Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 26 / 2

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.

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.