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Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$. The partial sums $A_N={\sum }_{n=0}^N{p}^{2n}{\zeta }_{p^{n+1}}$ form a Cauchy sequence. Each $A_N$ generates ${\mathbb{Q}}_p({\zeta }_{p^{N+1}})$: for a nonidentity cyclotomic automorphism with $r=v_p(a-1)$, the first changed term has valuation $2r+1/(p-1)$, strictly below every later changed term. Thus the conjugate distance is at least $p^{-2N-1/(p-1)}$. The tail bound is $p^{-2N-2}$, strictly smaller. If the limit were algebraic, Krasner's lemma would place every one of these unbounded-degree cyclotomic fields in one finite extension, a contradiction.

The completion ${\mathbb{C}}_p$ of an algebraic closure of ${\mathbb{Q}}_p$ is an algebraically closed field. Approximate a polynomial over ${\mathbb{C}}_p$ by one over ${\overline{\mathbb{Q}}}_p$, use continuity of roots over a non-Archimedean field to obtain a nearby algebraic root, and then use Krasner's lemma to show that the original root already belongs to ${\mathbb{C}}_p$.

A p-adic field has only finitely many extensions of a fixed degree up to isomorphism. There are finitely many possible residue degrees and a unique unramified extension of each such degree. Over each maximal unramified subfield, a totally ramified extension is generated by a root of an Eisenstein polynomial. The coefficient space of these polynomials of fixed degree is compact. Hensel lemma and Krasner's lemma show that sufficiently close polynomials generate isomorphic extensions, giving a finite cover by neighborhoods of a constant extension type.

Every finite extension of ${\mathbb{Q}}_p$ is isomorphic to the completion $F_{\mathfrak{p}}$ of some number field $F$ at a prime $\mathfrak{p}$ above $p$. Approximate a primitive element's minimal polynomial by a polynomial over $\mathbb{Q}$ and use Krasner's lemma to preserve the generated local field.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$, with ${\zeta }_{p^{n+2}}^p={\zeta }_{p^{n+1}}$. Use the slightly faster-converging series permitted by “or otherwise”:The terms have field absolute value $p^{-2n}$, so the partial sums are a Cauchy sequence in ${\overline{\mathbb{Q}}}_p$. We show that an algebraic limit would contradict the growth of cyclotomic extension degrees. The faster weights make the strict conjugate-distance estimate work uniformly also at $p=2$.

Put $L_N={\mathbb{Q}}_p({\zeta }_{p^{N+1}})$. Its degree is $p^N(p-1)$: the translated cyclotomic polynomial ${\Phi }_{p^{N+1}}(1+T)$ has constant term $p$ and reduces modulo $p$ to $T^{p^N(p-1)}$, so is Eisenstein. To verify the reduction, write it as ${\sum }_{j=0}^{p-1}(1+T{)}^{j{p}^N}$ and use characteristic $p$ to reduce this to ${\sum }_{j=0}^{p-1}(1+T^{p^N}{)}^j=T^{p^N(p-1)}$. All roots lie in $L_N$, and its Galois automorphisms are ${\sigma }_a(\zeta )={\zeta }^a$ for units $a$ modulo $p^{N+1}$.

For a nonidentity automorphism, put $r=v_p(a-1)$, choosing a representative with $0\le r\le N$. It fixes the terms with $n<r$. For $n\ge r$, the root ${\zeta }_{p^{n+1}}^{a-1}$ has order $p^{n+1-r}$, so the root-of-unity valuation formula, with $v_p(p)=1$, givesThis is strictly increasing as $n$ increases: successive differences are $2-p^{-(n-r+1)}>0$. Thus the first changed term is the unique term of smallest valuation and cannot cancel. ConsequentlyIn particular no nonidentity automorphism fixes $A_N$, so ${\mathbb{Q}}_p(A_N)=L_N$. Also the distance from $A_N$ to every distinct conjugate is at least $p^{-2N-1/(p-1)}$.

Suppose the Cauchy sequence converged to $A\in {\overline{\mathbb{Q}}}_p$. The tail estimate givesApply Krasner's lemma with $\alpha =A_N$, $\beta =A$. It forces $L_N={\mathbb{Q}}_p(A_N)\subseteq {\mathbb{Q}}_p(A)$ for every $N$. The right-hand side has finite degree, while $[L_N:{\mathbb{Q}}_p]=p^N(p-1)$ is unbounded, a contradiction. ThereforeThis proves algebraic closure of the p-adic numbers is not complete by an explicit algebraic Cauchy sequence and a fully quantified conjugate-separation argument.

Krasner's lemma states: let $K$ be complete for a Non-Archimedean absolute value, let $\alpha$ be a separable algebraic element over $K$, and let $\beta$ be algebraic over $K$, all in a fixed algebraic closure with the extended field absolute value. Ifthen $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is automatic and there is no conjugate-distance condition to check.

To prove it, put $F=K(\beta )$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a separable polynomial, so $N/F$ is Galois. The finite extension $F$ is complete, and uniqueness of extensions of an field absolute value from a complete non-Archimedean field makes every $\sigma \in \mathrm{Gal}(N/F)$ an isometry. It fixes $\beta$, soIf $\sigma (\alpha )\ne \alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The Fundamental theorem of Galois theory gives $\alpha \in F$, provingOnly $\alpha$ needs to be a separable algebraic element: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta )$.

True. Put $H=\mathrm{Gal}(K/{\mathbb{Q}}_p(\beta ))$. Every $\sigma \in H$ fixes $\beta$, permutes the polynomial's polynomial roots, and is an isometry by the uniqueness proved in part (a). ThereforeThe strict inequalities single out ${\alpha }_1$ as the unique closest polynomial root, so $\sigma ({\alpha }_1)={\alpha }_1$ for every $\sigma \in H$. The Fundamental theorem of Galois theory then givesThis closest-root descent in a Galois extension is the relevant form of Krasner's lemma. If polynomial roots repeat, the strict hypothesis excludes a second occurrence of ${\alpha }_1$; otherwise the same argument applies unchanged.

False. Use the polynomial $X(X^2-p)$, with splitting field $K={\mathbb{Q}}_p(\theta )$, ${\theta }^2=p$, and label the polynomial roots $0,\theta ,-\theta$. Take $\beta =p\theta$. ThenBoth $p-1$ and $p+1$ are p-adic units, also when $p=2$. Hence zero is the uniquely closest polynomial root, but $\beta \notin {\mathbb{Q}}_p(0)={\mathbb{Q}}_p$, because $\theta \notin {\mathbb{Q}}_p$. The polynomial need not be irreducible in the given setup. The inclusion established by Krasner's lemma goes in the direction in part (iii), not its converse.

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.

We count extensions up to $K$-isomorphism. An extension of degree $n$ has a maximal unramified extension $E/K$ inside it, with $[E:K]=f\mid n$, and $L/E$ is totally ramified of degree $e=n/f$. The unramified extension of each degree is unique, so there are only finitely many possibilities for $E$.

Fix $E$ and $e$. The coefficient space of monic degree-$e$ Eisenstein polynomials is compact: its nonconstant nonleading coefficients lie in the compact maximal ideal, and its constant coefficient lies in the compact set ${\mathfrak{m}}_E\setminus {\mathfrak{m}}_E^2$. Each such polynomial $F$ has a root $\alpha$ generating a degree-$e$ extension. For sufficiently close coefficients of a polynomial $G$, the numbers $G(\alpha )$ and $G^{\prime }(\alpha )-F^{\prime }(\alpha )$ are arbitrarily small, while $F^{\prime }(\alpha )\ne 0$ because the characteristic is zero. The strong form of Hensel lemma, applied in $E(\alpha )$, supplies a root $\beta$ of $G$ as close to $\alpha$ as desired. Choose the neighborhood so that the Krasner's lemma inequality holds. Then $E(\alpha )\subseteq E(\beta )$, and both degrees are $e$, so the extensions are equal. Compactness gives a finite cover by such neighborhoods, proving finiteness of degree-$n$ extensions. If extensions are realized as subfields of a fixed algebraic closure rather than counted up to isomorphism, each has only finitely many embeddings, so that count is finite too.

For odd $p$, write $K^{\times }={\pi }^{\mathbb{Z}}{\mathcal{O}}_K^{\times }$. The square map is bijective on $1+{\mathfrak{m}}_K$ by Hensel lemma, and the finite residue field has cyclic multiplicative group of even order. ThusThe quadratic extensions from square classes correspond to its three nontrivial classes. Representatives are $u,\pi ,u\pi$, where $u$ is a unit with nonsquare residue. There are $\boxed{3}$ quadratic extensions: $K(\sqrt{u})$ is unramified, while $K(\sqrt{\pi })$ and $K(\sqrt{u\pi })$ are totally ramified. For completeness, two nonsquare parameters give the same extension only if their ratio is a square: writing $\sqrt{b}=x+y\sqrt{a}$ and squaring forces $x=0$, hence $b/a=y^2$.

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.

Krasner's lemma states that if $K$ is complete, $\alpha$ is separable over $K$, and an algebraic element $\beta$ satisfiesfor every other $K$-conjugate ${\alpha }^{\prime }$ of $\alpha$, then $K(\alpha )\subseteq K(\beta )$.

Let $f\in {\mathbb{C}}_p[X]$ be nonconstant and let $\alpha$ be a root in an algebraic closure of ${\mathbb{C}}_p$. Because the characteristic is zero, replace $f$ by the separable minimal polynomial of $\alpha$. Approximate its coefficients arbitrarily closely by elements of ${\overline{\mathbb{Q}}}_p$. By continuity of roots over a non-Archimedean field, the approximating polynomial has a root $\beta$ arbitrarily close to $\alpha$. Since ${\overline{\mathbb{Q}}}_p$ is algebraically closed, $\beta \in {\overline{\mathbb{Q}}}_p\subset {\mathbb{C}}_p$.

Choose the approximation so that $\beta$ is closer to $\alpha$ than every other ${\mathbb{C}}_p$-conjugate of $\alpha$. Krasner's lemma givesso $\alpha \in {\mathbb{C}}_p$. Hence completion of an algebraic closure of a p-adic field is algebraically closed proves that ${\mathbb{C}}_p$ is algebraically closed.

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.