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The valuation ring is $\mathcal{O}=\{x\in K:|x|\le 1\}$, and $\mathfrak{m}=\{x\in K:|x|<1\}$ is an ideal by the ultrametric inequality. Its complement inside $\mathcal{O}$ consists exactly of elements of field absolute value one, whose inverses also lie in $\mathcal{O}$. Every proper ideal must avoid units and hence lie in $\mathfrak{m}$. Therefore $\mathfrak{m}$ is the unique maximal ideal: $\mathcal{O}$ is a local ring. Also its fraction field is $K$, since every nonzero element outside the ring has its inverse inside it.

If $z\in K$ is integral over $\mathcal{O}$, it satisfies a monic equation $z^n+a_{n-1}{z}^{n-1}+\cdots +a_0=0$ with $|a_i|\le 1$. Were $|z|>1$, the leading term would have strictly larger field absolute value than every other term, so the ultrametric inequality would prevent cancellation to zero. Thus $z\in \mathcal{O}$, proving the valuation ring is integrally closed.

For the ideal criterion, assume the valuation is nontrivial and write $v=-\log |\cdot |$. If its value group is discrete, normalize it to $\mathbb{Z}$ and choose a uniformizer $\pi$ of valuation one. In any nonzero ideal $I$, the nonnegative integer valuations have a least value $m$. Choose $a\in I$ of that value. For every $b\in I$, $v(b/a)\ge 0$, so $b\in a\mathcal{O}$. Hence $I=(a)=({\pi }^m)$ and $\mathcal{O}$ is a principal ideal domain.

Conversely, if $\mathcal{O}$ is a principal ideal domain, its nonzero maximal ideal is $(\pi )$. For every element $x$ of positive valuation, $x\in \mathfrak{m}$ implies $x=\pi y$ with $y\in \mathcal{O}$, so $v(x)\ge v(\pi )>0$. Thus $\gamma =v(\pi )$ is the least positive value. For any value $t$, subtract $\lfloor t/\gamma \rfloor \gamma$. The remainder belongs to the value group and lies in $[0,\gamma )$, so is zero. The value group is therefore $\gamma \mathbb{Z}$, and the valuation is discrete. We have proved the principal ideal criterion for a nontrivial rank-one valuation ring.

The nontriviality qualification matters under the standard definition of discrete valuation, which requires value group isomorphic to $\mathbb{Z}$. With the trivial field absolute value, $\mathcal{O}=K$ is still a principal ideal domain, but the value group is zero and there is no uniformizer. Thus the literal unrestricted equivalence has a field exception. If “discrete” instead includes the trivial value group as a discrete subgroup of $\mathbb{R}$, that exceptional case satisfies the equivalence too.

The discrete valuation identities $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ make $R$ closed under products and sums; it contains $0,1$ and additive inverses. As a subring of the field $K$, it is an integral domain.

Its units are exactly the elements of valuation zero: $x^{-1}\in R$ if and only if $v(x)=0$. The remaining elements form the ideal $\mathfrak{m}=\{x:v(x)>0\}$, and every proper ideal is contained in it, since an ideal containing a unit is the whole ring. Hence $R$ is a local ring with this unique maximal ideal.

For a nonzero ideal $I$, the set of valuations of its nonzero elements is a nonempty subset of ${\mathbb{Z}}_{\ge 0}$ and has a least value. Choose $x\in I$ attaining it. Every $y\in I$ satisfies $v(y/x)\ge 0$, so $y/x\in R$ and $I=(x)$. The zero ideal is also principal. This proves that $R$ is a discrete valuation ring, in particular a principal ideal domain.

Choose a uniformizer $\pi$, with $v(\pi )=1$. For $z\in K$, some $n\ge 0$ makes ${\pi }^{n}z\in R$, and $z=({\pi }^{n}z)/{\pi }^n$ is a quotient of elements of $R$. Thus $K$ is the field of fractions of $R$.

Put $\alpha =\sqrt[p]{p}$ and $\lambda ={\zeta }_p-1$. The polynomial $T^p-p$ is Eisenstein, so $[{\mathbb{Q}}_p(\alpha ):{\mathbb{Q}}_p]=p$, while the preceding part gives degree $p-1$ for $E$. Both numbers divide $[K:{\mathbb{Q}}_p]$ by the tower law, and the compositum degree is at most their product. Coprimality forces $[K:{\mathbb{Q}}_p]=p(p-1)$. Since $K$ contains all the roots ${\zeta }_p^a\alpha$ of $T^p-p$, it is the splitting field of a separable polynomial and is Galois.

Extend $v_p$ with $v_p(p)=1$. Then $v_p(\alpha )=1/p$, $v_p(\lambda )=1/(p-1)$, andThe value group therefore forces the ramification index to be at least $p(p-1)$; it cannot exceed the extension degree. Thus $K/{\mathbb{Q}}_p$ is totally ramified and $\varpi$ is a uniformizer for the normalized valuation $v_K=p(p-1)v_p$.

Write its automorphisms as ${\sigma }_{a,b}(\alpha )={\zeta }_p^a\alpha$ and ${\sigma }_{a,b}({\zeta }_p)={\zeta }_p^b$, with $a\in {\mathbb{F}}_p$, $b\in {\mathbb{F}}_p^{\times }$. The degree count makes all $p(p-1)$ pairs occur, giving $G\cong {\mathbb{F}}_p⋊{\mathbb{F}}_p^{\times }$. Their action on the uniformizer isIf $b\ne 1$, the residue of this ratio is $b$, so subtracting one gives a unit and $v_K({\sigma }_{a,b}(\varpi )-\varpi )=1$. If $b=1$ and $a\ne 0$, the ratio is ${\zeta }_p^{-a}$, giving displacement valuation $1+v_K({\zeta }_p^{-a}-1)=p+1$. The identity has infinite displacement valuation.

By the uniformizer criterion for lower ramification groups, $\sigma \in G_i$ precisely when that displacement valuation is at least $i+1$. Hence the ramification groups of the splitting field of Xp minus p over the p-adic numbers have sizesFor $1\le i\le p$ the group itself is $\{{\sigma }_{a,1}:a\in {\mathbb{F}}_p\}\cong C_p$. The lower wild ramification break is $p$.

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 $p$.

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.

Now suppose $\mathrm{char}K=p>0$. The Teichmuller lifts constructed in (b) are additive as well as multiplicative: if $a^q=a$ and $b^q=b$, then the characteristic-$p$ Frobenius identity gives $(a+b{)}^q=a^q+b^q=a+b$. Uniqueness of the lift with prescribed residue yields $[\bar{a}+\bar{b}]=[\bar{a}]+[\bar{b}]$. Their set is therefore a finite coefficient field in positive-characteristic local fields, and reduction identifies it with $k$.

Choose a uniformizer $\pi$. For $x\in {\mathcal{O}}_K$, let $a_0$ be the lift of its residue and write $x=a_0+\pi x_1$ with $x_1\in {\mathcal{O}}_K$. Repeat with $x_1$, then $x_2$, and so on. This givesThe remainder tends to zero, and completeness gives $x={\sum }_{j\ge 0}{a}_j{\pi }^j$. The expansion is unique: in a nonzero difference of two series the first different coefficient is a nonzero element of the coefficient field, hence a unit, and so gives exactly the valuation of that difference.

Consequently substitution $t\mapsto \pi$ defines a bijection $k[[t]]\to {\mathcal{O}}_K$. It is a ring homomorphism because the coefficient lift is a field homomorphism, polynomial operations agree before passage to limits, and multiplication is continuous. Every element of $K$ becomes integral after multiplication by a sufficiently large power of $\pi$. Allowing finitely many negative powers therefore extends substitution to the field isomorphismIt preserves the normalized discrete valuation, since the first nonzero coefficient determines the valuation on each side. The results used are the structural facts stated in (a), its proved Hensel lemma, the proved multiplicative lifting in (b), and the characteristic-$p$ identity $(a+b{)}^q=a^q+b^q$. No classification theorem for local fields is needed.

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.

Let $\varphi :C\setminus S\to {\mathbb{P}}^n$ be the given algebraic morphism of algebraic varieties. Its value at the generic point supplies homogeneous coordinates $[h_0:\cdots :h_n]$ with $h_i\in k(C)$, not all zero. Thus it defines a rational map of projective varieties on $C$.

Fix a missing point $p\in S$. The local ring of a smooth algebraic curve at $p$ is a discrete valuation ring $A={\mathcal{O}}_{C,p}$, with uniformizer $t$ and valuation $v_p$. PutZero coordinates remain zero. Every $a_i$ is in $A$, and some $a_j$ is a unit because its valuation is zero. Choose a neighborhood $U$ of $p$ on which all $a_i$ are regular and $a_j$ remains nonvanishing; this is possible because membership in the local ring means regularity on a neighborhood. The affine target-chart coordinatesare regular on $U$, so they define a morphism of algebraic varieties $U\to {\mathbb{P}}^n$. At $p$ its homogeneous value is $[a_0(p):\cdots :a_n(p)]$, well-defined because $a_j(p)\ne 0$.

Multiplication of all homogeneous coordinates by the same nonzero rational function does not change their projective point. Thus this local morphism agrees with the original one on the common dense open where their coordinates were chosen. They agree throughout the overlap: the target's diagonal is closed, being defined in homogeneous coordinate pairs by $T_i{U}_j-T_j{U}_i=0$ for all $i,j$. Thus the equalizer is closed and contains that dense open. Apply this construction at each point of the finite set $S$ and glue with the original morphism. The same closed-diagonal argument proves uniqueness. ThereforeThis gives the extension of a rational map from a smooth projective curve by explicit local coordinates. The mechanism is the one-dimensional smooth local ring: it permits removal of the smallest common pole order while leaving at least one unit coordinate.

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

For lifting residue-field embeddings of unramified extensions, use the generator $z$ and polynomial $h$ just constructed when $L/K$ is unramified. Given a $k_K$-field homomorphism $\iota :k_L\to k_M$, its value $\iota (\bar{z})$ is a simple root of $\bar{h}$. Hensel lemma in $M$ produces the unique root $z^{\prime }\in {\mathcal{O}}_M$ with that residue. The assignment $z\mapsto z^{\prime }$ defines a $K$-field homomorphism $K[T]/(h)\to M$; it induces $\iota$ because $\bar{z}$ generates $k_L$. Conversely any such field homomorphism preserves the unique extended valuation, hence induces a residue field homomorphism, and its image of $z$ is uniquely forced by Hensel uniqueness. This proves the asserted bijection. The same proof only requires that $M$ be a non-Archimedean local field containing $K$ with its compatible valuation; its finiteness over $K$ is not essential.

Now set $M=L$. The finite residue field extension has exactly $f=[L:K]$ automorphisms, cyclically generated by $a\mapsto a^{|k_K|}$. Their unique lifts give $[L:K]$ automorphisms of $L$. Consequently $L/K$ is a Galois extension, and reduction identifies its Galois group with that cyclic residue Galois group. Its generator is the arithmetic Frobenius:

For the compositum, choose a uniformizer ${\pi }_M$ of the totally ramified extension $M/K$. The Eisenstein generator of a totally ramified extension characterization, proved in the next solution, says that $M=K({\pi }_M)$ and its minimal polynomial of an algebraic element is Eisenstein over ${\mathcal{O}}_K$. A uniformizer of $K$ remains a uniformizer in the unramified extension $L$. All its Eisenstein coefficient conditions therefore remain true over ${\mathcal{O}}_L$. Thus the same Eisenstein polynomial is irreducible over $L$ and generates $LM=L({\pi }_M)$ as a totally ramified extension. This proves $LM/L$ is totally ramified, of degree $[M:K]$.

Here is an explicit example showing that the converse decomposition of an arbitrary extension can fail. Take $K={\mathbb{Q}}_3$ and $L=K(i)$ with $i^2=-1$. Since $T^2+1$ is irreducible and separable modulo three, the degree and residue-degree argument above shows that $L/K$ is unramified quadratic. PutThe element $1+i$ is a unit in ${\mathcal{O}}_L$, so $T^2-3(1+i)$ is Eisenstein over $L$. Hence $[N:L]=2$, $e(N/K)=2$, and $f(N/K)=2$. Moreover $i={\alpha }^2/3-1$ gives $N=K(\alpha )$, withWe verify its Galois group, since that is what rules out a totally ramified complement. By Hensel lemma there is $r\in K$ with $r^2=-2$: the residue root $1$ of $T^2+2$ is simple modulo three. Set $\beta =(1+i)/r$. ThenThe assignments $\sigma (i)=-i$ and $\sigma (\alpha )=\beta \alpha$ respect the defining relation, because $(\beta \alpha {)}^2=-3i(1+i)=3(1-i)$. They therefore define a $K$-automorphism of $N$. Its square satisfies ${\sigma }^2(i)=i$ and ${\sigma }^2(\alpha )=\bar{\beta }\beta \alpha =-\alpha$, so $\sigma$ has order four. There are thus four automorphisms of the degree-four extension, andThis is the cyclic quartic local extension without a totally ramified complement. Its unique quadratic subfield is $L$, fixed by ${\sigma }^2$, and is unramified. If $N=L^{\prime }{M}^{\prime }$ with $L^{\prime }/K$ unramified and $M^{\prime }/K$ totally ramified, then $[L^{\prime }:K]\le f(N/K)=2$ and $[M^{\prime }:K]\le e(N/K)=2$. To obtain the full degree four both would have degree two, forcing $M^{\prime }$ to be a totally ramified quadratic subfield. There is no such subfield. Thus $N/{\mathbb{Q}}_3$ is the required counterexample.

Let $d=[L:K]$, with normalized discrete valuations $v_K,v_L$. Suppose first that $L/K$ is totally ramified, so $e(L/K)=d$. Choose a uniformizer $x$ of $L$. The elements $1,x,\dots ,x^{d-1}$ are linearly independent over $K$: the nonzero terms $c_j{x}^j$ in any proposed relation have valuations $d{v}_K(c_j)+j$, distinct modulo $d$, and a sum with a unique least valuation cannot vanish by the ultrametric inequality. Hence $[K(x):K]\ge d$, proving $L=K(x)$.

Extend $v_K$ to an algebraic closure with values in $\mathbb{Q}$. Every conjugate of $x$ has valuation $1/d$, by uniqueness of the extended valuation on finite extensions of the complete field $K$. In its monic minimal polynomial of an algebraic element $g(T)=T^d+a_{d-1}{T}^{d-1}+\cdots +a_0$, the coefficient $a_{d-j}$ is a sum of products of $j$ conjugates, counted with multiplicity. Thus $v_K(a_{d-j})\ge j/d>0$ for $j\ge 1$, so every nonleading coefficient is in ${\mathfrak{m}}_K$. The constant coefficient is the product of all $d$ conjugates, so has valuation exactly one. This proves that $g$ is an Eisenstein polynomial. This argument also works for inseparable finite extensions, since the conjugates can be counted with their multiplicities.

Conversely let $g$ be an Eisenstein polynomial of degree $d$, with root $x$ and $L=K(x)$. The Eisenstein criterion proves irreducibility. For completeness, monic factors would have integral coefficients; their reductions would be positive-degree powers of $T$, so both constant coefficients would be divisible by the uniformizer of $K$, contradicting $v_K(g(0))=1$. To determine the root valuation, put $t=v_K(x)$ in the extended valuation. If $t<0$, the leading term is the unique term of least valuation; if $t=0$, the leading term again has uniquely least valuation zero. Both are impossible. If $t>0$, every nonconstant lower term has valuation at least $1+jt>1$. Cancellation in $g(x)=0$ therefore requires the leading term and constant term to have equal valuation:Since $t=v_L(x)/e(L/K)$ and $v_L(x)$ is a positive integer, $e(L/K)=d{v}_L(x)\ge d$. The opposite inequality follows from $[L:K]=e(L/K)f(L/K)=d$. Hence $e=d$, $f=1$, and $v_L(x)=1$. We have proved both the equivalence and the uniformizer assertion:

Now take $q=p^a$ with $a\ge 1$ and $\zeta ={\zeta }_{p^a}$. The translated cyclotomic polynomial isIn characteristic $p$, putting $U=T^{p^{a-1}}$ reduces this to ${\sum }_{j=0}^{p-1}(1+U{)}^j=U^{p-1}=T^{p^{a-1}(p-1)}$. Its constant coefficient over the integers is $p$. It is therefore Eisenstein at $p$, of degree $d=p^{a-1}(p-1)$. Consequently ${\mathbb{Q}}_p({\zeta }_{p^a})/{\mathbb{Q}}_p$ is totally ramified of degree $d$, with uniformizer ${\zeta }_{p^a}-1$. If $q=1$, the field is simply ${\mathbb{Q}}_p$, a trivial totally ramified extension.

For a general positive integer $n=p^{a}m$ with $p\nmid m$, the field generated by the prime-to-$p$ root of unity ${\zeta }_m$ is unramified. Indeed $T^m-1$ has distinct roots modulo $p$. In a finite unramified field whose residue field contains those roots, Hensel lemma lifts all of them uniquely. Reduction preserves their orders: a prime-to-$p$ root reducing to one must itself be one, by Hensel uniqueness at the root one of $T^m-1$. The arithmetic Frobenius therefore acts as ${\zeta }_m\mapsto {\zeta }_m^p$, and the extension degree is the multiplicative order of $p$ modulo $m$, with degree one when $m=1$.

The identityfollows by taking powers of a primitive nth root of unity and using the coprimality of $p^a,m$ in the opposite direction. The prime-to-$p$ field is unramified and the $p$-power field is totally ramified; unramified base change preserves total ramification proves that their compositum has the same residue degree as the prime-to-$p$ field. Thus the total ramification criterion for local cyclotomic extensions gives the full classification:This includes trivial extensions when $a=0$. For $p=2$ only $m=1$ is possible.

For a finite totally ramified Galois extension $L/K$ with Galois group $G$, define its lower ramification numbering byHere $G_0$ is the inertia group, which equals $G$, and $G_1$ is the wild inertia group. For a uniformizer $\pi$ the condition is equivalent to $v_L(\sigma (\pi )-\pi )\ge i+1$. To justify this uniformizer criterion for lower ramification groups, expand every $b\in {\mathcal{O}}_L$ as a convergent series ${\sum }_{j\ge 0}{c}_j{\pi }^j$ with the coefficients $c_j$ chosen from representatives of $k_K$ in ${\mathcal{O}}_K$. Such an expansion is obtained by successive subtraction of a residue representative and division by $\pi$, since total ramification gives $k_L=k_K$. The automorphism fixes each $c_j$, and $\sigma (\pi {)}^j-{\pi }^j$ is divisible by $\sigma (\pi )-\pi$. Thus the required valuation bound for $\pi$ implies it for every $b$, including the limit by completeness.

For $L={\mathbb{Q}}_p({\zeta }_{p^a})$, all primitive conjugates lie in $L$ and the Eisenstein degree is $\phi (p^a)$. HenceIf ${\sigma }_b\ne 1$, put $s=v_p(b-1)$, so $0\le s<a$. The element ${\zeta }^{b-1}$ is a primitive $p^{a-s}$th root of unity. Applying the already proved uniformizer result to its field, and multiplying its valuation by the ramification index of $L$ over that field, givesThe identity automorphism has infinite valuation. Therefore the lower ramification filtration of a prime-power cyclotomic extension is explicitlyThe order of the middle group is $p^{a-j}$. At odd $p$, $G_0/G_1$ has order $p-1$; at $p=2$, the condition $b\equiv 1(\mathrm{mod}2)$ is automatic, so $G_0=G_1$ when $a\ge 2$. The case $p=2,a=1$ has trivial Galois group, and the displayed formula remains valid. For $a=0$ all groups are trivial. The valuation calculation also provides the upper ramification numbering, if desired, by integrating the reciprocal indices with the Herbrand function; no change of numbering is implicit in the displayed answer.

The multiplicative Hilbert theorem 90 states that, for a finite Galois extension $L/K$ with Galois group $G$, every multiplicative cocycle $c:G\to L^{\times }$ satisfying $c_{\sigma \tau }=c_{\sigma }\sigma (c_{\tau })$ has the form $c_{\sigma }=\sigma (y)/y$ for some $y\in L^{\times }$. In the cyclic case $G=⟨\sigma ⟩$, it says

Here is the full existence argument. Distinct field homomorphisms $\sigma :L\to L$ are linearly independent as functions over $L$. To see this Artin independence theorem directly, assume a nonzero identically vanishing relation with the fewest nonzero coefficients. There are at least two terms. Choose two participating embeddings ${\sigma }_1,{\sigma }_j$ and an element $a$ with ${\sigma }_1(a)\ne {\sigma }_j(a)$. Evaluate the relation at $at$ and subtract ${\sigma }_j(a)$ times its value at $t$. The $j$th term disappears while the first remains nonzero, giving a shorter vanishing relation, a contradiction.

Apply this independence toThe coefficients are nonzero, so $A$ is not identically zero; choose $t$ with $y=A(t)\ne 0$. The cocycle identity gives $\tau (c_{\sigma }^{-1})=c_{\tau \sigma }^{-1}{c}_{\tau }$, and henceThis proves the general statement. In the cyclic case, given $N(x)=1$, set $c_1=1$ and $c_{{\sigma }^j}=x\sigma (x)\cdots {\sigma }^{j-1}(x)$ for $1\le j<n$. The norm-one identity makes this definition compatible with ${\sigma }^n=1$, and the cocycle identity follows by concatenating the products. The argument above gives $\sigma (y)/y=x$. Conversely that quotient has norm one because its conjugate product telescopes.

Finally let $L/K$ be unramified, and take its arithmetic Frobenius $\varphi$. For a unit $x$ of norm one, the cyclic theorem gives $b\in L^{\times }$ with $\varphi (b)/b=x$. Write $r=v_L(b)$ and choose the common uniformizer $\pi \in K$. Then $y={\pi }^{-r}b$ has valuation zero, and $\varphi (\pi )=\pi$ implies $\varphi (y)/y=\varphi (b)/b=x$. Thus the unit form of Hilbert theorem 90 isThe unramified hypothesis is what permits shifting an arbitrary integer valuation of $b$ by a power of a uniformizer belonging to $K$.

Let $\pi \in K$ be a uniformizer. Because $L/K$ is unramified, the same $\pi$ is a uniformizer of $L$, and $v_L{|}_K=v_K$. The conjugates of $z\in L^{\times }$ all have the same valuation, soThis proves that every field norm has valuation divisible by $n$. Conversely $N_{L/K}({\pi }^r)={\pi }^{nr}$ for every integer $r$, so it remains to prove surjectivity of unit norms in unramified extensions.