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Let $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.

For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also $p-1$ and, with the stated real-index convention,As a consistency check, ${\pi }_L$ satisfies the Eisenstein polynomial $X^p-t^{p-1}X-t$. Its derivative in characteristic $p$ is $-t^{p-1}$, giving different exponent $p(p-1)$, equal to ${\sum }_{i\ge 0}(|G_i|-1)=p(p-1)$ from the different exponent from ramification groups.

Write $q=\#k_K=p^f$. For every $a\in k_K$, the polynomial $F(T)=T^q-T$ has $a$ as a residue root and $\overline{F^{\prime }}=-1$. By the Hensel lemma there is a unique element $[a]\in {\mathcal{O}}_K$ such thatThis defines the Teichmuller lift, including $[0]=0$. The two displayed properties determine each value uniquely by Hensel's uniqueness statement.

They also imply the standard Teichmuller lift properties. The elements $0,1$ already satisfy their defining equations, so $[0]=0$ and $[1]=1$. The product $[a][b]$ reduces to $ab$ and satisfies $([a][b]{)}^q=[a][b]$, so uniqueness gives $[ab]=[a][b]$. Likewise $[a{]}^p$ reduces to $a^p$ and is fixed by its $q$th power, givingFor $a\ne 0$, division of $[a{]}^q=[a]$ by $[a]$ gives $[a{]}^{q-1}=1$. Thus the nonzero Teichmuller representatives are precisely the prime-to-$p$ roots of unity lifting the nonzero residue classes. This is a multiplicative section of reduction; additivity is not asserted in mixed characteristic.

For the convergence, use the normalized discrete valuation $v_K$ with $v_K(\pi )=1$, and put $e=v_K(p)\ge 1$. A useful binomial estimate isIndeed, writing $u=v+\delta$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$, so all terms of $(v+\delta {)}^p-v^p$ have valuation at least $\min \{e+r,pr\}$. The ultrametric inequality gives the bound for their sum.

Finite fields are perfect, so the successive $p$th roots $x_n$ exist uniquely and satisfy $x_n^{p^n}=x$. Let $a_n=[x_n]$. Since $y_n-a_n$ lies in the maximal ideal, iterating the estimate $n$ times givesMultiplicativity, or the Frobenius identity above, gives $a_n^{p^n}=[x_n^{p^n}]=[x]$. ThereforeThis proves the approximation of Teichmuller lifts by iterated pth powers for completely arbitrary choices of the lifts $y_n$, including $x=0$. The original PDF confirms $x\in k_K$, correcting the capital $K_K$ in the TeX transcription.

Suppose first that the valuation is discrete and the residue field $k$ is finite. For a uniformizer $\pi$, every quotient ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$ is finite, and completeness givesThis inverse limit is compact. Since ${\mathcal{O}}_K$ is a compact neighborhood of zero, $K$ is locally compact.

Conversely, local compactness gives a compact ball about zero, which can be rescaled to make ${\mathcal{O}}_K$ compact. Its distinct residue classes are disjoint open balls of radius below one, so compactness forces the residue field to be finite. Cover ${\mathcal{O}}_K$ by finitely many balls of some radius $r<1$. Applying the ultrametric inequality to centers lying in the maximal ideal produces $\rho <1$ such that every nonunit has absolute value at most $\rho$. Hence the value group has a largest value below one, and the valuation is discrete. This proves the local compactness criterion for a complete non-Archimedean field.

An algebraically closed valued field has an $n$th root of every element. If its valuation were discrete and $\pi$ were a uniformizer, then $v(\sqrt[n]{\pi })=1/n$ would contradict discreteness. Therefore an algebraically closed non-Archimedean field cannot be locally compact.

One strong form of Hensel lemma is this: if $K$ is complete for a discrete valuation $v$, $f\in {\mathcal{O}}_K[X]$, and $a_0\in {\mathcal{O}}_K$ satisfiesthen there is a unique root $\alpha$ satisfying

Define the Newton iteration over a valued fieldTaylor expansion and the ultrametric inequality show inductively that $v(f^{\prime }(a_n))=v(f^{\prime }(a_0))$ andThus the valuations of the corrections $a_{n+1}-a_n$ tend to infinity, so $(a_n)$ is a Cauchy sequence. Completeness gives a limit $\alpha$, and continuity gives $f(\alpha )=0$. If $\beta$ is another root in the stated ball, Taylor expansion of $f(\beta )-f(\alpha )$ shows that the linear term has strictly smaller valuation than all higher terms unless $\beta =\alpha$, proving uniqueness.

Krasner's lemma says that if $K$ is complete, $\alpha$ is separable over $K$, andthen $K(\alpha )\subseteq K(\beta )$.

To prove it, let $\tau$ be an automorphism of $\overline{K}$ fixing $K(\beta )$. The absolute value has a unique extension to every finite extension of the complete field $K$, so it is invariant under $\tau$. If $\tau (\alpha )\ne \alpha$, the ultrametric inequality and the displayed strict inequality giveBut $\tau (\beta )=\beta$ and invariance gives $|\tau (\alpha )-\beta |=|\alpha -\beta |$, a contradiction. Every automorphism fixing $K(\beta )$ therefore fixes $\alpha$, which proves the field inclusion by Galois correspondence.

Now let $L/{\mathbb{Q}}_p$ be finite. By the primitive element theorem, write $L={\mathbb{Q}}_p(\alpha )$ with separable minimal polynomial $f\in {\mathbb{Q}}_p[X]$. Approximate the coefficients of $f$ closely by those of a polynomial $g\in \mathbb{Q}[X]$ of the same degree. Continuity of roots over a non-Archimedean field gives a root $\beta$ of $g$ arbitrarily close to $\alpha$. Choose it close enough for Krasner's lemma; thenThe degree bound from $\mathrm{deg}g=\mathrm{deg}f$ forces equality. If $F=\mathbb{Q}(\beta )$ and $\mathfrak{p}$ is the prime selected by the embedding $F\hookrightarrow {\overline{\mathbb{Q}}}_p$, its completion isThus every finite extension of ${\mathbb{Q}}_p$ is the completion of a number field at a prime ideal.

Assume first that $K$ is complete. If $x$ is integral over ${\mathcal{O}}_K$, its monic equation and the ultrametric inequality imply $|x{|}_L\le 1$, so $x\in {\mathcal{O}}_L$. Conversely, if $|x{|}_L\le 1$, part (i) gives $|\sigma (x){|}_L\le 1$ for every $K$-embedding $\sigma$. The coefficients of the minimal polynomial of $x$ are elementary symmetric polynomials in its conjugates, so they all lie in ${\mathcal{O}}_K$. Thus $x$ is integral over ${\mathcal{O}}_K$, proving the integral closure in a finite extension of a complete discretely valued field identity ${\mathcal{O}}_L={\overline{{\mathcal{O}}_K}}^L$.

Completeness is necessary. Give $\mathbb{Q}$ its $5$-adic absolute value, take $L=\mathbb{Q}(i)$, and choose the extension corresponding to the prime $(2+i)$ above $5$. Thenhas nonnegative valuation at $(2+i)$, so it belongs to the chosen valuation ring ${\mathcal{O}}_L$. At the conjugate prime $(2-i)$ it has negative valuation, so it does not lie in the integral closure of ${\mathbb{Z}}_{(5)}$ in $\mathbb{Q}(i)$. Hence the chosen valuation ring can be strictly larger than the integral closure when the base field is not complete.

Suppose $x\notin {\mathcal{O}}_K$, so $v(x)<0$. The ultrametric inequality gives $v(1+x)=v(x)$. If $1+x=y^n$, thenFor a discrete valuation, only finitely many positive integers $n$ divide the fixed nonzero integer $v(x)$. Therefore $x$ cannot belong to $S$, proving $S\subseteq {\mathcal{O}}_K$.

One direction follows by restriction. Conversely, suppose $|\cdot |$ is a Non-Archimedean absolute value. Then $|m{|}_L=|m|\le 1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality giveTaking $n$th roots and letting $n\to \infty$ yields the ultrametric inequality for $|\cdot {|}_L$. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives

This can be false: take $y=-x\ne 0$. Then $|x+y|=0<|x|=\min (|x|,|y|)$. The upper inequality is always the ultrametric inequality.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

The ultrametric inequality, also called the strong triangle inequality, bounds a direct distance by the larger of the two distances through an intermediate point:A metric space satisfying this rule is an ultrametric space. For a Non-Archimedean absolute value, the corresponding inequality is $|a+b|\le \max \{|a|,|b|\}$, and the associated metric is $d(x,y)=|x-y|$. The ordinary triangle inequality follows because the maximum of two nonnegative real numbers is at most their sum.

For example, the 2-adic absolute value gives the three points $0,1,8$ distancesEach distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an ultrametric space has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the ultrametric inequality. The drawn lengths of a triangle on paper need not represent these distances.

The ultrametric inequality gives a useful exact rule when $|a|\ne |b|$:To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\le |a|$; applying the ultrametric inequality to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1{|}_2=|1{|}_2=1$, but $|1+1{|}_2=1/2$. This makes a uniquely largest term decisive in a sum, a principle used in the Newton polygon and in arguments about roots of a polynomial.

An ultrametric space is a metric space whose metric satisfies the ultrametric inequality. The rational numbers with $d_p(x,y)=|x-y{|}_p$, and their P-adic number completion, are examples. Their geometry organizes points into nested clusters rather than measuring separation along a number line.

For integers $a,b$ and $k\ge 0$, the definition of the P-adic valuation givesThus a closed ball of radius $p^{-k}$ in the integers is a congruence class modulo $p^k$. In the p-adic integers the same closed ball is $a+p^k{\mathbb{Z}}_p$. More shared low-order digits mean a smaller p-adic absolute value of the difference.

The binary tree first splits the sixteen integers into even and odd congruence classes, then refines modulo $4$, $8$, and $16$. Each node fixes one more low-order digit. The leaves $5$ and $13$ share the class $5$ modulo $8$, then separate modulo $16$, so their distance is exactly $1/8$. The leaves are deliberately ordered by their low-order digits; horizontal spacing and edge lengths are layout choices, not the metric. The singleton leaves describe this finite sample; a congruence class modulo $16$ in all p-adic integers remains infinite.

The ultrametric inequality also explains the unusual behavior of balls. Every point of an open ball can serve as its center: if $d(a,b)<r$, then $d(a,x)<r$ if and only if $d(b,x)<r$. Two open balls that intersect are nested, since a point of their intersection serves as a common center and their radii can be compared. The same nesting rule holds for closed balls and for a pair consisting of one open ball and one closed ball.

Every open ball is a clopen set. For a point $y$ outside $B(a,r)$, the open ball $B(y,d(a,y))$ stays outside $B(a,r)$: if $d(y,z)<d(a,y)$, the ultrametric inequality forces $d(a,z)=d(a,y)\ge r$. Hence the complement is an open set. Also every closed ball of positive radius is an open set, because $B(b,r)$ lies in the closed ball of radius $r$ around $a$ whenever $d(a,b)\le r$. A radius-zero closed ball is a singleton and need not be an open set. Distinct points can therefore be separated by a clopen set, proving that an ultrametric space is totally disconnected. These ball properties turn distance thresholds into a hierarchy of compatible clusters. ^[ref]