The residue field $k_K={\mathbb{F}}_q$ has a unique degree-$n$ extension ${\mathbb{F}}_{q^n}$. Choose a monic irreducible polynomial $\bar{f}$ defining it and lift $\bar{f}$ to a monic $f\in {\mathcal{O}}_K[X]$. Hensel's lemma shows that a root generates an unramified extension $L/K$ of degree $n$ with that residue field. Any two such extensions embed into a common algebraic closure and have the same Teichmuller lifts of ${\mathbb{F}}_{q^n}$, which generate them; hence they coincide. This proves existence and uniqueness of the unramified extension.

For odd $p$, ${\mathbb{Q}}_p$ contains exactly the $p-1$ Teichmuller roots of unity; for $p=2$ it contains $\{\pm 1\}$, so there are two. For odd $p$, adjoining ${\zeta }_p$ adds the $p$ roots of unity of $p$-power order and no primitive $p^2$th root, because the latter would enlarge the degree by $p$. Combining the coprime-order groups givesFor $p=2$, ${\zeta }_2=-1$ already lies in ${\mathbb{Q}}_2$, so the answer remains two.

Since ${\bar{a}}^{q-1}=1$ in the residue field, $u=a^{q-1}\in 1+\pi {\mathcal{O}}_K$. Powers $u^{q^n}$ tend to one, by the binomial theorem initially and the $p$-adic logarithm once they enter its convergence domain. Therefore $a^{q^n}$ is Cauchy; let its limit be $\omega$. Reduction modulo $\pi$ gives $\bar{\omega }=\bar{a}$, whileThis is the Teichmuller representative of $\bar{a}$.

For sufficiently large $r$, the convergent p-adic logarithm and p-adic exponential series are inverse homomorphismsTheir identities $\log (xy)=\log x+\log y$ and $\exp (x+y)=\exp x\exp y$ follow first formally and then by convergence. Multiplication by ${\pi }^r$ identifies $({\mathcal{O}}_K,+)$ with $({\pi }^r{\mathcal{O}}_K,+)$. Since $(1+{\pi }^r{\mathcal{O}}_K)$ has finite index in ${\mathcal{O}}_K^{\ast }$, the conclusion follows.

The Laurent series field $k((t))$ consists of series ${\sum }_{n\ge N}{a}_n{t}^n$. The mapis a discrete valuation. A Cauchy sequence has each coefficient eventually constant, and these stabilized coefficients define its limit, proving completeness.

For $K={\mathbb{F}}_p((t))$, write $K^{\ast }=t^{\mathbb{Z}}\times {\mathbb{F}}_p^{\ast }\times (1+t{\mathbb{F}}_p[[t]])$. If $p$ is odd, Hensel's lemma makes squaring an automorphism of the last factor, so the square-class group has order four. If $p=2$, Frobenius sends $\sum a_n{t}^n$ to $\sum a_n^2{t}^{2n}$, and the classes of units with arbitrarily placed odd-degree terms give infinitely many square classes. Hence the group is finite exactly for odd $p$.

For $f(X)=X^2-X-2m$, $f(0)\equiv 0(\mathrm{mod}2)$ and $f^{\prime }(0)\equiv 1(\mathrm{mod}2)$. Hensel's lemma gives $x\in {\mathbb{Z}}_2$ with $x^2-x=2m$, hence

For $v=\infty$, signs give two square classes and $4/|2{|}_{\infty }=2$. For odd $p$, parity of $v_p$ gives two classes and ${\mathbb{F}}_p^{\ast }/({\mathbb{F}}_p^{\ast }{)}^2$ gives two more, while Hensel makes every unit congruent to $1$ modulo $p$ a square; hence there are four. For $p=2$, valuation parity gives two classes and odd units modulo squares are represented by $1,3,5,7\mathrm{mod}8$, giving eight. Thus in every case

One form of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[X]$, and $f(a_1)\equiv 0(\mathrm{mod}\pi )$ while $f^{\prime }(a_1)\not\equiv 0(\mathrm{mod}\pi )$, then there is a unique $\alpha \in R$ with $f(\alpha )=0$ and $\alpha \equiv a_1(\mathrm{mod}\pi )$.

Inductively, if $f(a_n)\equiv 0(\mathrm{mod}{\pi }^n)$, choose $t$ modulo $\pi$ so thatand put $a_{n+1}=a_n+{\pi }^{n}t$. The unit $f^{\prime }(a_n)$ makes $t$ unique. The resulting sequence is Cauchy, so completeness gives a root $\alpha$. Applying the same first-order congruence to two roots proves uniqueness.

The restriction to $\mathbb{Q}$ cannot be trivial: otherwise the infinitely many integers would be pairwise distance one inside a bounded ball, contradicting local compactness. By part (c), it induces the $p$-adic topology for one prime $p$. Since a locally compact valued field is complete, the embedding $\mathbb{Q}\hookrightarrow K$ extends to a closed embedding ${\mathbb{Q}}_p\hookrightarrow K$.

Now $K$ is a locally compact topological vector space over the nondiscrete local field ${\mathbb{Q}}_p$. Such a vector space is finite-dimensional: a compact neighbourhood, together with a maximal linearly independent subset chosen at a fixed separation scale, is totally bounded only if that subset is finite, and its span is then open and closed; maximality makes it all of $K$. Thus $[K:{\mathbb{Q}}_p]<\infty$.

Two absolute values are equivalent when $|\cdot {|}_2=|\cdot {|}_1^c$ for some $c>0$; equivalently, they induce the same topology. The nontrivial non-Archimedean absolute values on $\mathbb{Q}$ are, up to equivalence, exactly the p-adic absolute value $|\cdot {|}_p$.

Indeed $|n|\le 1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$. If $(m,p)=1$, choose $a_r,b_r\in \mathbb{Z}$ with $a_{r}m+b_r{p}^r=1$. For large $r$, $|b_r{p}^r|<1$, so the ultrametric inequality forces $|m|=1$. HenceIf no prime has absolute value below one, the absolute value is trivial. This proves the non-Archimedean part of Ostrowski theorem.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfying $|x|=0\iff x=0$, $|xy|=|x||y|$, and $|x+y|\le |x|+|y|$. It is non-Archimedean when the stronger inequality $|x+y|\le \max (|x|,|y|)$ holds.

If $\mathrm{char}K=p>0$, thenThe ordinary triangle inequality givesLetting $r\to \infty$ proves the strong triangle inequality.

Fix a hyperbolicity constant ${\delta }_0>0$ for ${\mathbb{H}}^2$. For a prescribed $\delta >0$, scale its distance byAll distances in every geodesic triangle, including its thinness constant, scale by $\delta /{\delta }_0$. Henceis $\delta$-hyperbolic. Multiplication of a metric by a fixed positive constant is a bilipschitz equivalence and hence a quasi-isometry. Therefore $X_{\delta }$ is quasi-isometric to ${\mathbb{H}}^2$ and cannot be a quasi-tree. This supplies an example for every $\delta >0$.

The hyperbolic plane ${\mathbb{H}}^2$ is a geodesic Gromov-hyperbolic metric space for some universal constant ${\delta }_0$. It is not a quasi-tree. Indeed, for every $C>0$, choose two points on opposite sides of a large closed metric ball centred at the midpoint of their joining geodesic. The complement of that ball in ${\mathbb{H}}^2$ is path connected, so the endpoints can be joined by a continuous path that stays more than $C$ from the midpoint. Thus ${\mathbb{H}}^2$ fails the bottleneck property, whereas part (b) shows that every quasi-tree satisfies it.

Retain the $(\lambda ,\epsilon )$-quasi-isometry $f:X\to T$ and the Morse lemma for quasi-geodesics constant $R$. Given $x,y\in X$, let $m$ be the midpoint of a geodesic $[x,y]$. There is a point $z$ on the tree geodesic $[f(x),f(y)]$ with

For any continuous path $\alpha$ from $x$ to $y$, choose a partition fine enough that consecutive $\alpha (t_i)$ are at distance at most one. Consecutive images under $f$ are then at distance at mostRemoving $z$ separates $f(x)$ from $f(y)$ in the tree along their geodesic, so this finite $D$-chain must contain an element within $D$ of $z$. For the corresponding $t_i$,The lower quasi-isometry inequality yieldsThis constant depends only on the chosen quasi-isometry, so every quasi-tree has the bottleneck property.

Let $f:X\to T$ be a $(\lambda ,\epsilon )$-quasi-isometry to a tree. The image under $f$ of every geodesic segment in $X$ is a $(\lambda ,\epsilon )$-quasigeodesic in $T$. By the Morse lemma for quasi-geodesics, it lies within a constant $R=R(\lambda ,\epsilon )$ of the tree geodesic with the same endpoints.

Consider a geodesic triangle in $X$. A point $p$ on one side maps within $R$ of the corresponding side of the comparison triangle in $T$. Every geodesic triangle in a tree is $0$-thin, so that comparison side is contained in the other two sides. Those two tree sides are in turn within $R$ of the images of the other two sides of the original triangle. Hence some point $q$ on one of those sides satisfiesThe lower quasi-isometry inequality givesThus $X$ is Gromov-hyperbolic metric space with $\delta =\lambda (2R+\epsilon )$.

Let $T^{Z(G)}$ be the common fixed subtree of the centre, which is nonempty by hypothesis and is $G$-invariant because $Z(G)$ is central. Restrict the action to this subtree. There $x$ and $y$ act pointwise trivially, so the action factors throughIn particular, the induced tree isometries satisfy

Assume, as usual for a combinatorial tree action, that edge inversions have been removed by barycentric subdivision. The finite-order elements $a,b,c$ are then elliptic. Moreoverso each pairwise product is elliptic. Serre lemma for tree actions implies that the fixed subtrees of each pair intersect. Convex subtrees of a tree have the Helly property, soSince $a,b,c$ generate $G/⟨x,y⟩$, $G$ fixes a vertex of $T^{Z(G)}$. Thus the action of $G$ on $T$ is trivial in the tree-action sense.

A hyperbolic isometry of a tree has a unique invariant axis of a tree isometry, on which it acts by a nonzero translation. Let $L$ be the axis of $x$. By part (c), $x$ is central, so for every $g\in G$,Thus $gL$ is another axis of $x$. Uniqueness gives $gL=L$, and hence the whole group $G$ preserves the line $L$.