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For $K/{\mathbb{Q}}_p$ finite, let $x_n^{p^n}=x$ in its finite residue field, and choose arbitrary lifts $y_n\in {\mathcal{O}}_K$. With normalized discrete valuation,Applying this binomial estimate repeatedly to $y_n$ and $[x_n]$ proves the displayed bound, and hence convergence to $[x]$.

For a normalized discrete valuation, a uniformizer $\pi$, and a chosen set $A$ of representatives of the residue field, every existing nonzero element has a unique displayed expansion, with its first digit not in the maximal ideal. Starting from $y_0={\pi }^{-v(x)}x$, choose $a_j$ representing $y_j$ modulo the maximal ideal and put $y_{j+1}=(y_j-a_j)/\pi$. The error after $r$ digits has valuation at least $v(x)+r$, so the sums converge to $x$ even without completeness. Conversely arbitrary digit sequences need not converge in an incomplete field: in $\mathbb{Q}$ with its three-adic valuation, the digits of $\sqrt{10}\in {\mathbb{Q}}_3$ give a counterexample. Hensel lemma gives this square root, but it is not rational.

A discretely valued field is a field equipped with a discrete valuation. It is complete when every Cauchy sequence for the induced metric converges.

For a smooth irreducible algebraic curve, send a divisor on an algebraic curve $D={\sum }_P{n}_{P}P$ to the line bundle associated to a divisor defined locally by ${\mathcal{O}}_X(D{)}_P=t_P^{-n_P}{\mathcal{O}}_{X,P}$. A nonzero rational section of a line bundle gives its inverse construction, and changing that section changes its divisor by a principal divisor. Thus the divisor class group is the Picard group. The quotient of the sheaf of nonzero rational functions on an irreducible variety by the sheaf of units of the structure sheaf has stalk $\mathbb{Z}$ at each closed point, via the discrete valuation. Its global sections are finite sums of points. The corresponding long exact sequence in sheaf cohomology, and the vanishing for the rational-function sheaf, identify both groups with $H^1(X,{\mathcal{O}}_X^{\ast })$.

If a non-Archimedean local field $K$ has uniformizer $\pi$, valuation ring ${\mathcal{O}}_K$, maximal ideal ${\mathfrak{m}}_K$ and residue field ${\mathbb{F}}_q$, thenThe first factor records the discrete valuation, the Teichmuller representatives split reduction ${\mathcal{O}}_K^{\times }\to {\mathbb{F}}_q^{\times }$, and $1+{\mathfrak{m}}_K$ is the group of principal units.

Put $\mathcal{D}={\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, the quotient sheaf of abelian groups. At a closed point $P$, the discrete valuation gives an isomorphismIndeed, a local parameter on a smooth algebraic curve has valuation one, and the valuation kernel is precisely the unit group of the discrete valuation ring. Thus a local section represented by $f$ determines its local principal divisor, and changing $f$ by a regular function which is a unit does not change those coefficients.

These local valuations identify $\Gamma (X,\mathcal{D})$ with $\mathrm{Div}(X)$. To see that they give finite support, represent a section locally by finitely many rational functions on an algebraic variety, using Noetherian Zariski topology to take a finite cover. Each representative has only finitely many zeros and poles, so their combined support is finite. Conversely, for $D=\sum n_{P}P$, choose at each support point a local parameter $t_P$ and shrink its neighbourhood to exclude every other zero or pole of $t_P$ and every other support point. The local rational equation $t_P^{n_P}$ represents the desired valuations there; on the complement of the support use $1$. On overlaps these equations differ by units, so their classes glue in the quotient sheaf. Finally, a section with every valuation zero has zero germ everywhere and is the identity section. Hence the identification is both surjective and injective.

The map on global sections sends $f\in k(X{)}^{\ast }$ to $\mathrm{div}(f)$. Therefore its cokernel is

Let $V\subseteq U$ be open subsets, and let $s\in {\mathcal{F}}_3(V)$. The assumed flasque-kernel section-lifting lemma lifts $s$ to $\tilde{s}\in {\mathcal{F}}_2(V)$. Since ${\mathcal{F}}_2$ is a flabby sheaf, extend $\tilde{s}$ to a section on $U$, then project to ${\mathcal{F}}_3(U)$. Its restriction is $s$. Thus every restriction map of ${\mathcal{F}}_3$ is surjective, proving that the quotient of two flabby sheaves in a short exact sequence is flabby.

For the construction of sheaf cohomology, embed a sheaf of abelian groups $\mathcal{F}$ into the sheafby sending a section to all its germs. This is injective because a section with every germ zero is zero. Arbitrary families of stalk elements glue point by point, and restriction maps are projections; hence $C(\mathcal{F})$ is a flabby sheaf. Set ${\mathcal{Q}}^0=\mathcal{F}$, ${\mathcal{I}}^j=C({\mathcal{Q}}^j)$ and ${\mathcal{Q}}^{j+1}={\mathcal{I}}^j/{\mathcal{Q}}^j$. Compose the quotient map with the next embedding to obtain the flasque resolutionThe sheaf cohomology groups are the cohomology groups of the resulting cochain complex of global sections:The usual comparison of resolutions identifies this construction with the right derived functors of global sections; the construction above is the Godement resolution.

If $\mathcal{F}$ is itself a flabby sheaf, the quotient argument shows inductively that every ${\mathcal{Q}}^j$ is flabby. The assumed lifting result therefore makesan exact sequence for every $j$. The global-section complex is consequently exact in every positive degree. In particular,This establishes the vanishing from the section-lifting property rather than assuming it in advance.

For the smooth algebraic curve $X$, a divisor on an algebraic curve is a finite formal sum $D={\sum }_P{n}_{P}P$ of closed points with integer coefficients. At each closed point, the local ring of a smooth algebraic curve is a discrete valuation ring; write its normalized discrete valuation as $v_P$. For $f\in k(X{)}^{\ast }$ the principal divisor is $\mathrm{div}(f)={\sum }_P{v}_P(f)P$. Only finitely many coefficients are nonzero: zeros and poles are proper closed subsets in the Noetherian Zariski topology of a curve. The divisor class group is the abelian groupEquivalently, its elements are divisors on an algebraic curve modulo linear equivalence of divisors.

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 $R$.

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.

For the finite Galois extension, normalize the discrete valuation $v_L$ to have value group $\mathbb{Z}$. The inertia group isThe lower ramification numbering isThe wild inertia group is $P=G_1$, and the tame inertia quotient is $I/P$. We will prove that $P$ is a $p$-group, where $p=\mathrm{char}{k}_K$, and that $I/P$ is cyclic of order prime to $p$. Tame inertia here is a quotient, not a specified canonical subgroup complementary to $P$.

First reduce the inertia calculation to a totally ramified extension. Choose $\bar{b}$ generating the finite residue extension $k_L/k_K$. Its minimal polynomial is separable. Lift that monic polynomial to $f\in {\mathcal{O}}_K[X]$, and use Hensel lemma in ${\mathcal{O}}_L$ to obtain a root $b$ with residue $\bar{b}$. Irreducibility of the reduction implies irreducibility of $f$ over $K$, by the monic form of the Gauss lemma for polynomials over a discrete valuation ring. Thus $K_0=K(b)$ has degree $[k_L:k_K]$ and residue field $k_L$; the degree formula makes $K_0/K$ unramified. An automorphism fixes $b$ precisely when it fixes $\bar{b}$: the nontrivial direction is uniqueness in Hensel lemma. Hence $I=\mathrm{Gal}(L/K_0)$, and $L/K_0$ is totally ramified.

Use the permitted integral-generation fact ${\mathcal{O}}_L={\mathcal{O}}_{K_0}[{\pi }_L]$. Every element of $I$ fixes the coefficients in this expression. The identity $\sigma ({\pi }_L^j)-{\pi }_L^j=(\sigma {\pi }_L-{\pi }_L){\sum }_{h=0}^{j-1}(\sigma {\pi }_L{)}^h{\pi }_L^{j-1-h}$ consequently proves the uniformizer criterion for lower ramification groups:Necessity follows by testing $x={\pi }_L$, and the displayed power identity proves sufficiency for every integral polynomial in ${\pi }_L$.

Define the uniformizer character of inertiaIt is multiplicative: writing $u_{\tau }=\tau {\pi }_L/{\pi }_L$, one has $\sigma \tau {\pi }_L/{\pi }_L=\sigma (u_{\tau })(\sigma {\pi }_L/{\pi }_L)$, and $\sigma$ acts trivially on residue units. Its kernel is exactly $v_L(\sigma {\pi }_L-{\pi }_L)\ge 2$, namely $G_1$. Changing the uniformizer to $u{\pi }_L$ leaves the residue character unchanged, since $\overline{\sigma (u)/u}=1$. ThereforeEvery finite subgroup of a field's multiplicative group is cyclic: if its exponent is $m$, every element is a root of $X^m-1$, so its order is at most $m$; a finite abelian group has an element of order its exponent, constructed by multiplying elements of maximal prime-power orders, so its order is at least $m$. Equality makes it cyclic. Here $k_L^{\times }$ has order $|k_L|-1$, prime to $p$. We have proved

To verify the assertion about wild inertia as well, for $i\ge 1$ defineWrite $\sigma {\pi }_L={\pi }_L+a{\pi }_L^{i+1}$ and $\tau {\pi }_L={\pi }_L+b{\pi }_L^{i+1}$. Composition gives $\sigma \tau {\pi }_L\equiv {\pi }_L+(a+b){\pi }_L^{i+1}(\mathrm{mod}{\pi }_L^{i+2})$, because inertia fixes residues and $i\ge 1$. Thus ${\theta }_i$ is additive with kernel $G_{i+1}$. Thus positive ramification quotients are elementary abelian: each $G_i/G_{i+1}$ is a subgroup of the additive finite field, hence an elementary abelian $p$-group.

Finally the filtration terminates: a nonidentity element of $I$ cannot fix ${\pi }_L$, since it already fixes $K_0$ and ${\mathcal{O}}_L={\mathcal{O}}_{K_0}[{\pi }_L]$. Its displacement has finite valuation, and there are only finitely many automorphisms. Iterating the finite $p$-group quotients gives that $P$ is a $p$-group. It is normal in $I$ and has prime-to-$p$ quotient, so it is its unique Sylow $p$-subgroup, justifying the wild inertia terminology.

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.

First include the trivial absolute value if that convention is allowed. For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value on a field at most one, and some nonzero integer has absolute value on a field less than one: otherwise every nonzero rational number, being a quotient of two integers, would have absolute value on a field one. Factoring that integer shows that some prime number $p$ has $|p|<1$.

There cannot be two such prime numbers. If $|p|<1$ and $|q|<1$ for distinct prime numbers, choose integers $a,b$ with $ap+bq=1$. Then the ultrametric inequality givesEvery other prime number therefore has absolute value on a field one. Factoring numerator and denominator of a rational number givesThus the nontrivial equivalence classes are precisely the p-adic absolute values, one for each prime number.

Now let $F$ be a number field, with ring of integers of a number field $A$. If the restriction of an absolute value on a field to $\mathbb{Q}$ is trivial, a monic polynomial over $\mathbb{Q}$ satisfied by $x\in F$ forces $|x|\le 1$: if $|x|>1$, its leading term would strictly dominate the other terms. Applying the same argument to $x^{-1}$ forces $|x|=1$ for $x\ne 0$. Thus a nontrivial absolute value on a field defined on $F$ has a nontrivial restriction, associated with some prime number $p$.

Every $a\in A$ has $|a|\le 1$, by applying the same leading-term argument to its monic integral equation. Consequentlyis a proper prime ideal: it is an ideal by the ultrametric inequality, and $|ab|<1$ with $|a|,|b|\le 1$ implies that at least one factor has absolute value on a field less than one. It is nonzero because it contains $p$.

The localization $A_{\mathfrak{p}}$ lies in the valuation ring of the absolute value on a field: a denominator outside $\mathfrak{p}$ has absolute value on a field one. Since $A$ is a Dedekind domain, $A_{\mathfrak{p}}$ is a discrete valuation ring. Choose its uniformizer $\pi$. Each $x\in F^{\times }$ has a unique expressionBoth $u$ and $u^{-1}$ lie in the valuation ring, so $|u|=1$. Also $\pi$ belongs to its maximal ideal, hence $|\pi |<1$. Therefore $|x|=|\pi {|}^{v_{\mathfrak{p}}(x)}$.

Conversely each nonzero prime ideal of $A$ gives the Non-Archimedean absolute value $|x|=c^{v_{\mathfrak{p}}(x)}$ for any $0<c<1$, using the discrete valuation of $A_{\mathfrak{p}}$. All choices of $c$ are equivalent, and different prime ideals are distinguished by the elements of $A$ having absolute value on a field less than one. We have proved Non-Archimedean absolute values on a number field:The trivial class is additional when trivial absolute values on a field are admitted.

True. Each element of the Galois group preserves the extending absolute value on a field, because composing it with that absolute value on a field gives another extension, equal to the original by part (a). Thus all conjugates of $x$ have the same absolute value on a field. Writing $d=[K:{\mathbb{Q}}_p]$, the field norm givesA unit field norm has absolute value on a field one, forcing $|x|=1$. This means both $x$ and $x^{-1}$ are in the valuation ring, so $x$ is a unit.

Equivalently, with normalized discrete valuations and residue degree $f$,This is unit norm detects units in a local extension.

False. This is an example of integral trace need not imply integrality. Take $K={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the splitting field of $X^2-p$. Its two conjugates are $\theta ,-\theta$, and its normalized discrete valuation satisfies $v_K(\theta )=1$, $v_K(p)=2$, by the Eisenstein polynomial criterion. Forthe field trace is zero, but $v_K(x)=-1$. Therefore $x$ is not in the valuation ring, despite having integral field trace. Cancellation in a sum of conjugates explains why a field trace test fails.

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

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

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

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

For $L={\mathbb{Q}}_p({\zeta }_p)$, put $\pi ={\zeta }_p-1$. The polynomialis Eisenstein at $p$. It gives $[L:{\mathbb{Q}}_p]=e=p-1$, total ramification, and $v_L(\pi )=1$. Also ${\mathcal{O}}_L={\mathbb{Z}}_p[\pi ]$: in the power basis, terms $c_i{\pi }^i$ have valuations $(p-1)v_p(c_i)+i$ in different residue classes modulo $p-1$, so cannot cancel at the minimum. Integrality forces all $c_i$ integral.

Use part (b) with ${\Phi }_p(T)$ and differentiate the quotient at ${\zeta }_p$:The root of unity is a unit, while $v_L(p)=p-1$ and $v_L(\pi )=1$. Hence the cyclotomic local different exponent isThis includes $p=2$: that extension is trivial and the exponent is zero.

For $m=p^2-1$, construct an unramified quadratic extension $E/{\mathbb{Q}}_p$ by lifting an irreducible quadratic polynomial over ${\mathbb{F}}_p$. Its residue field is ${\mathbb{F}}_{p^2}$. Each of its nonzero residue elements is a simple polynomial root of $T^m-1$, so Hensel's lemma lifts all $m$ polynomial roots into $E$. A primitive residue element lifts to a primitive mth polynomial root: its reduction already has order $m$, and its mth power is one.