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For a field with a Non-Archimedean absolute value, define the coefficient norm $\Vert f{\Vert }_G={\max }_i|a_i|$ for $f(T)={\sum }_i{a}_i{T}^i$. Then $\Vert fg{\Vert }_G=\Vert f{\Vert }_G\Vert g{\Vert }_G$. For nonzero polynomials, scale each by one of its coefficients of maximum absolute value; the resulting polynomials have integral coefficients and at least one unit coefficient. Their reductions in the residue field are nonzero, so their product is nonzero. The product therefore also has coefficient norm $1$, proving the formula after reversing the scaling. This compares binary-polynomial coefficient heights with the heights of their roots.

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

For a profinite group $G$ and a complete discrete valuation ring $I$ with a continuous ${\mathbb{Z}}_p$-algebra structure, an $I$-valued measure is a bounded $I$-linear functional on continuous $I$-valued functions, with the uniform norm. Equivalently it assigns finitely additive values in $I$ to clopen sets. Locally constant functions are uniformly dense, so integration extends uniquely from finite sums. Countable additivity in the usual real-measure sense is not imposed. The coefficient ring may have infinite residue field and need not be compact.

Use $R={\mathcal{O}}_{k_{\mathfrak{p}}}$, the valuation ring of the completion of a number field at a prime ideal, with uniformizer $\pi$ and residue field $\kappa$. This completeness is essential for coprime-factor Hensel lifting. Normalize the two residue factors to be monic first; if the original factors have reciprocal leading units, undo the normalization by multiplying the eventual factors by reciprocal lifts of those units.

Let the two monic residue degrees be $d_1,d_2$. Start with monic lifts $g_1,h_1$ satisfying $f\equiv g_1{h}_1(\mathrm{mod}\pi )$. Suppose monic lifts $g_r,h_r$ of these degrees satisfy $f-g_r{h}_r\in {\pi }^{r}R[X]$. Put $E_r=(f-g_r{h}_r)/{\pi }^r$, of degree less than $d_1+d_2$. Over $\kappa$, the linear mapis injective: a zero image makes ${\varphi }_1$ divide $u$, by coprimality, forcing $u=0$ by its degree bound, and then $v=0$. The dimensions agree, so this map is a bijection. Choose lifts $u_r,v_r$ whose image is ${\overline{E}}_r$, and setTheir product agrees with $f$ modulo ${\pi }^{r+1}$; the new cross term has factor ${\pi }^{2r}$ and $2r\ge r+1$. The degree bounds preserve monicity. Each coefficient is a Cauchy sequence in the complete discrete valuation ring $R$, so the limits $f_1,f_2$ satisfyIf ${\mathfrak{o}}_{\mathfrak{p}}$ were interpreted only as an ordinary localization rather than the completed valuation ring, the assertion would be false. For example $X^2-2$ splits modulo seven but is irreducible over $\mathbb{Q}$. The completion notation in the next part provides the intended local-field interpretation.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

Only the prime two can ramify. More precisely, every field $\mathbb{Q}(P_n)$ is unramified at every odd prime; actual ramification at two depends on $P$ and the chosen division point. Use $[2^n]P_n=P$, as printed in the original PDF.

Fix an odd prime $p$ and put $N=2^n$. The elliptic-curve discriminant $-2^8$ gives good reduction at $p$, and $N$ is a unit in ${\mathbb{Z}}_p$. The unramified division torsors at good primes imply that all division points of an elliptic curve above $P$ are defined over finite unramified extensions of ${\mathbb{Q}}_p$. Here is a direct lifting proof, which does not require $P$ to have integral affine coordinates.

Properness gives a reduction $\tilde{P}$. Choose a point $\overline{Q}$ with $[N]\overline{Q}=\tilde{P}$ over the algebraic closure of the residue field, and a finite residue extension containing it. Pass to the corresponding finite unramified extension $K/{\mathbb{Q}}_p$. The surjectivity of good reduction over a local field lifts $\overline{Q}$ to $Q_0\in E(K)$. Its error $[N]Q_0-P$ lies in the kernel of reduction of an elliptic curve. The prime-to-residue-characteristic multiplication on a formal group makes $[N]$ bijective on that kernel. Choose the unique kernel point $R$ with $[N]R=[N]Q_0-P$; then $Q=Q_0-R$ satisfies $[N]Q=P$ and has reduction $\overline{Q}$.

Every algebraic division point has some such reduction and is the unique lift in its division fibre: the difference of two lifts lies in the reduction kernel, which has no $N$-torsion. Thus every chosen $P_n$ lies in an unramified local extension. For every prime of its possibly non-Galois coordinate field above $p$, the ramification index is one. Globally,One cannot universally assert that two actually ramifies: $P=O$ and $P_n=O$ give the field $\mathbb{Q}$. It can ramify. For example, a half of $T=(-1,0)$ isThe elliptic curve group law verifies this by doubling. Its coordinate field contains $\mathbb{Q}(\sqrt{2})$, which has ramification in the quadratic extension generated by the square root of two, so the only possible bad prime does occur for this choice.

Choose an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. For $P\in E({\mathbb{Q}}_p)$, choose homogeneous coordinates $[X:Y:Z]$ in ${\mathbb{Z}}_p$ with at least one unit by multiplying all three coordinates by a common power of $p$. DefineTwo primitive integral representatives differ by a unit, so this is well-defined. Every point of the reduced cubic is a nonsingular point of an algebraic curve, and the identity reduces to $[0:1:0]$. This definition includes points whose affine coordinates are nonintegral.

To prove the good-reduction map respects elliptic addition, use the smooth proper integral model. The elliptic curve group law is a morphism on this model, constructed by addition of degree-zero divisor classes; it is defined even where an affine chord formula has a vanishing denominator. Passing to the residue field commutes with this morphism. Henceso reduction is a homomorphism of abelian groups.

For surjectivity, the simple-root Hensel lemma says that if $g\in {\mathbb{Z}}_p[t]$, $g(t_0)\equiv 0(\mathrm{mod}p)$ and $g^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, then there is a unique $t\in {\mathbb{Z}}_p$ with $t\equiv t_0(\mathrm{mod}p)$ and $g(t)=0$. The point at infinity lifts to $O$. At an affine point $(\overline{x},\overline{y})$ of the reduced equation $F(x,y)=0$, being a nonsingular point of an algebraic curve means at least one partial derivative $F_x,F_y$ is nonzero. If $F_y\ne 0$, fix any integral lift $x_0$ of $\overline{x}$ and apply the Hensel lemma to $g(t)=F(x_0,t)$ to lift $\overline{y}$. If $F_x\ne 0$, fix $y_0$ and lift $\overline{x}$ instead. This proves the surjectivity of good reduction over a local field and gives

Write $\mathcal{O}$ for the valuation ring of $K$, $\pi$ for a uniformizer, $v(\pi )=1$, and $k={\mathbb{F}}_q$ for the residue field. A useful strong form of Hensel lemma is the following: if $f\in \mathcal{O}[X]$ and $a_0\in \mathcal{O}$ satisfythen there is a root $a\in \mathcal{O}$ with $v(a-a_0)>v(f^{\prime }(a_0))$, unique among roots in that ball. If $f(a_0)\ne 0$, its distance satisfies $v(a-a_0)=v(f(a_0))-v(f^{\prime }(a_0))$.

Here is a proof by successive correction. Put $s=v(f^{\prime }(a_0))$, $d_j=v(f(a_j))$, andStop if a root has already been reached. Otherwise the correction has valuation $d_j-s>s$. Integral Taylor expansions show that $v(f^{\prime }(a_{j+1}))=s$ andThus $d_j-2s$ grows at least geometrically, the corrections tend to zero, and completeness gives a limit $a$ with $f(a)=0$. The first correction has strictly smaller valuation than every later nonzero correction, proving the asserted distance. If $x,y$ are in the ball $v(x-a_0),v(y-a_0)>s$, thenThe second factor cannot vanish. Two roots in the ball must therefore coincide. This proves Hensel lemma; in particular a simple root modulo $\pi$ has a unique lift with that residue.

We use this to establish existence and uniqueness of unramified local extensions. For each integer $d\ge 1$, choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $d$ and lift it to a monic $f\in \mathcal{O}[X]$. Reduction and Gauss lemma show that $f$ remains irreducible over $K$. LetThe ring $A$ is finite free and complete over $\mathcal{O}$, and $A/\pi A\cong {\mathbb{F}}_{q^d}$. An element outside $\pi A$ is invertible: lift an inverse from the residue field and invert the remaining $1+\pi z$ by a convergent geometric series. Every nonzero element of $A$ is ${\pi }^t$ times such a unit, since ${\bigcap }_t{\pi }^{t}A=0$ in a finite free module. Thus $A$ is a discrete valuation ring, its fraction field is $L$, andThis constructs an unramified extension of each degree.

Conversely, let $M/K$ be an unramified extension of degree $d$. Its residue field is the unique degree-$d$ finite field extension ${\mathbb{F}}_{q^d}/{\mathbb{F}}_q$. Choose a residue root of $\bar{f}$ in it. The root is simple, so Hensel lemma lifts it to a root of $f$ in $M$. This embeds $L$ into $M$, and equality of degrees makes it an isomorphism. All $d$ residue conjugates lift in $L$ itself; hence $L/K$ is Galois, and reduction identifies its Galois group with the cyclic Galois group of the residue field extension. Its distinguished Frobenius automorphism reduces to $x\mapsto x^q$.

There is exactly one unramified extension $K_d$ of each degree inside a fixed algebraic closure, andFor the middle equality, every nonzero element of ${\mathbb{F}}_{q^d}$ lifts uniquely to a $(q^d-1)$st root of unity, because that exponent is prime to $p$. A primitive such residue generates the finite field extension, so its lift generates $K_d$. Taking the union gives the maximal unramified extension, with Galois group $\hat{\mathbb{Z}}$.

Fix a prime $\mathfrak{q}$ of $E$ over $\mathfrak{p}$, and a prime $\mathfrak{r}$ of $LE$ over $\mathfrak{q}$. Let $\mathfrak{P}$ be its restriction to $L$. Inside the completed compositum, putThe assumption means $U/F$ is an unramified extension. Its residue degree is some $f$; the completed extension $(LE{)}_{\mathfrak{r}}/J$ is the corresponding compositum $UJ/J$. It suffices to prove that this is unramified, regardless of the ramification of $J/F$.

Here is a proof of unramified extensions remain unramified under base change using the classification already established. Choose an integral generator $\alpha$ of $U/F$ whose monic minimal polynomial $g\in {\mathcal{O}}_F[X]$ has irreducible separable reduction of degree $f$. Write $k_F={\mathbb{F}}_q$ and $k_J={\mathbb{F}}_{q^r}$. Let $J_f/J$ be the degree-$f$ unramified extension. Its residue field ${\mathbb{F}}_{q^{rf}}$ contains ${\mathbb{F}}_{q^f}$, so $\bar{g}$ has a simple root there. Hensel lemma lifts this root to a root of $g$ in $J_f$. This gives an $F$-embedding $U\hookrightarrow J_f$.

An unramified extension of a local field is Galois, so all $F$-embeddings of $U$ into a fixed algebraic closure have the same image. The original $U$ therefore lies in $J_f$, and $UJ$ is an intermediate field of the unramified extension $J_f/J$. Its ramification index is one. Since $\mathfrak{q}$ and $\mathfrak{r}$ were arbitrary,This proof also explains why no linear-disjointness assumption on $L$ and $E$, and no assumption that $E/K$ is unramified, is needed.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

Normalize each finite-place valuation to have image $\mathbb{Z}$. For a radical class $[a]$, the residue of $v(a)$ modulo $n$ is well-defined, because replacing $a$ by $a{b}^n$ adds $nv(b)$. A suitable finite set isOne can compute the last union from any finite generating set of $\Delta$. Each representative has only finitely many nonzero valuations, so this indeed defines a finite set depending only on $n$ and the classes, not on a choice of representatives. Including all infinite places safely allows possible real-to-complex ramification.

Fix $v\notin S$ and work over the completion $K_v$. A representative has $v(a)=nk$, so $a={\pi }^{nk}u$ with $u$ a unit. Adjoining an $n$th root of $a$ is the same as adjoining an $n$th root of $u$. Since $n$ is invertible in the residue field, $X^n-\bar{u}$ is separable: its derivative $n{X}^{n-1}$ has no common zero with it.

Take a finite extension of the finite residue field containing all roots of all these finitely many reduced polynomials. The corresponding finite unramified extension of $K_v$ exists by the classification proved in Question 2. Hensel's lemma lifts each simple residue root to a root of the original polynomial in that unramified extension. Thus all the required radicals, and every conjugate root, lie in an unramified local extension. Every completion of the radical field at $v$ is therefore unramified. This proves finite ramification support for radical extensions.

The local argument does not assume ${\mu }_n\subset K$: adjoining all roots proves the stronger unramified assertion for the splitting field, and hence for any field obtained by choosing the radicals. For $n=1$ the radical extension is trivial; the displayed set is still valid.

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.

For $a$ prime to $p$, putFor positive $a$, $b_a(X)=X^{1-a}{\sum }_{j=0}^{a-1}{X}^{2j}$ is a Laurent polynomial. For negative $a$, use $b_a=-b_{-a}$. Hence it is regular at $X=1$ with value $a$, and $f_a\in {\mathbb{Z}}_p[[T]{]}^{\times }$ has constant coefficient $a^{p-1}\equiv 1(\mathrm{mod}p)$. Evaluation at ${\zeta }_n$ shows that $c_n(a)$ is a unit and reduces to one in the residue field, so $c_n(a)\in U_n$.

Because $p$ is odd, ${\prod }_{\xi \in {\mu }_p}\xi =1$. Both maps $\xi \mapsto {\xi }^2$ and $\xi \mapsto {\xi }^{2a}$ permute ${\mu }_p$, and thereforeThis rational-function identity also holds for negative $a$ and extends across the removable point $X=1$. It implies $N{f}_a=f_a$. Part 1(b) now proves all the required norm compatibilities, and $c(a)\in U_{\infty }$, with Coleman series $f_a$.

Use the additive formal coordinate $z=\log (1+T)$, in which $D=d/dz$. ThenIts logarithm after normalization by $a^{p-1}$ is even, so all odd derivatives at zero vanish. For the even derivatives, the hyperbolic cotangent generating function for Bernoulli numbers givesHere the Bernoulli numbers use $B_1=-1/2$. Differentiating $k-1$ more times yields the answer for every index:These rational expressions lie in ${\mathbb{Z}}_p$ because they equal the integral logarithmic-derivative definition in part (a). In particular ${\delta }_1=0$; substituting $B_1$ into the even-index formula would give an incorrect answer. For $a=\pm 1$ the sequence is one and every derivative vanishes, as the formula also shows.