Absolute Values Completions and P Adic Numbers

Since $1_F\mathrm{\ne }0$, multiplicativity in [L1] gives $\mathrm{\mid }{1}_F\mathrm{\mid }=\mathrm{\mid }{1}_F{\mathrm{\mid }}^2$, hence $\mathrm{\mid }{1}_F\mathrm{\mid }=1$. Also $\mathrm{\mid }-1{\mathrm{\mid }}^2=\mathrm{\mid }1\mathrm{\mid }=1$, so $\mathrm{\mid }-1\mathrm{\mid }=1$.

Assume conversely that $\mathrm{\mid }n{1}_F\mathrm{\mid }\le 1$ for every integer $n$. Fix $x,y\in F$ and put $M:=\max \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\}$. If $M=0$ there is nothing to prove, so scale by a nonzero element and reduce to $M=1$. Then every binomial coefficient has absolute value at most $1$, so $\mathrm{\mid }(x+y{)}^m\mathrm{\mid }\le {\max }_{0\le k\le m}\mid \left(\genfrac{}{}{0px}{}{m}{k}\right)\mid \mathrm{\mid }x{\mathrm{\mid }}^k\mathrm{\mid }y{\mathrm{\mid }}^{m-k}\le 1$ for every $m\ge 1$. Hence $\mathrm{\mid }x+y{\mathrm{\mid }}^m\le 1$ for all $m$, so $\mathrm{\mid }x+y\mathrm{\mid }\le 1=M$. Undoing the scaling gives $\mathrm{\mid }x+y\mathrm{\mid }\le \max \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\}$.

Assume first that $\mathrm{\mid }\cdot \mathrm{\mid }$ is nonarchimedean. For $n\ge 1$, induction using step 1.1 and $\mathrm{\mid }(n+1)1_F\mathrm{\mid }\le \max \{\mathrm{\mid }n{1}_F\mathrm{\mid },\mathrm{\mid }{1}_F\mathrm{\mid }\}$ gives $\mathrm{\mid }n{1}_F\mathrm{\mid }\le 1$. For negative $n$, $\mathrm{\mid }n{1}_F\mathrm{\mid }=\mathrm{\mid }-1\mathrm{\mid }\mathrm{\mid }(-n)1_F\mathrm{\mid }\le 1$, and $\mathrm{\mid }0\mathrm{\mid }=0\le 1$.

Thus the integer bound implies the strong triangle inequality, and step 2.1 proved the reverse implication.

If $\mathrm{\mid }x{\mathrm{\mid }}_2=\mathrm{\mid }x{\mathrm{\mid }}_1^c$ for some $c>0$, then $t\mapsto t^c$ is increasing on ${\mathbb{R}}_{>0}$, so the conditions $\mathrm{\mid }x-a{\mathrm{\mid }}_1<r$ and $\mathrm{\mid }x-a{\mathrm{\mid }}_2<r^c$ define the same neighborhoods of every point $a$. Hence the two absolute values induce the same topology.

Assume conversely that the two topologies agree. Then $\mathrm{\mid }x{\mathrm{\mid }}_1<1⟺x^n\to 0\text{ in }\mathrm{\mid }\cdot {\mathrm{\mid }}_1⟺x^n\to 0\text{ in }\mathrm{\mid }\cdot {\mathrm{\mid }}_2⟺\mathrm{\mid }x{\mathrm{\mid }}_2<1,$ and the same argument with $x^{-1}$ gives $\mathrm{\mid }x{\mathrm{\mid }}_1>1⟺\mathrm{\mid }x{\mathrm{\mid }}_2>1$ for every nonzero $x$.

Choose $a\in F^{\times }$ with $\mathrm{\mid }a{\mathrm{\mid }}_1>1$; then step 1.2 gives $\mathrm{\mid }a{\mathrm{\mid }}_2>1$. Set $c:=\frac{\log \mathrm{\mid }a{\mathrm{\mid }}_2}{\log \mathrm{\mid }a{\mathrm{\mid }}_1}>0.$ Fix $x\in F^{\times }$. If $m\mathrm{/}n<\log \mathrm{\mid }x{\mathrm{\mid }}_1\mathrm{/}\log \mathrm{\mid }a{\mathrm{\mid }}_1$, then $\mathrm{\mid }x{\mathrm{\mid }}_1^n<\mathrm{\mid }a{\mathrm{\mid }}_1^m$, so $\mathrm{\mid }{x}^n{a}^{-m}{\mathrm{\mid }}_1<1$ and therefore $\mathrm{\mid }{x}^n{a}^{-m}{\mathrm{\mid }}_2<1$ by step 1.2. This gives $n\log \mathrm{\mid }x{\mathrm{\mid }}_2<m\log \mathrm{\mid }a{\mathrm{\mid }}_2$. Reversing the inequality yields the opposite bound. Rational approximation therefore forces $\frac{\log \mathrm{\mid }x{\mathrm{\mid }}_2}{\log \mathrm{\mid }a{\mathrm{\mid }}_2}=\frac{\log \mathrm{\mid }x{\mathrm{\mid }}_1}{\log \mathrm{\mid }a{\mathrm{\mid }}_1},$ so $\mathrm{\mid }x{\mathrm{\mid }}_2=\mathrm{\mid }x{\mathrm{\mid }}_1^c$.

Step 1.1 proves one direction and step 2.1 proves the other, so the two topologies agree exactly when the absolute values are equivalent.

If one of $x$ or $y$ is zero then $\mathrm{\mid }xy{\mathrm{\mid }}_p=0=\mathrm{\mid }x{\mathrm{\mid }}_p\mathrm{\mid }y{\mathrm{\mid }}_p$. If both are nonzero, [L1] and [L2] give $\mathrm{\mid }xy{\mathrm{\mid }}_p=p^{-v_p(xy)}=p^{-v_p(x)-v_p(y)}=\mathrm{\mid }x{\mathrm{\mid }}_p\mathrm{\mid }y{\mathrm{\mid }}_p\mathrm{.}$

If $x+y=0$, then $\mathrm{\mid }x+y{\mathrm{\mid }}_p=0\le \max \{\mathrm{\mid }x{\mathrm{\mid }}_p,\mathrm{\mid }y{\mathrm{\mid }}_p\}$. If $x,y,x+y$ are nonzero, [L1] gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\},$ so exponentiating by $p^{-t}$, which reverses order, yields $\mathrm{\mid }x+y{\mathrm{\mid }}_p\le \max \{\mathrm{\mid }x{\mathrm{\mid }}_p,\mathrm{\mid }y{\mathrm{\mid }}_p\}\mathrm{.}$

The multiplicative law is step 1.1 and the strong triangle inequality is step 1.2, so $\mathrm{\mid }\cdot {\mathrm{\mid }}_p$ is nonarchimedean.

Assume as the first case that $\mathrm{\mid }n\mathrm{\mid }\le 1$ for every integer $n$. Then [L1] makes $\mathrm{\mid }\cdot \mathrm{\mid }$ nonarchimedean. Because the absolute value is nontrivial, there is an integer $m$ with $\mathrm{\mid }m\mathrm{\mid }\mathrm{\ne }1$; the inequality $\mathrm{\mid }m\mathrm{\mid }\le 1$ forces $\mathrm{\mid }m\mathrm{\mid }<1$, so some prime divisor $p$ of $m$ has $\mathrm{\mid }p\mathrm{\mid }<1$ by [L4]. If also $\mathrm{\mid }q\mathrm{\mid }<1$ for a different prime $q$, then [L3] gives $up+vq=1$, and the nonarchimedean inequality yields $1=\mathrm{\mid }1\mathrm{\mid }\le \max \{\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }q\mathrm{\mid }\}<1$, impossible because integers all have absolute value at most $1$. Thus there is a unique prime $p$ with $\mathrm{\mid }p\mathrm{\mid }<1$.

Assume as the second case that $\mathrm{\mid }n\mathrm{\mid }>1$ for some integer $n$. Choosing a prime divisor of $n$ and using [L4], at least one prime $p$ satisfies $\mathrm{\mid }p\mathrm{\mid }>1$. Put $\alpha :=\log \mathrm{\mid }p\mathrm{\mid }\mathrm{/}\log p>0$ and, for each integer $t\in \{0,\dots ,p-1\}$, let $C_p:=\max \mathrm{\mid }t\mathrm{\mid }$. For every positive integer $m$, write the base-$p$ expansion $m=a_0+a_{1}p+\cdots +a_r{p}^r$ with $0\le a_i<p$. The triangle inequality gives $\mathrm{\mid }m\mathrm{\mid }\le {\sum }_{i=0}^r\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }p{\mathrm{\mid }}^i\le C_p(r+1)\mathrm{\mid }p{\mathrm{\mid }}^r\le C_p(1+{\log }_{p}m)m^{\alpha },$ because $p^r\le m<p^{r+1}$. Applying the same estimate to $m^k$ and taking $k$-th roots gives $\mathrm{\mid }m\mathrm{\mid }\le (C_p(1+k{\log }_{p}m){)}^{1\mathrm{/}k}{m}^{\alpha },$ so letting $k\to \mathrm{\infty }$ yields $\mathrm{\mid }m\mathrm{\mid }\le m^{\alpha }$. Now fix $m>1$ and put ${\alpha }_m:=\log \mathrm{\mid }m\mathrm{\mid }\mathrm{/}\log m$. Writing $p^k$ in base $m$ and repeating the same argument with $m$ in place of $p$ gives $\mathrm{\mid }p{\mathrm{\mid }}^k\le C_m(1+k{\log }_{m}p)p^{k{\alpha }_m}$ for some constant $C_m$, hence $\mathrm{\mid }p\mathrm{\mid }\le p^{{\alpha }_m}$ after taking $k$-th roots and letting $k\to \mathrm{\infty }$. Therefore $\alpha \le {\alpha }_m$. Since $\mathrm{\mid }m\mathrm{\mid }\le m^{\alpha }$ is exactly ${\alpha }_m\le \alpha$, we get ${\alpha }_m=\alpha$ for every $m>1$. Thus $\mathrm{\mid }m\mathrm{\mid }=m^{\alpha }$ for every positive integer $m$, and then $\mathrm{\mid }a\mathrm{/}b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\mathrm{/}\mathrm{\mid }b\mathrm{\mid }=\mathrm{\mid }a\mathrm{/}b{\mathrm{\mid }}_{\mathrm{\infty }}^{\alpha }$ for every nonzero rational.

For any prime $q\mathrm{\ne }p$, step 1.1 and [L3] applied to $p$ and $q$ give $\mathrm{\mid }q\mathrm{\mid }=1$. Hence if $x=\pm p^{k}a\mathrm{/}b$ with $p\nmid ab$, uniqueness of factorisation [L4] gives $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }p{\mathrm{\mid }}^k$. Writing $c:=-\log \mathrm{\mid }p\mathrm{\mid }\mathrm{/}\log p>0$, this becomes $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }x{\mathrm{\mid }}_p^c$, so $\mathrm{\mid }\cdot \mathrm{\mid }$ is equivalent to $\mathrm{\mid }\cdot {\mathrm{\mid }}_p$ by [L5].

Step 2.1 gives the nonarchimedean case and step 1.2 gives the archimedean case, and the two cases are disjoint because [L2] says every $p$-adic absolute value is nonarchimedean. Therefore every nontrivial absolute value on $\mathbb{Q}$ is equivalent either to the usual absolute value or to a unique $p$-adic one.

Write $x=\pm {\prod }_{i=1}^r{p}_i^{e_i}$ with distinct primes $p_i$ and integers $e_i$, as supplied by [L2]. Then $\mathrm{\mid }x{\mathrm{\mid }}_{p_i}=p_i^{-e_i}$ for the listed primes and $\mathrm{\mid }x{\mathrm{\mid }}_q=1$ for every other prime $q$.

The archimedean factor is $\mathrm{\mid }x{\mathrm{\mid }}_{\mathrm{\infty }}={\prod }_{i=1}^r{p}_i^{e_i}$, so $\mathrm{\mid }x{\mathrm{\mid }}_{\mathrm{\infty }}{\prod }_{i=1}^r\mathrm{\mid }x{\mathrm{\mid }}_{p_i}=\left({\prod }_{i=1}^r{p}_i^{e_i}\right)\left({\prod }_{i=1}^r{p}_i^{-e_i}\right)=1.$ Since the omitted prime factors are all $1$, this is exactly the full product formula.

Reorder the places so that $v_1,\dots ,v_s$ are the finite places $p_1,\dots ,p_s$ and, if the archimedean place occurs, it is $v_r=\mathrm{\infty }$. Choose integers $N_i\ge 1$ with $p_i^{-N_i}<{\epsilon }_i$ for $1\le i\le s$. Let $D_0$ be a common positive denominator of the finite targets $a_1,\dots ,a_s$, and replace it by $D:=T{D}_0$, where $T$ is a large integer coprime to every $p_i$; this keeps every $D{a}_i$ integral and lets us later make $M\mathrm{/}D$ as small as we wish.

Put $M:={\prod }_{i=1}^s{p}_i^{N_i+v_{p_i}(D)}\mathrm{.}$ By [L2], there is an integer $y$ such that $y\equiv D{a}_i(\mathrm{m}\mathrm{o}\mathrm{d}{p}_i^{N_i+v_{p_i}(D)})(1\le i\le s)\mathrm{.}$ Then for $x_0:=y\mathrm{/}D$ one has $v_{p_i}(x_0-a_i)\ge N_i$, hence $\mathrm{\mid }{x}_0-a_i{\mathrm{\mid }}_{p_i}\le p_i^{-N_i}<{\epsilon }_i$ for every finite place in the list.

If $\mathrm{\infty }$ is not among the chosen places, then $x:=x_0$ works. Otherwise every number of the form $x_k:=x_0+kM\mathrm{/}D$ has the same finite-place congruence conditions as $x_0$, because $v_{p_i}(M\mathrm{/}D)=N_i(1\le i\le s)\mathrm{.}$ By taking $T$ in step 1.1 so large that $M\mathrm{/}D<{\epsilon }_r$, the arithmetic progression $x_0+\mathbb{Z}(M\mathrm{/}D)$ has mesh smaller than ${\epsilon }_r$, so some integer $k$ satisfies $\mathrm{\mid }{x}_k-a_r{\mathrm{\mid }}_{\mathrm{\infty }}<{\epsilon }_r$.

The chosen $x$ satisfies all requested inequalities. The density formulation is the same statement with the local targets first approximated by rational elements in each factor.

By the specific construction fixed in [L3] and supplied by [L1], an element of ${\mathbb{Q}}_p$ is represented by a $d_p$-Cauchy sequence $(x_n)$ in $\mathbb{Q}$, and the distance between classes is ${\widehat{d}}_p([x],[y])={\lim }_n\mathrm{\mid }{x}_n-y_n{\mathrm{\mid }}_p\mathrm{.}$ Define $[x]+[y]:=[x_n+y_n],[x][y]:=[x_n{y}_n],\mathrm{\mid }[x]{\mathrm{\mid }}_p:={\lim }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}_p\mathrm{.}$

The nonarchimedean inequality in [L2] shows that sums and differences of Cauchy sequences are Cauchy. A Cauchy sequence in a nonarchimedean metric is bounded, so products of Cauchy sequences are again Cauchy, and equivalent representatives give equivalent sums and products because $x_n{y}_n-x_n^{\mathrm{\prime }}{y}_n^{\mathrm{\prime }}=x_n(y_n-y_n^{\mathrm{\prime }})+y_n^{\mathrm{\prime }}(x_n-x_n^{\mathrm{\prime }})\mathrm{.}$ Passing to the limit through [L2] proves representative independence of the extended absolute value as well.

Let $[x]\in {\mathbb{Q}}_p$ be nonzero. Then $\mathrm{\mid }[x]{\mathrm{\mid }}_p>0$, and step 1.1 says $\mathrm{\mid }{x}_n{\mathrm{\mid }}_p\to \mathrm{\mid }[x]{\mathrm{\mid }}_p$ in $\mathbb{R}$. So there are a real constant $c>0$ and an index $N$ with $\mathrm{\mid }{x}_n{\mathrm{\mid }}_p\ge c$ for all $n\ge N$. In particular $x_n\mathrm{\ne }0$ eventually. For $n,m\ge N$, $\mathrm{\mid }{x}_n^{-1}-x_m^{-1}{\mathrm{\mid }}_p=\frac{\mathrm{\mid }{x}_n-x_m{\mathrm{\mid }}_p}{\mathrm{\mid }{x}_n{\mathrm{\mid }}_p\mathrm{\mid }{x}_m{\mathrm{\mid }}_p}\le c^{-2}\mathrm{\mid }{x}_n-x_m{\mathrm{\mid }}_p,$ and the right-hand side tends to $0$ because $(x_n)$ is Cauchy. Thus $(x_n^{-1})$ is eventually defined and Cauchy, so every nonzero class has an inverse.

Multiplicativity and the strong triangle inequality on ${\mathbb{Q}}_p$ follow by taking limits of the corresponding rational identities from [L2]. Completeness is already part of [L1]. Thus ${\mathbb{Q}}_p$ is a complete nonarchimedean valued field.

If $x$ lies in the open ball $B(a,r)$, then $\mathrm{\mid }x-a{\mathrm{\mid }}_p<r$. For any $y$ with $\mathrm{\mid }y-x{\mathrm{\mid }}_p<r$, [L1] gives $\mathrm{\mid }y-a{\mathrm{\mid }}_p\le \max \{\mathrm{\mid }y-x{\mathrm{\mid }}_p,\mathrm{\mid }x-a{\mathrm{\mid }}_p\}<r,$ so $y\in B(a,r)$. Thus every point of an open ball is again a center, hence every open ball is open and every closed ball is open by the same argument with $\le r$.

If $\mathrm{\mid }y-x{\mathrm{\mid }}_p<\mathrm{\mid }x-a{\mathrm{\mid }}_p$, then [L1] forces $\mathrm{\mid }y-a{\mathrm{\mid }}_p=\mathrm{\mid }x-a{\mathrm{\mid }}_p$. Hence, if $x\mathrm{\notin }B(a,r)$, then $\mathrm{\mid }x-a{\mathrm{\mid }}_p\ge r$ and so $y\mathrm{\notin }B(a,r)$; and if $x$ lies outside the closed ball of radius $r$ around $a$, then $\mathrm{\mid }x-a{\mathrm{\mid }}_p>r$ and so $y$ lies outside that closed ball as well. Thus the complement of either the open or the closed ball around $a$ is a union of open balls. Therefore every open or closed ball is also closed.

If $B(a,r)$ and $B(b,s)$ intersect, choose $x$ in the intersection. For any $y\in B(a,r)$, $\mathrm{\mid }y-b{\mathrm{\mid }}_p\le \max \{\mathrm{\mid }y-x{\mathrm{\mid }}_p,\mathrm{\mid }x-b{\mathrm{\mid }}_p\}<\max \{r,s\}=s,$ so $y\in B(b,s)$. The same proof works for closed balls. Hence intersecting comparable balls are nested.

Let $R\subseteq {\mathbb{Q}}_p$ be the set of classes having a $p$-adically Cauchy representative $(z_n)$ with every $z_n\in \mathbb{Z}$. For $x=[(z_n)]\in R$ and each $m\ge 1$, the sequence $(z_n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p^m)$ is eventually constant, because $z_n-z_{n^{\mathrm{\prime }}}\in p^m\mathbb{Z}$ for all large $n,n^{\mathrm{\prime }}$. Thus $x$ determines a compatible residue system and hence a map $\mathrm{\Phi }:R⟶{\mathbb{Z}}_p\mathrm{.}$ Equivalent integer representatives give the same eventual residues, so $\mathrm{\Phi }$ is well defined and is a ring homomorphism.

Conversely, let $(a_m{)}_m\in {\mathbb{Z}}_p$ be a compatible residue system and choose the standard lift $b_m\in \{0,\dots ,p^m-1\}$ of $a_m$. Compatibility means $b_{m+1}\equiv b_m(\mathrm{m}\mathrm{o}\mathrm{d}{p}^m)$, so $\mathrm{\mid }{b}_{m+1}-b_m{\mathrm{\mid }}_p\le p^{-m}$. Hence $(b_m)$ is a $p$-adic Cauchy sequence of integers and defines an element of $R$. This construction inverts $\mathrm{\Phi }$, so $R\cong {\mathbb{Z}}_p$ as rings.

The subring $R$ is exactly the closed unit ball. One inclusion is immediate because every integer has $p$-adic absolute value at most $1$. Conversely, let $x=[(x_n)]$ satisfy $\mathrm{\mid }x{\mathrm{\mid }}_p\le 1$. After passing to a tail, take $\mathrm{\mid }{x}_n{\mathrm{\mid }}_p\le 1$ and write $x_n=a_n\mathrm{/}{b}_n$ in lowest terms. Then $p\nmid b_n$ by [L4]. For each $n$, [L3] gives an integer $c_n$ with $b_n{c}_n\equiv a_n(\mathrm{m}\mathrm{o}\mathrm{d}{p}^n)$. Hence $\mathrm{\mid }{c}_n-x_n{\mathrm{\mid }}_p={\mid \frac{b_n{c}_n-a_n}{b_n}\mid }_p\le p^{-n}\mathrm{.}$ Thus $(c_n)$ is an integer Cauchy sequence equivalent to $(x_n)$, so $x\in R$. Therefore $R=\{x\in {\mathbb{Q}}_p:\mathrm{\mid }x{\mathrm{\mid }}_p\le 1\}\mathrm{.}$

Every element of ${\mathbb{Q}}_p$ is a fraction of elements of $R$. Indeed, if $x=[(x_n)]$, the Cauchy sequence $(x_n)$ is bounded, so for some integer $N\ge 0$ one has $\mathrm{\mid }{p}^N{x}_n{\mathrm{\mid }}_p\le 1$ eventually. Passing to the limit and using step 2.1 gives $p^{N}x\in R$, while $p^N\in R$ is nonzero. Hence $x=\frac{p^{N}x}{p^N}\mathrm{.}$ Since $R$ is a subring of the field ${\mathbb{Q}}_p$, its fraction field is all of ${\mathbb{Q}}_p$.

Transporting step 2.1 through the ring isomorphism $R\cong {\mathbb{Z}}_p$ from step 1.2 identifies ${\mathbb{Q}}_p$ canonically with $\mathrm{Frac}({\mathbb{Z}}_p)$ and identifies ${\mathbb{Z}}_p$ with the closed unit ball.

By [L1], an element of ${\mathbb{Q}}_p$ lies in the image of ${\mathbb{Z}}_p$ exactly when its $p$-adic absolute value is at most $1$.

Thus the image is exactly $\{x\in {\mathbb{Q}}_p:\mathrm{\mid }x{\mathrm{\mid }}_p\le 1\}$, which is the valuation ring of the valued field ${\mathbb{Q}}_p$.

An element $x\in {\mathbb{Z}}_p$ is a unit exactly when $\mathrm{\mid }x{\mathrm{\mid }}_p=1$: if $\mathrm{\mid }x{\mathrm{\mid }}_p=1$ then $\mathrm{\mid }{x}^{-1}{\mathrm{\mid }}_p=1$ as well, so $x^{-1}\in {\mathbb{Z}}_p$ by [L1]; if $\mathrm{\mid }x{\mathrm{\mid }}_p<1$ then $\mathrm{\mid }{x}^{-1}{\mathrm{\mid }}_p>1$, so $x^{-1}\mathrm{\notin }{\mathbb{Z}}_p$. Therefore the nonunits are precisely the elements with $\mathrm{\mid }x{\mathrm{\mid }}_p<1$, which is the principal ideal $p{\mathbb{Z}}_p$.

In the compatible-residue model of [L2], multiplication by $p$ is exactly the condition that the first residue coordinate is $0$. Therefore the quotient by $p{\mathbb{Z}}_p$ remembers only the first residue class, giving a canonical map ${\mathbb{Z}}_p\mathrm{/}p{\mathbb{Z}}_p\to \mathbb{Z}\mathrm{/}p\mathbb{Z}={\mathbb{F}}_p\mathrm{.}$ This map is bijective because every residue class lifts to a compatible system and two systems differ by an element of $p{\mathbb{Z}}_p$ exactly when their first coordinates agree.

Since the quotient by $p{\mathbb{Z}}_p$ is the field ${\mathbb{F}}_p$, the ideal $p{\mathbb{Z}}_p$ is maximal, and step 1.1 shows it contains every nonunit, so it is the unique maximal ideal.

If $x=0$, take every digit $a_n=0$. Now assume $x\mathrm{\ne }0$. By the density statement in [L3], choose $q\in \mathbb{Q}$ with $\mathrm{\mid }x-q{\mathrm{\mid }}_p<\mathrm{\mid }x{\mathrm{\mid }}_p$. The ultrametric inequality from [L4] then gives $\mathrm{\mid }q{\mathrm{\mid }}_p=\mathrm{\mid }x{\mathrm{\mid }}_p$. Write $q=p^{N}a\mathrm{/}b$ with $N\in \mathbb{Z}$ and integers $a,b$ not divisible by $p$; then $\mathrm{\mid }q{\mathrm{\mid }}_p=p^{-N}$, so $\mathrm{\mid }x{\mathrm{\mid }}_p=p^{-N}$. Put $u:=p^{-N}x$. Then $\mathrm{\mid }u{\mathrm{\mid }}_p=1$, so $u\in {\mathbb{Z}}_p$ by [L2], and also $\mathrm{\mid }{u}^{-1}{\mathrm{\mid }}_p=1$, so $u^{-1}\in {\mathbb{Z}}_p$. Hence $u\in {\mathbb{Z}}_p^{\times }$.

In the compatible-residue description of ${\mathbb{Z}}_p$ from [L1], for each $m\ge 1$ choose the unique digits $a_0,\dots ,a_{m-1}\in \{0,\dots ,p-1\}$ such that $u\equiv a_0+a_{1}p+\cdots +a_{m-1}{p}^{m-1}(\mathrm{m}\mathrm{o}\mathrm{d}{p}^m)\mathrm{.}$ Because $u$ is a unit, its residue modulo $p$ is nonzero, so $a_0\mathrm{\ne }0$. Writing $s_m:={\sum }_{j=0}^{m-1}{a}_j{p}^j$, one has $u-s_m\in p^m{\mathbb{Z}}_p$, hence $\mathrm{\mid }u-s_m{\mathrm{\mid }}_p\le p^{-m}$. So $(s_m)$ is Cauchy and converges to $u$ by [L4]. Multiplying by $p^N$ gives $x={\sum }_{m=0}^{\mathrm{\infty }}{a}_m{p}^{m+N};$ renaming the digits by index shift yields the claimed expansion with leading digit $a_N=a_0\mathrm{\ne }0$.

For uniqueness, suppose ${\sum }_{n=N}^{\mathrm{\infty }}{a}_n{p}^n={\sum }_{n=M}^{\mathrm{\infty }}{b}_n{p}^n$ with digits in $\{0,\dots ,p-1\}$ and nonzero leading digits. Equality of absolute values forces $N=M$. If $a_n\mathrm{\ne }{b}_n$ first occurs at $n=k$, then the difference of the two series equals $(a_k-b_k)p^k+p^{k+1}y$ for some $y\in {\mathbb{Z}}_p$. Since $a_k-b_k$ is not divisible by $p$, that difference has $p$-adic absolute value $p^{-k}$ and cannot be $0$. Therefore all digits agree.