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The Mahler coefficients tend to zero, so the displayed series defines a continuous function and has value zero at zero. The Pascal's identity and uniform convergence give $\Delta Jf=f$. Thus the forward difference operator is surjective on continuous functions on the p-adic integers. Its kernel consists of constants: period one implies agreement on the dense nonnegative integers, and continuity then implies constancy. This selects the unique discrete antiderivative vanishing at zero.

There is also a direct construction on locally constant functions. If $h$ has period $M=p^r$, put $a_j=h(j)$, $S={\sum }_{i=0}^{M-1}{a}_i$, and $P_j={\sum }_{i<j}{a}_i$. For $x=j+My$ with $0\le j<M$ and $y\in {\mathbb{Z}}_p$, setThis is a continuous function on each residue class. Increasing $j$ gives $Jh(x+1)-Jh(x)=a_j$; at $j=M-1$, the next point has residue zero and quotient $y+1$, giving the same identity. The ultrametric inequality gives $\Vert Jh{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$, and $Jh(0)=0$. The construction is independent of the chosen period, since two normalized discrete antiderivatives agree on the nonnegative integers and then on the p-adic integers by continuity. It is linear on the locally constant functions, which form a dense subset in the supremum norm. Completeness therefore extends it to every continuous function, retaining $\Delta J=I$ and the norm bound. This proves surjectivity without first using the Mahler theorem.

Let $K$ be complete for a Non-Archimedean absolute value and let $V$ be a finite-dimensional vector space with a norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$ and the ultrametric inequality. All such norms are equivalent to the maximum of coordinates in any basis. The upper bound is immediate. For the lower bound, induct on dimension. The span $W$ of all but the last basis vector is complete by induction, hence closed for the given norm. Thus $\delta ={\inf }_{w\in W}\Vert e_n-w\Vert >0$. For $v=w+a{e}_n$, $\Vert v\Vert \ge |a|\delta$; also $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e_n\Vert )$. The induction bound on coordinates of $w$ now bounds all coordinates by a constant times $\Vert v\Vert$. This proof needs no local compactness.

An absolute value on a field is non-Archimedean if and only if $|n\cdot 1|\le 1$ for every positive integer $n$. Necessity follows from the ultrametric inequality. Conversely the binomial theorem gives $|x+y{|}^m\le (m+1)\max (|x|,|y|{)}^m$, since every binomial coefficient has absolute value at most one. Take $m$th roots and let $m\to \infty$. In particular an extension of an absolute value that is non-Archimedean stays non-Archimedean.

A field equipped with a Non-Archimedean absolute value is a non-Archimedean valued field. Its metric is $d(x,y)=|x-y|$ and satisfies the ultrametric inequality. The P-adic numbers give the basic mixed-characteristic examples. Completeness means that every Cauchy sequence for this metric has a limit in the field.

The sets $R=\{x:|x|\le 1\}$ and $\mathfrak{m}=\{x:|x|<1\}$ are its valuation ring and maximal ideal. Multiplicativity and the strong triangle inequality prove that $R$ is a ring and $\mathfrak{m}$ is an ideal. A nonzero element of $R$ is invertible in $R$ exactly when its absolute value is one, since $|x^{-1}|=|x{|}^{-1}$. Thus all nonunits are precisely $\mathfrak{m}$, proving that it is the unique maximal ideal and that $R/\mathfrak{m}$ is a field. These definitions concern a valuation, independently of whether the field admits an ordering.

In a complete Non-Archimedean absolute value field, a series converges if and only if its terms tend to zero: the ultrametric inequality bounds every finite tail by its largest term. The displayed sequence tends to zero in ${\mathbb{Q}}_p$. But each block $p^k\le n<p^{k+1}$ contributes $p-1$ to the real sum of its p-adic absolute values, so that real sum diverges.

The p-adic metric on the P-adic numbers is induced by the p-adic absolute value. Its ultrametric inequality is $d_p(x,z)\le \max (d_p(x,y),d_p(y,z))$. On p-adic integers, digit agreement and p-adic distance expresses metric precision as shared low-order digits.

Choose the blocksIf $p^k\le n<p^{k+1}$, then $|x_n{|}_p=p^{-k}$, so $x_n\to 0$ in ${\mathbb{Q}}_p$. By the ultrametric inequality, for $N\ge M$,The partial sums are Cauchy sequence, and completeness of ${\mathbb{Q}}_p$ proves convergence of $\sum x_n$.

In the real series of p-adic absolute values, the block $p^k\le n<p^{k+1}$ contains $(p-1)p^k$ equal terms of size $p^{-k}$. Its sum is $p-1$, independently of $k$. Infinitely many such positive block sums force divergence. This exhibits p-adic convergence without summable absolute values: the criterion is that the terms tend to zero, rather than real absolute summability.

The constant and linear Taylor terms cancel in the Newton step, leavingThus residuals strictly decrease whenever the current residual is nonzero, and the steps tend to zero doubly exponentially. By the ultrametric inequality, the distance between any two sufficiently late iterates is bounded by the largest intervening step. The sequence is Cauchy and converges in the complete valuation ring to $a$. Continuity gives $f(a)=0$, andThe literal infinite chain of strict inequalities needs the root-hit exception: if $f(a_0)=0$, or a later iterate is already a root, it cannot keep strictly decreasing below zero. The construction is eventually constant in that case; all existence and distance conclusions remain valid.

For the equality case, suppose $|f(a_0)|=d^2$ with $0<d<1$. The first Newton step has norm $|h_0|=d$. Write the Taylor coefficient of $h^2$ as $c_2(a_0)=f^{\prime \prime }(a_0)/2$. Its norm is strictly below one by hypothesis; all higher coefficients have norm at most one. HenceThe derivative changes by $2c_2(a_0)h_0+O(h_0^2)$, of norm strictly below $d$, so it still has norm $d$. The strict Hensel argument applies from this first iterate onward, and the root remains within distance $d=|f(a_0)|/d$ of $a_0$. If characteristic two is admitted, interpret the displayed half-second-derivative as the second Taylor coefficient ${\sum }_j\left(\genfrac{}{}{0ex}{}{j}{2}\right)c_j{b}^{j-2}$; this is the division-free form of the required hypothesis.

For the final local equation put $F(t)=t^3+3t$. Its values modulo nine are $0,4,5$, so the sum of two such values is confined to $0,1,4,5,8$. Conversely choose initial pairs for these five classes asrespectively. Fix the second coordinate $y_0$ and apply the equality-case Hensel result to $g(X)=F(X)+F(y_0)-n$ at the first coordinate $x_0$. Its residual is divisible by nine, $g^{\prime }(x_0)=3(x_0^2+1)$ has 3-adic norm $1/3$, and $g^{\prime \prime }(b)/2=3b$ has norm below one for every $b\in {\mathbb{Z}}_3$. The strict or equality case therefore supplies a root in ${\mathbb{Z}}_3$. Thusis the necessary and sufficient condition.

Let $d=|f^{\prime }(a_0)|>0$ and $q_0=|f(a_0)|/d^2<1$. At any step before reaching a root, setAssuming inductively that $|f^{\prime }(a_i)|=d$ and $q_i=|f(a_i)|/d^2<1$, this givesAll derivatives of a polynomial over $R$ are integral on $R$, so $d\le 1$. Hence $|h_i|<1$, and the ultrametric inequality keeps $a_{i+1}$ in $R$. If a root is reached, retain that value for subsequent terms, and the asserted weak step inequality still holds.

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.

A Non-Archimedean absolute value on a field is a function $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ with $|x|=0$ exactly for $x=0$, $|xy|=|x||y|$, and the ultrametric inequality $|x+y|\le \max (|x|,|y|)$. It is nontrivial if some nonzero element has value different from one. Equivalent absolute values induce the same topology; for nontrivial absolute values this is equivalent to $|x{|}_2=|x{|}_1^c$ for some $c>0$.

Here is also a proof of the latter equivalence in the case needed. A positive power plainly preserves the open neighborhoods of zero. Conversely, if the topologies coincide, then $|x{|}_1<1$ if and only if $x^n\to 0$ in the first topology, hence if and only if $|x{|}_2<1$. Choose $a$ with $0<|a{|}_1<1$. For every $x\ne 0$, $m\in \mathbb{Z}$ and $n\ge 1$, whether $|x^n/a^m|<1$ agrees for the two values. Comparison with all rational numbers $m/n$ therefore identifies the two ratios $\log |x{|}_j/\log |a{|}_j$. Thus $|x{|}_2=|x{|}_1^c$ with $c=\log |a{|}_2/\log |a{|}_1>0$.

Now let $|\cdot |$ be nontrivial and non-Archimedean on $\mathbb{Q}$. Since $|1|=1$ and every positive integer is a sum of ones, $|n|\le 1$ for every integer. If all nonzero integers had value one, multiplicativity would make the absolute value trivial on all rationals. Hence some integer has value less than one; its prime factorization shows that some prime $p$ has $|p|<1$.

There cannot be two distinct such primes $p,q$. A Bezout identity $ap+bq=1$ with $a,b\in \mathbb{Z}$ would giveThus every other prime has value one. Prime factorization of the numerator and denominator of $x\in {\mathbb{Q}}^{\times }$ now givesConversely $|x{|}_p=p^{-v_p(x)}$ is multiplicative and satisfies the ultrametric inequality, since $v_p(x+y)\ge \min (v_p(x),v_p(y))$. Distinct primes give inequivalent values: $p^m\to 0$ for the $p$-adic absolute value but $|p^m{|}_q=1$ for $q\ne p$. We have proved the non-Archimedean part of Ostrowski theorem:There is exactly one equivalence class for each prime.

The two cosets of $2{\mathbb{Z}}_2$ are clopen sets in the 2-adic integers. The function is constant on each, so it is a locally constant function and therefore continuous. Its values on the nonnegative integers are $f(j)=(-1{)}^j$.

The Mahler coefficient at index $n$ is the nth forward difference operator applied at zero. HereThus the Mahler expansion of the parity function isTo verify the expansion directly, each binomial polynomial $\left(\genfrac{}{}{0ex}{}{x}{n}\right)$ takes values in ${\mathbb{Z}}_2$: it does so at nonnegative integers, which form a dense subset, and polynomial evaluation is continuous into ${\mathbb{Q}}_2$. Hence the nth summand has 2-adic absolute value at most $2^{-n}$ throughout the domain. The ultrametric inequality proves uniform convergence, and the sum is continuous. At each nonnegative integer $m$, the series becomes the finite binomial theorem identity $(1-2{)}^m=(-1{)}^m$. Equality on a dense subset and continuity prove equality everywhere. Finally, the coefficients in any such expansion are determined successively by its values at $0,1,2,\dots$, since $\left(\genfrac{}{}{0ex}{}{m}{n}\right)=0$ for $n>m$ and $\left(\genfrac{}{}{0ex}{}{m}{m}\right)=1$. This also verifies uniqueness without needing the general Mahler theorem.

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.

Two absolute values on a field are equivalent absolute values when they induce the same topology. For nontrivial absolute values on a field this is equivalent tofor a fixed real number $c>0$. The trivial absolute value is equivalent only to itself.

Restriction of a Non-Archimedean absolute value plainly preserves the ultrametric inequality. Conversely, if the restriction is non-Archimedean, every integer, interpreted in either field, has absolute value on a field at most one. In particular every binomial coefficient has absolute value on a field at most one. The triangle inequality applied to the binomial expansion gives, for $x,y\in L$ and every positive integer $N$,Raise both sides to the power $1/N$ and let $N\to \infty$. Since $(N+1{)}^{1/N}\to 1$, this yieldsThis is the bounded-integer criterion for a non-Archimedean absolute value. The proof works for arbitrary field extensions, not just algebraic ones, and in positive characteristic as well.

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.

An extending absolute value on a field is non-Archimedean by Question 1(a). We prove finite-dimensional non-Archimedean norm equivalence over a complete field, using completeness rather than local compactness.

Fix a basis $e_1,\dots ,e_d$ of a finite-dimensional vector space over $K$, and let $\Vert \cdot \Vert$ be a norm homogeneous for the given absolute value on a field and satisfying the ultrametric inequality. Write $\Vert \sum a_i{e}_i{\Vert }_0={\max }_i|a_i|$. The upper boundis immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\dots ,e_{d-1}$ is complete for its restricted norm by the induction comparison with the coordinate norm and completeness of $K$. A complete vector subspace of a metric space is closed, soFor $v=w+a{e}_d$ with $a\ne 0$, scaling by $a$ gives $\Vert v\Vert \ge |a|\delta$; the same coefficient bound is trivial when $a=0$. AlsoThe induction bound controls every coordinate of $w$ by a constant times $\Vert w\Vert$. Together with the bound on $a$, this proves $\Vert v{\Vert }_0\le C^{\prime }\Vert v\Vert$.

Two extending absolute values on a field on $L$ are norms of this kind on its finite-dimensional vector space over $K$. Hence there is $B>0$ with $|x{|}_2\le B|x{|}_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:Letting $N\to \infty$ gives $|x{|}_2\le |x{|}_1$. Interchanging the two norms gives equality. ThusThe proof covers the trivial base absolute value on a field too. It proves uniqueness, without assuming an existence theorem or compactness of the coordinate sphere $\{v:\Vert v{\Vert }_0=1\}$.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfyingA Non-Archimedean absolute value satisfies the stronger ultrametric inequality $|x+y|\le \max (|x|,|y|)$. Two equivalent absolute values induce the same topology, or equivalently differ by a positive real power. The trivial absolute value on a field takes value one on every nonzero element. The rational classification and the compactness criterion below concern nontrivial absolute values on a field; the trivial exceptions are given explicitly.

Here the additive valuation has real values. For any $c>1$, the mutually inverse constructions areMultiplicativity becomes $v(xy)=v(x)+v(y)$, and the ultrametric inequality becomes $v(x+y)\ge \min (v(x),v(y))$. Equivalent real-valued valuations differ by positive scaling, so these constructions give the required bijection on equivalence classes. Changing $c$ merely rescales the valuation.

To justify the topology formulation, $|a|<1$ is equivalent to $a^n\to 0$. Thus two nontrivial absolute values on a field with the same topology give the same strict positivity relation on their additive valuations. Fix $t$ with $v_1(t)>0$. Comparing the signs of $m{v}_j(t)-n{v}_j(x)=v_j(t^m{x}^{-n})$, for integers $m$ and positive integers $n$, shows that $v_1(x)/v_1(t)$ and $v_2(x)/v_2(t)$ have identical rational cuts. They are equal, proving $v_2=c{v}_1$ for $c>0$. If one allows valuations in arbitrary ordered groups, the nontrivial classes arising this way are precisely rank-one valuations: higher-rank ordered value groups do not embed order-preservingly in $\mathbb{R}$.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, $|n|\le 1$ for every integer $n$, by repeatedly applying the ultrametric inequality to sums of ones. Some prime $p$ must have $|p|<1$, otherwise prime factorization and multiplicativity would make every nonzero rational have value one. There is at most one such prime: if both $|p|,|q|<1$, a Bezout identity $ap+bq=1$ contradicts the ultrametric inequality. If $p\nmid m$, another Bezout identity gives $|m|=1$. HenceThis proves the non-Archimedean part of the Ostrowski theorem. If the trivial absolute value on a field is admitted, it supplies one additional class and is not equivalent to any p-adic absolute value.

The valuation ring, its maximal ideal, and its residue field areSuppose the absolute value on a field is nontrivial and $R$ is compact. The ideal $\mathfrak{m}$ is an open additive subgroup of $R$, so $k$ is discrete; as a continuous image of a compact space it is finite. The subgroup $\mathfrak{m}$ is also closed, since all its cosets are open, and is therefore compact. The continuous function $|\cdot |$ attains a maximum $\rho$ on $\mathfrak{m}$, with $0<\rho <1$. Choose $\pi$ with $|\pi |=\rho$. Then the positive values of $v=-\log |\cdot |$ have least element $-\log \rho$. Division with remainder in this additive subgroup of $\mathbb{R}$ proves $v(K^{\times })=v(\pi )\mathbb{Z}$. Thus $K$ is a discretely valued field and $\pi$ is a uniformizer.

Conversely, normalize the discrete valuation by $v(\pi )=1$. If $k$ has $q$ elements, $R/{\pi }^{N}R$ has $q^N$ elements. Each quotient therefore supplies a finite cover by balls of radius tending to zero. The ring $R$ is a closed subset of the complete metric field $K$, hence complete and totally bounded, so it is compact. Equivalently,This is the local compactness criterion for a complete non-Archimedean field, with nontriviality understood. For the trivial absolute value on a field, $R=K$ has the discrete topology and is compact exactly when $K$ is a finite field; its value group is zero rather than a nonzero discrete cyclic group.

For $q\in \mathbb{Q}\setminus \{0\}$ write $q=p^{k}u/v$, where $k\in \mathbb{Z}$ and neither integer $u$ nor integer $v$ is divisible by $p$. Define the P-adic valuation $v_p(q)=k$ and the p-adic absolute value $|q{|}_p=p^{-k}$, with $|0{|}_p=0$. The p-adic metric isFactoring out the smaller power of $p$ from two rational numbers shows $v_p(q+r)\ge \min \{v_p(q),v_p(r)\}$, with the usual convention $v_p(0)=+\infty$. Consequently the ultrametric inequality givesTogether with symmetry, positivity and separation of points, this verifies that $d_p$ is a metric.

For the given geometric partial sums, use the finite geometric series identity. Since $p$ does not divide $p-1$,Thus the limit is the rational number $\boxed{1/(1-p)}$, even though the partial sums diverge in the ordinary real metric.

If $|a{|}_p>|b{|}_p$, the ultrametric inequality gives $|a+b{|}_p\le |a{|}_p$. Applying it to $a=(a+b)-b$ gives $|a{|}_p\le \max \{|a+b{|}_p,|b{|}_p\}$. Since $|b{|}_p<|a{|}_p$, this forces $|a+b{|}_p=|a{|}_p$. Interchanging $a,b$ handles the other unequal case, provingFinally, let $b$ be outside the open ball $B(a,\delta )$, so $d_p(a,b)\ge \delta$. If $d_p(b,c)<\delta$, the preceding unequal-norm identity gives $d_p(a,c)=d_p(a,b)\ge \delta$. Therefore $B(b,\delta )$ also lies outside $B(a,\delta )$. Its complement is open, so every p-adic open ball is also closed, that is, a clopen set.

A suitable version of the Hensel lemma is this: if $R$ is a complete discrete valuation ring, $h\in R[X]$, and $\overline{a}$ is a root of $\overline{h}$ with ${\overline{h}}^{\prime }(\overline{a})\ne 0$, there is a unique $a\in R$ reducing to $\overline{a}$ with $h(a)=0$. Starting with any lift, Newton iteration over a valued field $a_{j+1}=a_j-h(a_j)/h^{\prime }(a_j)$ keeps the derivative a unit and doubles the error valuation, giving convergence. Uniqueness follows by factoring $h(u)-h(v)$ when two roots have the same residue.

The field in the rest of this question is not assumed complete, so that version cannot simply be applied to it. Instead use the stated uniqueness of the extension of an absolute value. Let ${\alpha }_i,{\alpha }_j$ be any two roots of a polynomial $f$, which is irreducible. The field embedding sending ${\alpha }_i$ to ${\alpha }_j$ identifies $K({\alpha }_i)$ with $K({\alpha }_j)$. Pulling the latter field's absolute value back gives an extension on $K({\alpha }_i)$, which must coincide with the given one. More generally, for every $h\in K[X]$,This proves equal absolute values of algebraic conjugates without assuming separability or a transitive action of a Galois group; repeated roots cause no difficulty.

Every root of the monic $f\in {\mathcal{O}}_K[X]$ has absolute value at most one. Indeed, if $|\alpha |>1$, then each lower term satisfies $|c_j{\alpha }^j|\le |\alpha {|}^j<|\alpha {|}^n$. The ultrametric inequality makes the leading term dominate their sum, contradicting $f(\alpha )=0$. Thus

All roots lie in the valuation ring ${\mathcal{O}}_M$, so their residues are defined in the residue field $k_M$. Let $\phi \in k_K[X]$ be the monic minimal polynomial of an algebraic element of ${\overline{\alpha }}_1$; this residue is algebraic because it satisfies the monic $\overline{f}$. Choose a coefficientwise lift $\Phi \in {\mathcal{O}}_K[X]$. Then $|\Phi ({\alpha }_1)|<1$, and the conjugacy argument gives $|\Phi ({\alpha }_i)|<1$ for every $i$. Hence every ${\overline{\alpha }}_i$ is a root of $\phi$.

Reduction of the splitting-field factorization, with multiplicities, givesEvery monic irreducible factor of $\overline{f}$ over $k_K$ has a root among these residues. That root also satisfies $\phi$, so its minimal polynomial is $\phi$. Unique factorization therefore proves the pure-power reduction of a monic irreducible polynomial:This does not assert that $\phi$ is separable or that the reduction is square-free.

Finally, factor the monic $g$ over $K$ as $g={\prod }_j{h}_j^{e_j}$ with distinct monic irreducible polynomials $h_j$. The same leading-term argument bounds every root of $g$ by one. Coefficients of any $h_j$ are elementary symmetric polynomials in some of these roots, so they lie in ${\mathcal{O}}_K$. By the preceding result each ${\overline{h}}_j$ is a power of a single monic irreducible residue polynomial ${\phi }_j$.

Since the prescribed ${\overline{g}}_1,{\overline{g}}_2$ are coprime, every ${\phi }_j$ occurs on exactly one side. Assign the entire factor $h_j^{e_j}$ to that side and form the two products. Comparing irreducible-factor multiplicities in the reduction proves the coprime factor lifting from valuation-extension uniqueness:Empty products are one, covering constant prescribed factors. The construction proves the required factor-lifting form of the Hensel lemma directly from extension uniqueness, without adding completeness as an unstated hypothesis.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.