The p-adic completion agrees with the fraction field of Z_p

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.