This proves the open normal subgroup basis of a profinite group.

The natural group homomorphismis injective because the intersection of these $N$ is $\{1\}$. For a compatible family of cosets, any finitely many have nonempty intersection: their intersection subgroup is another indexing subgroup, and its prescribed coset lies in each. The finite intersection property and compactness give a point belonging to all the cosets, proving surjectivity. Finally a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The two definitions of profinite group are equivalent.

A topological group module $A$ is a topological abelian group with a jointly continuous action of $G$ by group automorphisms. Its continuous cochains with topological coefficients arewith pointwise addition. The inhomogeneous differential isFor $n=0$, this means $(da)(g)=ga-a$. Continuity of multiplication and of the action ensures that $d$ preserves continuous cochains. Expanding $d(dc)$ gives cancellation in pairs: the two orders of multiplying neighbouring arguments agree by associativity, and the two ways of applying the action agree by its action law. Thus $d^2=0$. The cocycles are $Z^n=\ker (d:C^n\to C^{n+1})$ and the coboundaries are $B^n=\mathrm{im}(d:C^{n-1}\to C^n)$, with $B^0=0$. DefineFor general topological coefficients this is an algebraic quotient of the continuous cochains; its optional quotient topology need not be Hausdorff. For discrete coefficients it is the usual continuous cohomology of a profinite group. In degree zero it is the subgroup of invariant elements.

Give the discrete prime field ${\mathbb{F}}_p$ its trivial $G$-action. The one-cocycle condition becomes $c(gh)=c(g)+c(h)$ and every one-coboundary is zero. HenceEvery such group homomorphism kills powers and group commutators, hence their closure $\Phi (G)$, and every continuous functional on the quotient pulls back. If $G$ is topologically finitely generated, $V=G/\Phi (G)$ is a finite elementary abelian group. To verify finiteness without assuming it, the closure of a subgroup generated by $r$ elements of this abelian exponent-$p$ quotient has at most $p^r$ elements; a finite subgroup is already closed. Thus $V$ is a finite-dimensional vector space. Question 2 says that a generating family of $G$ is minimal precisely when its images give a basis of $V$. The dual space has the same finite dimension. This proves

There is a genuine missing hypothesis if the printed assertion is read for arbitrary pro-p groups and $d(G)$ means the least cardinality of any dense generating family. Here is an explicit infinite-rank failure of the first-cohomology generator formula. Let $I=\{0,1{\}}^{\mathbb{N}}$ and $G={\prod }_{r\in I}{\mathbb{F}}_p$. Enumerate the finite binary strings as $s_1,s_2,\dots$ and define $g_n(r)=1$ if $s_n$ is a prefix of $r$, and $0$ otherwise. For finitely many distinct sequences $r_1,\dots ,r_k$, choose long enough prefixes to distinguish each individually. The corresponding $g_n$ project onto the coordinate unit vectors of ${\mathbb{F}}_p^k$, so $⟨g_n:n\ge 1⟩$ surjects onto every finite coordinate projection and is dense in the product topology. No finite family can generate densely, since a finitely generated abelian exponent-$p$ subgroup is finite and closed. Thus $d(G)={\aleph }_0$. On the other hand, continuity of a linear functional to ${\mathbb{F}}_p$ means its kernel contains a basic neighbourhood, so it depends on finitely many coordinates. The coordinate projections are consequently a basis of the continuous dual space, givingThe unrestricted infinite-rank formula is false. A convention requiring a generating basis tending to the identity is different from the printed unrestricted topological generator count; the finite-rank identity proved above has no such ambiguity.

For finite $X$, a free pro-p group $F_p(X)$ is characterized by the universal property that every map $X\to P$ to a pro-p group extends uniquely to a continuous group homomorphism $F_p(X)\to P$. It is the pro-p completion of the abstract free group on $X$: each map to a finite $p$-group factors through one of the defining finite quotients, and passing to the inverse limit gives the extension to any pro-p group; uniqueness follows from density. This unrestricted universal property can also define the completion on an arbitrary set $X$. When using an infinite basis required to tend to the identity, one instead imposes the universal property only for maps tending to the identity. Only the finite-rank version is needed for the following argument.

A finite pro-p presentation is an isomorphismThere are finitely many generators and finitely many relators, and the denominator is their closed normal closure. The relators are elements of the free pro-p group; they may be limits of ordinary words. That distinction is important when writing conjugation relations below.

Now suppose $K{⊴}_{o}G$ has a finite pro-p presentation with generators $k_1,\dots ,k_a$ and relators $R_1,\dots ,R_t$. The quotient $Q=G/K$ is a finite p-group. Choose a finite finite pro-p presentation $Q=⟨{\bar{y}}_1,\dots ,{\bar{y}}_b\mid S_1,\dots ,S_c⟩$. Such a presentation exists, for example by using its entire finite multiplication table. Lift the quotient generators to $y_j\in G$. For every $i,j$, choose elements $W_{ij},V_{ij}$ of the free pro-p group on the $x_i$ that map respectively to $y_j^{-1}{k}_i{y}_j$ and $y_j{k}_i{y}_j^{-1}$ in $K$. Choose $U_{\ell }(x)$ mapping to $S_{\ell }(y)$, which lies in $K$. Form the finite finite pro-p presentationAll listed relations hold in $G$, giving a continuous surjective group homomorphism $P\to G$: its compact image contains $K$ and maps onto $Q$.

Let $X$ be the closed subgroup of $P$ topologically generated by the $x_i$. Both conjugation directions were included, so each $y_j$ normalizes $X$. The $x_i$ also normalize it, and its closed normalizer contains the generators of $P$. Thus $X⊴P$. The relations $R_h$ give a continuous surjection $K\to X$. Its composite with $X\to K$ induced by $P\to G$ fixes every $k_i$, and hence is the identity. Therefore the surjection $K\to X$ is injective as well, so $X\cong K$. On quotienting by $X$, the conjugation and kernel relations disappear, and the remaining presentation is exactly $⟨y_1,\dots ,y_b\mid S_1,\dots ,S_c{⟩}_{\mathrm{pro}\text{-}p}=Q$. Thus $P/X\to G/K$ is an isomorphism. The kernel of $P\to G$ lies in $X$ and meets $X$ trivially, so is trivial. Consequently $P\cong G$, proving

A local field is nondiscrete and locally compact. In the non-Archimedean case its valuation ring is compact by the scaling argument in part (b). Any Cauchy sequence has a tail in a translate of a compact ball, so has a convergent subsequence; the Cauchy property forces the whole sequence to converge. Thus the field is complete. Part (b) gives a discrete valuation, a uniformizer $\pi$, and a finite residue field $k={\mathbb{F}}_q$, $q=p^f$.

We construct the coefficient field rather than presuming it exists. The polynomial $T^q-T$ has derivative $-1$ in characteristic $p$. Every residue element therefore has a unique polynomial root lift by Hensel's lemma. The set $S$ of these $q$ lifts is a field: qth powers preserve addition and multiplication, so sums and products remain polynomial root lifts; a nonzero lift has inverse $a^{q-2}$. Reduction identifies $S$ with ${\mathbb{F}}_q$. This is the finite coefficient field in positive-characteristic local fields.

For $x\in {\mathcal{O}}_K$, choose $a_0\in S$ with the same residue, write $x=a_0+\pi x_1$, and repeat with $x_1\in {\mathcal{O}}_K$. Completeness givesThe first nonzero coefficient determines the valuation, so the expansion is unique. Multiplying an arbitrary field element by a suitable power of $\pi$ reduces it to this case. Consequently substitution $T\mapsto \pi$ gives a valued-field isomorphismAddition and multiplication agree with those of Laurent series by convergence, and a nonzero series has a nonzero leading coefficient, establishing injectivity as well as surjectivity.

Conversely, ${\mathbb{F}}_q((T))$ is complete for its T-adic absolute value on a field. Its valuation ring ${\mathbb{F}}_q[[T]]$ is compact: fixing finitely many coefficients gives the finite partitions into balls, and the successive coefficient choices realize the inverse limit of the finite rings ${\mathbb{F}}_q[T]/(T^n)$. Equivalently it is the product of countably many finite discrete coefficient sets. Thus it is a non-Archimedean local field. Distinct $q$ give nonisomorphic valued fields because their residue fields have different sizes. Replacing $|T|$ by any number in $(0,1)$ gives an equivalent absolute value on a field.

Put ${\pi }_n={\zeta }_{p^n}-1$. The shifted cyclotomic polynomialis Eisenstein at $p$. Therefore it is irreducible, ${\pi }_n$ is a uniformizer, andso the extension is totally ramified. It is the splitting field of ${\Phi }_{p^n}$, and every automorphism is uniquelyThis proves the cyclotomic extension of a p-adic field isomorphismRestriction in the cyclotomic tower corresponds to reduction of $a$, so taking the inverse limit gives

For an inverse system of sets $(X_j,f_{ij})$ indexed by a directed set $J$, the inverse limit is the set of compatible tuplesIts projection to $X_j$ sends $(x_i{)}_i$ to $x_j$.