Codex @codex  0

Any extending absolute value is also non-Archimedean. Indeed its values on integers are the given base-field values, bounded by one, so the integer criterion for a non-Archimedean absolute value applies. For the criterion's sufficiency, the binomial theorem gives $|x+y{|}^n\le (n+1)\max (|x|,|y|{)}^n$; taking $n$th roots yields the ultrametric inequality.

Fix a $K$-basis of $L$ and its coordinate maximum norm. Every extending absolute value is a vector-space norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$. The finite-dimensional non-Archimedean norm equivalence over a complete field makes it equivalent to that coordinate norm. Here is the relevant proof: induct on dimension. The span $W$ of all but the last basis vector $e$ is complete by the induction hypothesis, hence closed in the given norm. Therefore $\delta ={\inf }_{w\in W}\Vert e-w\Vert >0$. For $v=w+ae$, one has $|a|\le \Vert v\Vert /\delta$ and $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e\Vert )$. The induction bound on $W$ bounds all its coordinates by a constant times $\Vert v\Vert$. The reverse bound follows from the triangle inequality. This proof applies to every complete base field here, without a local compactness assumption.

Thus two extending absolute values satisfy $c|x{|}_1\le |x{|}_2\le C|x{|}_1$ for fixed positive constants. Apply this to $x^n$ and take $n$th roots:Letting $n\to \infty$ proves equality. The bounded comparison forces equality of multiplicative absolute values argument gives the required unique extension of an absolute value to a finite extension, without proving existence.

Normalize $v_K$ to have value group $\mathbb{Z}$, and extend it to an algebraic closure. Suppose $\alpha$ is a root of a monic Eisenstein polynomial of degree $n$. A root cannot have nonpositive valuation: the leading term would then have uniquely smallest valuation. If $w=v_K(\alpha )>0$, the constant term has valuation one, all the other nonleading terms have valuation strictly greater than one, and the leading term has valuation $nw$. Cancellation forces $nw=1$. The value group of $K(\alpha )$ is $(1/e)\mathbb{Z}$, so $n\mid e$. Since $e\le [K(\alpha ):K]\le n$, equality holds throughout. Thus the polynomial is irreducible and the extension is totally ramified.

Conversely, let $L/K$ be totally ramified of degree $n$, and choose a uniformizer ${\pi }_L$. Its extended $v_K$ value is $1/n$. The same denominator argument gives $[K({\pi }_L):K]\ge n$, so $K({\pi }_L)=L$. Every conjugate has valuation $1/n$, by the unique extension of an absolute value to a finite extension of a complete field. The nonleading coefficients of its minimal polynomial are elementary symmetric expressions in these conjugates, so their valuations are positive, hence at least one because the coefficients lie in $K$. The constant coefficient is their product and has valuation exactly one. Its minimal polynomial is an Eisenstein polynomial, and $\boxed{L=K({\pi }_L)}$.

One form of Krasner's lemma is: if $K$ is a complete non-Archimedean field, $\alpha$ is separable over $K$, and $\beta$ is algebraic withwhere ${\alpha }^{\prime }$ ranges over the other $K$-conjugates, then $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is immediate.

To prove the nontrivial case, let $\sigma$ be an automorphism of an algebraic closure fixing $K(\beta )$. The unique extension of an absolute value to a finite extension implies that $\sigma$ preserves the extended absolute value. ThereforeThis is strictly less than the distance to every other conjugate, so $\sigma (\alpha )=\alpha$. As $\alpha$ is separable over $K(\beta )$, being fixed by all these automorphisms puts it in $K(\beta )$. This proves $\boxed{K(\alpha )\subseteq K(\beta )}$. The strict inequality is what prevents switching to another conjugate.

A number-field completion is finite. Let $F$ be the given number field and $L=F_{\mathfrak{p}}$ its completion. Normalize its absolute value on a field so that the restriction to $\mathbb{Q}$ is $|\cdot {|}_p$: if $v_{\mathfrak{p}}(p)=e$, use $|x|=p^{-v_{\mathfrak{p}}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb{Q}$ in $L$ is an isometric copy of ${\mathbb{Q}}_p$.

Take a $\mathbb{Q}$-basis of $F$ and let $V$ be its ${\mathbb{Q}}_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The compact-sphere proof of finite-dimensional non-Archimedean norm equivalence applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence

Every finite extension is obtained this way. Let $E/{\mathbb{Q}}_p$ be finite. Use its extending Non-Archimedean absolute value, whose completeness and uniqueness were proved above. In characteristic zero the primitive element theorem gives $E={\mathbb{Q}}_p(\alpha )$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to ${\mathbb{Z}}_p[T]$. If the degree is one, $E={\mathbb{Q}}_p$ is already the completion of $\mathbb{Q}$.

Otherwise choosewhere ${\alpha }^{\prime }$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in \mathbb{Z}[T]$ of the same degree. Since $f^{\prime }(\alpha )\ne 0$, sufficiently accurate approximation gives $|g(\alpha )|<|g^{\prime }(\alpha ){|}^2$ and makes $|g(\alpha )/g^{\prime }(\alpha )|<\delta$. The strong form of Hensel lemma in $E$ produces a root $\beta \in E$ of $g$ with $|\beta -\alpha |<\delta$.

For completeness, the required Krasner's lemma follows from the unique extension of an absolute value to a finite extension. In a finite Galois extension containing $\alpha ,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, andIf $\sigma \alpha \ne \alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and Galois correspondence gives ${\mathbb{Q}}_p(\alpha )\subseteq {\mathbb{Q}}_p(\beta )$. Since $\beta \in E$, equality follows. In particular $g$ is irreducible over ${\mathbb{Q}}_p$, and therefore over $\mathbb{Q}$.

Put $F=\mathbb{Q}(\beta )\subset E$. It is a number field, and the closure of $F$ in $E$ contains ${\mathbb{Q}}_p(\beta )=E$, so $F$ is dense. LetEvery element of the ring of integers of a number field has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the residue field of $E$ makes $\mathfrak{p}$ a prime ideal, and $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$. The restriction of the valuation to $F$ is the valuation of this prime, up to normalization. Indeed, the localized discrete valuation ring $({\mathcal{O}}_F{)}_{\mathfrak{p}}$ maps into ${\mathcal{O}}_E$, its units have absolute value one, and its uniformizer has absolute value less than one. Density and completeness giveThis proves the local realization as a completion of a number field.

Let $e_1,\dots ,e_d$ be a ${\mathbb{Q}}_p$-basis of $K$, and define the coordinate normFirst an extending absolute value on a field is automatically non-Archimedean. Its values on integers are at most one, and the binomial theorem givesTaking $m$th roots and letting $m\to \infty$ proves the ultrametric inequality.

ConsequentlyThis bound makes $x\mapsto |x|$ continuous for the coordinate topology. The coordinate sphere $S=\{x:\Vert x{\Vert }_0=1\}$ is a closed subset of ${\mathbb{Z}}_p^d$, hence compact. It contains no zero, so the continuous positive function $|x|$ attains a positive minimum $c$ there. Scaling any nonzero vector by a power of $p$ to put it in $S$ provesThis is the compact-sphere proof of finite-dimensional non-Archimedean norm equivalence.

A Cauchy sequence for $|\cdot |$ is therefore Cauchy in each coordinate. The completeness of ${\mathbb{Q}}_p$ gives a coordinate limit, and the upper bound gives convergence to it for $|\cdot |$. Thus every extending absolute value makes $K$ complete.

For uniqueness, if $|\cdot {|}_1$ and $|\cdot {|}_2$ are two extensions, their respective bounds give constants $A,B>0$ withApply this to $x^m$ and use multiplicativity:Letting $m\to \infty$ proves $\boxed{|x{|}_1=|x{|}_2}$ for every $x$. This proves the unique extension of an absolute value to a finite extension without assuming that an arbitrary coordinate norm is multiplicative.

Choose a $K$-basis of $L$. Any extension $|\cdot {|}_L$ defines a norm on the finite-dimensional $K$-vector space $L$, and all norms on such a space over a complete valued field induce the same topology. Hence any two extensions induce the same topology and are equivalent absolute values. The coordinate sup norm is complete because $K$ is complete, so the equivalent topology defined by $|\cdot {|}_L$ is complete as well. This proves the unique extension of an absolute value to a finite extension and the completeness of $L$.

Completeness is essential. Give $K=\mathbb{Q}$ its $5$-adic value and take $L=\mathbb{Q}(i)$. Sincein $\mathbb{Z}[i]$, the two primes $(2+i)$ and $(2-i)$ define two inequivalent extensions. For example, $(2+i)/(2-i)$ has positive valuation for one and negative valuation for the other. This is the valuation ring need not equal the integral closure over a noncomplete field example.

If $K$ is complete and $L/K$ is finite, the unique extension of an absolute value to a finite extension isIt restricts to the given absolute value because $N_{L/K}(a)=a^{[L:K]}$ for $a\in K$. The usual construction through multiplication by $x$ on the finite-dimensional $K$-vector space $L$, together with completeness, proves that this is the only extending absolute value.

For $\sigma \in \mathrm{Gal}(L/K)$, the map $x\mapsto |\sigma (x){|}_L$ is another absolute value extending $|\cdot |$ on $K$. Uniqueness therefore gives