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Use arithmetic Frobenius automorphisms: residues are raised to the size of the base residue field. Let $S$ contain the finite primes ramified in the finite abelian extension $L/K$. The Artin map is the group homomorphismwhere $I_K^S$ is the group of fractional ideals supported outside $S$. The Artin symbol ${\mathrm{Frob}}_{\mathfrak{p}}$ is independent of the prime above $\mathfrak{p}$, since the extension is abelian. Commutativity makes the displayed extension from prime ideals well defined. Defining this map does not yet assert the Artin reciprocity law, which describes its congruence group kernel.

For an intermediate field $K\subseteq E\subseteq L$, restriction givesTo prove it, start with an unramified prime $\mathfrak{p}$ and a prime of $L$ above it. Its Frobenius automorphism acts on the residue of every integral element of $E$ by raising it to $N\mathfrak{p}$. Its restriction is therefore the Frobenius automorphism for $E/K$ at the restricted prime. The identity for all allowed fractional ideals follows by the group homomorphism property. On Galois groups, restriction is the quotient by $\mathrm{Gal}(L/E)$, so passing to a subextension quotients the Artin map accordingly.

Now let $E/K$ be any finite extension and put $M=LE$. This is the translated extension: $M/E$ is abelian, and restriction identifies $\mathrm{Gal}(M/E)$ with $\mathrm{Gal}(L/(L\cap E))$. A prime $\mathfrak{q}$ of $E$ above a prime $\mathfrak{p}\notin S$ is unramified in $M/E$, because forming a compositum with $E$ preserves an unramified extension. Put $r=f(\mathfrak{q}/\mathfrak{p})$. The residue-field sizes satisfy $N\mathfrak{q}=(N\mathfrak{p}{)}^r$. Consequently the restriction to $L$ of Frobenius at $\mathfrak{q}$ acts on its integral residues by raising them to $(N\mathfrak{p}{)}^r$, givingBut the ideal norm is $N_{E/K}\mathfrak{q}={\mathfrak{p}}^r$. Multiplying this prime-by-prime identity provesThis is restriction and norm compatibility of the Artin map. The fractional ideal $\mathfrak{a}$ must be supported over primes unramified in $L/K$. No assumption that $E/K$ itself is unramified or Galois is needed. With geometric rather than arithmetic Frobenius, both maps are inverted and the compatibility identities remain unchanged.

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

For $a,b\in {\mathbb{Q}}_p^{\times }$, the quadratic Hilbert symbol $(a,b{)}_p$ is one if $b$ is a field norm from ${\mathbb{Q}}_p(\sqrt{a})$, and minus one otherwise; a square $a$ gives the trivial extension and symbol one. It depends only on the two square classes. Equivalently, the conic $z^2=a{x}^2+b{y}^2$ has a nonzero rational point precisely when the symbol is one. Interchanging $a$ and $b$ in the conic gives $(a,b{)}_p=(b,a{)}_p$. The Hilbert norm residue symbol is a group homomorphism in each argument; for nonsquare $a$, the cyclic local norm index makes its group kernel a subgroup of index two. The resulting pairing on square classes is nondegenerate. Alsosince $N(\sqrt{a})=-a$. In the language of local class field theory, $(a,b{)}_p={\mathrm{Art}}_p(b)(\sqrt{a})/\sqrt{a}$; in the Brauer group, it records whether the quaternion cyclic algebra $(a,b)$ splits. These are compatible norm-obstruction descriptions.

An unramified extension of local fields of degree two has norm group consisting precisely of elements with even valuation. Indeed,and the norms on units are surjective. For the latter assertion, the norm on the finite residue fields is surjective. On successive higher principal-unit groups, the congruenceallows one to correct a proposed unit norm one digit at a time: the residue field trace is surjective, and completeness makes the corrections converge. An unramified quadratic field is unique up to isomorphism. Therefore, if $d$ is its defining nonsquare class,

For odd $p$, choose the nonsquare unit $u_0$ from part (i). The polynomial $T^2-u_0$ has irreducible separable reduction, so it defines the unramified quadratic extension. ThusThe trivial class is included in ${\Delta }_2$, since a degree-one extension is unramified. The two units pair trivially, while $(u_0,p{)}_p=-1$. Any class outside ${\Delta }_1$ has odd valuation and consequently pairs nontrivially with $u_0$. This proves directly

For $p=2$, the unit square classes from part (i) give ${\Delta }_1=\{1,-1,5,-5\}$. The element $\omega =(1+\sqrt{5})/2$ satisfies $T^2-T-1$; its reduction $T^2+T+1$ is irreducible and separable over ${\mathbb{F}}_2$. This shows that ${\mathbb{Q}}_2(\sqrt{5})$ is unramified of degree two, givingFor a fully explicit pairing check, use the generators $(-1,5,2)$. We have $(-1,5{)}_2=1$ and $(-1,2{)}_2=1$, since $5=N(1+2i)$ and $2=N(1+i)$ from ${\mathbb{Q}}_2(i)$. But $(-1,-1{)}_2=-1$: a norm equal to $-1$ would be a sum of two squares. If either summand had negative valuation, the sum would have negative valuation (equal negative valuations give valuation $1-2r<0$); if both were integral, their squares could not sum to $-1\equiv 3(\mathrm{mod}4)$. Finally, the unramified norm criterion gives $(5,2{)}_2=-1$. The identity $(a,b{)}_2=(b,a{)}_2$ and $(a,a{)}_2=(a,-1{)}_2$ determine the remaining entries. Writing $(a,b{)}_2=(-1{)}^{B(a,b)}$, the exponent matrix isFor $b=(-1{)}^s{5}^t{2}^r$, annihilating both generators $-1,5$ of ${\Delta }_1$ requires $s=r=0$, so $b\in \{1,5\}$. Annihilating $5$ alone requires $r=0$, so $b$ is a unit class. HenceThese are the Hilbert-symbol annihilators of units and unramified square classes. Annihilators here are taken with respect to this pairing, or equivalently in the character group of a finite abelian group after identifying square classes with their Hilbert-symbol characters.

We first prove the simple-root Hensel lemma directly. Put $x_1=x$. Suppose $x_n\equiv x(\mathrm{mod}\pi )$ and $f(x_n)\equiv 0(\mathrm{mod}{\pi }^n)$. The derivative remains a unit since its reduction agrees with $f^{\prime }(x)$. Choose $t_n\in {\mathcal{O}}_K$ whose residue solvesand set $x_{n+1}=x_n+{\pi }^n{t}_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+{\pi }^n{t}_n{f}^{\prime }(x_n)+{\pi }^{2n}B$ with $B\in {\mathcal{O}}_K$. Since $2n\ge n+1$, the new value vanishes modulo ${\pi }^{n+1}$. The sequence is Cauchy because $v(x_{n+1}-x_n)\ge n$, and completeness supplies $y\in {\mathcal{O}}_K$ with $y\equiv x(\mathrm{mod}\pi )$. Continuity gives $f(y)=0$.

For uniqueness, if $y,z$ are two roots in that residue class, thenThe factor in parentheses is a unit, since $f^{\prime }(z)$ is a unit and $y-z\in \pi {\mathcal{O}}_K$. Thus $y=z$. This proves existence and uniqueness of the specified lifted root.

Now suppose the residue field has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^{\times }$ is a simple root of $T^{q-1}-1$. The Hensel lemma therefore lifts its $q-1$ roots uniquely to a subgroup of the roots of unity in $K$. These are the Teichmuller representatives and give all the prime-to-$p$ torsion. To verify the latter assertion, if a root of unity $u\equiv 1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\left(\genfrac{}{}{0ex}{}{m}{2}\right)t^2+\cdots$, the first term has valuation $v(t)$ and all later terms have greater valuation, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.

Any root of unity has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge 1$, reduction in characteristic $p$ gives $\bar{\zeta }=1$. The cyclotomic identityshows that all factors have the same valuation: $({\zeta }^j-1)/(\zeta -1)=1+\zeta +\cdots +{\zeta }^{j-1}$ reduces to $j\ne 0$. Consequently the valuation of a primitive p-power root of unity isThe left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. ThusIf $e<p-1$, even a primitive $p$th root would violate the valuation formula, so $s=0$.

At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_{+}={\mathbb{Q}}_p({\zeta }_p)$. The polynomial ${\Phi }_p(1+T)=p+\left(\genfrac{}{}{0ex}{}{p}{2}\right)T+\cdots +T^{p-1}$ is Eisenstein, so $K_{+}/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$ and $v_{K_{+}}(p)=p-1$. It contains a primitive $p$th root, while the valuation formula excludes primitive $p^2$th roots. Thus $s=1$.

For the other example, take $K_0={\mathbb{Q}}_p(\pi )$ with ${\pi }^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and residue field ${\mathbb{F}}_p$. If a primitive $p$th root belonged to $K_0$, the valuation formula would give ${\zeta }_p-1=u\pi$ with $u$ a unit. Substitute this into ${\Phi }_p(1+({\zeta }_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+{\bar{u}}^{p-1}=0$ in ${\mathbb{F}}_p$. But ${\bar{u}}^{p-1}=1$, contradicting $2\ne 0$. Thus $s=0$ in $K_0$. This proves both cases of roots of unity at the tame ramification boundary for odd primes.

The printed boundary claim needs the qualification $p>2$. When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the valuation formula excludes order four and higher, forcing $s=1$. For example $K={\mathbb{Q}}_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.

Here $F$ is the usual one-dimensional commutative formal group law. Normalize the valuation $v$ by $v(\pi )=1$, and put $e=v(p)$. We show explicitly that for any integer $m>e/(p-1)$,This subgroup will have finite index in $F(\pi {\mathcal{O}}_K)$.

Set $a(T)={\partial }_{Y}F(T,0)$. Since $F(X,Y)=X+Y+$ terms of total degree at least two, $a(T)\in 1+T{\mathcal{O}}_K[[T]]$. Its inverse has integral coefficients, say $a(T{)}^{-1}={\sum }_{j\ge 0}{c}_j{T}^j$ with $c_0=1$. Define the formal logarithm over $K$ byDifferentiate associativity $F(F(X,Y),Z)=F(X,F(Y,Z))$ with respect to $Z$ at zero. This gives $a(F(X,Y))={\partial }_{Y}F(X,Y)a(Y)$, soEvaluating at $Y=0$ proves the formal identity $\ell (F(X,Y))=\ell (X)+\ell (Y)$.

The coefficients satisfy $v(c_{n-1}/n)\ge -e{v}_p(n)$. Thus the series converges on $\pi {\mathcal{O}}_K$, since for $v(T)\ge 1$ the valuations of its terms are at least $n-e{v}_p(n)\to \infty$. More quantitatively,Indeed, if $k=v_p(n)$ then $n\ge p^k\ge 1+k(p-1)$. If $v(T)\ge m>e/(p-1)$, every nonlinear term has valuation strictly greater than $v(T)$. Write $h(T)=\ell (T)-T$. For $S,T\in {\pi }^m{\mathcal{O}}_K$, use $T^n-S^n=(T-S){\sum }_{j=0}^{n-1}{T}^{n-1-j}{S}^j$ to obtainThe extra valuation is integral and positive because $(n-1)m-e{v}_p(n)>0$ for all $n\ge 2$.

For any $b\in {\pi }^m{\mathcal{O}}_K$, solve $\ell (T)=b$ by iterating $T\mapsto b-h(T)$. This map preserves the complete ball ${\pi }^m{\mathcal{O}}_K$ and improves differences by at least one valuation step. The contraction mapping theorem gives a unique solution. Therefore $\ell$ is a bijection of this ball onto itself, and the formal identity, now evaluated in convergent series, makes it a group isomorphism. Multiplication by ${\pi }^{-m}$ then identifies the additive target with ${\mathcal{O}}_K$.

Integral coefficients in $F$ and its formal inverse ensure that every $F({\pi }^j{\mathcal{O}}_K)$ is a subgroup. Reduction modulo ${\pi }^m$ respects the group law; two parameters represent the same coset of $F({\pi }^m{\mathcal{O}}_K)$ exactly when they are congruent modulo ${\pi }^m$. If the residue field has $q_K$ elements, the quotient consequently has $q_K^{m-1}$ elements. This proves both the finite-index assertion and the claimed additive structure, with an explicit depth that also works in ramified extensions and when $p=2$.

The Weak Mordell-Weil theorem asserts that $E(K)/mE(K)$ is finite for an elliptic curve over a number field $K$ and an integer $m\ge 2$. The role of (i) is to make division by $m$ possible in the local formal group of an elliptic curve whenever the residue characteristic does not divide $m$.

Let $v$ be a finite place with good reduction of an elliptic curve and $v\nmid m$. The kernel of reduction of an elliptic curve is parametrized by $t=-x/y$ in the maximal ideal of the local valuation ring. Multiplication by $m$ is represented by an integral formal power seriesIts linear coefficient is a local unit. Part (i) supplies an integral compositional inverse. Both series converge on the maximal ideal, so multiplication by $m$ is a bijection on the reduction kernel, over the local field and over every finite unramified extension.

This implies unramified division torsors at good primes. For $P\in E(K)$, first divide its reduction by $m$ over the algebraic closure of the residue field. Multiplication by $m$ on the reduced curve is surjective and separable, so the chosen division point is defined over some finite residue extension. Smooth lifting gives a point $Q_0$ over the corresponding unramified local extension with the required reduction. The difference $P-m{Q}_0$ lies in the reduction kernel; the inverse series supplies $Q_1$ with $m{Q}_1=P-m{Q}_0$. Thus $Q=Q_0+Q_1$ divides $P$ over an unramified extension. Applying the same reasoning to all reduced $m$-torsion points lifts all of $E[m]$, so every point of $[m{]}^{-1}P$ lies in the maximal unramified extension. This excludes ramification outside a fixed finite set $S$ consisting of primes of bad reduction and primes over $m$.

For completeness, the remaining global finiteness step in the Kummer-theoretic proof of the weak Mordell-Weil theorem is as follows. Choose $Q\in E(\overline{K})$ with $mQ=P$. The cocycle $\sigma \mapsto \sigma Q-Q$ defines the Kummer map of an elliptic curveChanging $Q$ by an $m$-torsion point changes the cocycle by a coboundary. A zero class makes $Q$ rational after that change, so the map is injective.

Let $L=K(E[m])$, fixed independently of $P$. Since the torsion is rational over $L$, the cocycle restricted to its Galois group is a homomorphism into $E[m]$. Consequently $L(Q)/L$ is Galois of degree at most $m^2$, contains the entire division fibre, and is unramified outside $S$. Hence $[L(Q):K]\le m^2[L:K]$. There are only finitely many extensions of bounded degree unramified outside a fixed finite set: local extensions of bounded degree have bounded field discriminant exponents, and the Hermite–Minkowski theorem then applies to the bounded global field discriminants. Their finite Galois compositum $M$ contains every such division fibre; each $L(Q)$ is stable over $K$ because it contains $E[m]$ and the whole fibre above the rational point $P$. All the Kummer classes consequently factor through the fixed finite group $\mathrm{Gal}(M/K)$ with values in the fixed finite group $E[m]$. There are only finitely many such cocycles, provingThus (i) provides the local division step that bounds ramification; the global arithmetic finiteness completes the argument.

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 $p$.

Put $R={\mathcal{O}}_{X,x}$, $\mathfrak{m}={\mathfrak{m}}_x$ and $d=\dim R$. Smoothness makes $R$ a regular local ring, with residue field $k$. A list of local parameters $t_1,\dots ,t_d$ is a regular system of parameters: its images form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$. Set $J=(t_1,\dots ,t_d)$. The basis condition says $\mathfrak{m}=J+{\mathfrak{m}}^2$, so the finitely generated module $\mathfrak{m}/J$ equals $\mathfrak{m}(\mathfrak{m}/J)$. The Nakayama lemma gives

We now construct the requested formal power series, rather than assert that a local function has a convergent expansion. The associated graded ring of a regular local ring isIndeed these initial forms generate the graded ring, since the $t_i$ generate $\mathfrak{m}$. The Dimension theorem for Noetherian local rings gives $\dim {\mathrm{gr}}_{\mathfrak{m}}R=d$. A nonzero kernel in the surjection from the polynomial ring would make its quotient have dimension at most $d-1$, so that kernel is zero.

For $f\in R$, choose $P_0\in k$ to be its residue at $x$. Inductively suppose homogeneous polynomials $P_0,\dots ,P_{n-1}$ have been chosen such thatThe graded-ring isomorphism gives a unique homogeneous polynomial $P_n(T_1,\dots ,T_d)$ of degree $n$ representing the class of $r_n$ in ${\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}$. Subtracting $P_n(t)$ makes the next remainder lie in ${\mathfrak{m}}^{n+1}$. HenceThis equality is interpreted in the adic completion of a module $\hat{R}=\underleftarrow{\lim }R/{\mathfrak{m}}^n$. The same induction applies to every compatible sequence of residues in this inverse limit, so evaluation $T_i\mapsto t_i$ defines an isomorphism $k[[T_1,\dots ,T_d]]\cong \hat{R}$. It is injective because the lowest nonzero homogeneous term of a series has nonzero initial form. The Krull intersection theorem gives ${\bigcap }_n{\mathfrak{m}}^n=0$, so $R$ embeds in its completion and the series determines $f$. This is the formal expansion at a smooth point; it depends on the chosen regular system of parameters, and does not claim that the infinite sum converges inside $R$. If $d=0$, the local ring is just $k$, and only the constant term is needed.

The structural facts used about a non-Archimedean local field are that it is complete for a normalized discrete valuation $v:K^{\times }\to \mathbb{Z}$, its valuation ring ${\mathcal{O}}_K=\{x:v(x)\ge 0\}$ is a discrete valuation ring, its maximal ideal is ${\mathfrak{m}}_K=\pi {\mathcal{O}}_K$ for a uniformizer $\pi$, and its residue field is a finite field $k={\mathbb{F}}_q$. Polynomial operations are continuous for this valuation topology.

A simple-root version of Hensel lemma is: if $f\in {\mathcal{O}}_K[X]$, $a_1\in {\mathcal{O}}_K$, $f(a_1)\in {\mathfrak{m}}_K$, and $f^{\prime }(a_1)\notin {\mathfrak{m}}_K$, then there is a unique $a\in {\mathcal{O}}_K$ with $f(a)=0$ and $a\equiv a_1(\mathrm{mod}{\mathfrak{m}}_K)$.

Construct $a_n$ with $f(a_n)\in {\pi }^n{\mathcal{O}}_K$. Given $a_n$, put $a_{n+1}=a_n+{\pi }^n{b}_n$. Taylor expansion over the valuation ring givesThe omitted terms contain ${\pi }^{2n}$, and $2n\ge n+1$ for $n\ge 1$. The residue of $f^{\prime }(a_n)$ remains the nonzero residue of $f^{\prime }(a_1)$, so choose $b_n$ withThis makes $f(a_{n+1})\in {\pi }^{n+1}{\mathcal{O}}_K$. The congruences $a_{n+1}\equiv a_n(\mathrm{mod}{\pi }^n)$ make $(a_n)$ a Cauchy sequence; completeness gives its limit $a\in {\mathcal{O}}_K$. Continuity and $v(f(a_n))\ge n$ give $f(a)=0$, and the first congruence gives its specified residue class.

For uniqueness, if $a,b$ are two roots in that class, polynomial division of their difference givesThe second factor is a unit, since $f^{\prime }(b)$ is a unit and $a-b\in {\mathfrak{m}}_K$. Thus $a=b$. This proves both existence and uniqueness in the stated Hensel lemma.

Write the finite residue field as $k={\mathbb{F}}_q$, with $q=p^f$ and residue characteristic $p$. Every $\bar{a}\in k$ satisfies ${\bar{a}}^q=\bar{a}$. The polynomial $F(X)=X^q-X$ has derivative $q{X}^{q-1}-1$, whose reduction is $-1$. Hence Hensel lemma gives a unique lift $[\bar{a}]\in {\mathcal{O}}_K$ such thatThis includes $[0]=0$ and $[1]=1$. If $\bar{a}\ne 0$, its lift is a unit and satisfies $[\bar{a}{]}^{q-1}=1$.

The product $[\bar{a}][\bar{b}]$ also satisfies $X^q=X$ and has residue $\bar{a}\bar{b}$. Uniqueness therefore gives $[\bar{a}\bar{b}]=[\bar{a}][\bar{b}]$. Reduction is a left inverse of this map, so the map is injective. Restricting to nonzero residues proves the requested multiplicative Teichmuller lift:Its image is the group of $(q-1)$st roots of unity in $K$. Indeed a root of unity of that order is a unit satisfying $X^q=X$, hence is the unique lift of its residue. The only finite-field fact used here is $a^q=a$; for $a\ne 0$ it follows from Lagrange's theorem in the multiplicative group of order $q-1$.

Choose algebra generators $b_1,\dots ,b_n$ for $B$ over $A$. Because the extension is integral, each $b_i$ satisfies a monic equationRepeatedly reduce any exponent at least $e_i$ using this equation. Since the generators commute, every polynomial in them becomes an $A$-linear combination of the finitely many monomialsThus $B$ is a finitely generated module over $A$, with at most $D={\prod }_i{e}_i$ generators. This proves the relevant subalgebra generated by finitely many integral elements assertion rather than merely invoking it.

For a fixed prime ideal $\mathfrak{p}\subset A$, form the fiber ring $C=B{\otimes }_A\kappa (\mathfrak{p})$. Tensoring the finite generating list shows ${\dim }_{\kappa (\mathfrak{p})}C\le D$: its pure tensors with one span the fiber over the residue field $\kappa (\mathfrak{p})$. Part (b) therefore gives only finitely many primes in $C$. By the canonical bijection in part (a), these are precisely the primes of $B$ contracting to $\mathfrak{p}$. HenceNo Noetherian hypothesis on $A$ is needed. For an empty algebra-generating list, $B=A$ and the empty product $D=1$ gives the same conclusion.

Use the simple-root form of Hensel lemma. Let $R$ be a complete discrete valuation ring, let $\pi$ be a uniformizer, and let $F\in R[T]$. If $\bar{a}\in R/(\pi )$ is a root of $\bar{F}$ with ${\bar{F}}^{\prime }(\bar{a})\ne 0$, then there is exactly one root $a\in R$ reducing to $\bar{a}$.

Choose a lift $a_1$ of $\bar{a}$. Suppose $F(a_r)\in {\pi }^{r}R$. To construct a lift satisfying the next congruence, write $a_{r+1}=a_r+{\pi }^r{t}_r$. The polynomial expansion givesSince $F^{\prime }(a_r)$ is a unit, its residue is invertible in the residue field. There is a unique residue class of $t_r$ making this zero modulo ${\pi }^{r+1}$. Thus $a_{r+1}\equiv a_r(\mathrm{mod}{\pi }^r)$ and $F(a_{r+1})\equiv 0(\mathrm{mod}{\pi }^{r+1})$. This produces a Cauchy sequence in the complete discrete valuation ring. Its limit $a$ satisfies $F(a)=0$ by continuity of polynomial evaluation, and $a\equiv a_1(\mathrm{mod}\pi )$.

For uniqueness, suppose $a,b$ are two such roots. Polynomial division of their difference givesBecause $b-a\in \pi R$, the second factor is a unit. Hence $b-a=0$. The simple-root hypothesis is used both for constructing successive lifts and for this uniqueness argument.

All the local fields here carry normalized discrete valuations. Write ${\mathcal{O}}_F$ for the valuation ring, ${\mathfrak{m}}_F$ for its maximal ideal, and $k_F$ for its residue field. We use the basic degree identity $[B:A]=e(B/A)f(B/A)$ for finite extensions of non-Archimedean local fields, with $e$ the ramification index and $f$ the residue degree.

First construct the maximal unramified subextension of a local field extension inside an arbitrary finite extension $L/K$. Choose a primitive element of a field extension $\bar{z}$ for the finite residue field extension $k_L/k_K$. Let $\bar{h}$ be its monic irreducible minimal polynomial of an algebraic element, of degree $f$, and lift its coefficients to obtain a monic $h\in {\mathcal{O}}_K[T]$ of the same degree. Finite finite fields are perfect fields, so ${\bar{h}}^{\prime }(\bar{z})\ne 0$. By Hensel lemma applied in $L$, there is $z\in {\mathcal{O}}_L$ reducing to $\bar{z}$ with $h(z)=0$. For $U=K(z)$ we haveEvery inequality is therefore an equality. In particular $U/K$ is unramified, $k_U=k_L$, and $h$ is the minimal polynomial of an algebraic element $z$.

If $L/K$ is unramified, then $[L:K]=f$, so this construction gives $U=L$. Taking $x=z$ supplies the required generator, and $h^{\prime }(x)$ is a unit. Conversely suppose $L=K(x)$ with $x$ integral and its monic minimal polynomial of an algebraic element $g$ satisfying $g^{\prime }(x)\notin {\mathfrak{m}}_L$. Its coefficients lie in ${\mathcal{O}}_K$, since all conjugates of an integral element are integral and $K$ is complete. In the constructed unramified extension $U$ the element $\bar{x}\in k_U=k_L$ is a simple root of $\bar{g}$. Hensel lemma in $U$ produces a root $x^{\prime }\in U$ with residue $\bar{x}$. In $L$ both $x$ and $x^{\prime }$ lift this same simple residue root, so Hensel uniqueness gives $x=x^{\prime }$. Thus $L\subseteq U\subseteq L$, proving