Absolute Values Completions and P Adic Numbers -- Examples

The $N$th partial sum is $s_N=(1-p^{N+1})\mathrm{/}(1-p)$. By [L1], $\mathrm{\mid }{p}^{N+1}{\mathrm{\mid }}_p=p^{-(N+1)}\to 0$, so $s_N\to 1\mathrm{/}(1-p)$ in ${\mathbb{Q}}_p$.

In the usual absolute value, $\mathrm{\mid }{p}^N{\mathrm{\mid }}_{\mathrm{\infty }}=p^N$ does not tend to $0$, so the same series cannot converge there.

The $N$th partial sum is $(p-1)(1+p+\cdots +p^N)=p^{N+1}-1.$ Its difference from $-1$ is $p^{N+1}$, whose $p$-adic absolute value tends to $0$. Hence the series converges to $-1$.

Every digit is $p-1$, so by the uniqueness part of [L1] this is the digit expansion of $-1$.

Modulo $5$ one has $2^2\equiv -1$. Since $-1$ is a unit, [L1] shows that $-1$ is a square in ${\mathbb{Q}}_5$.

Concretely, $f(2)=5\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}5)$ and $f^{\mathrm{\prime }}(2)=4\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}5)$, so [L2] produces a unique $5$-adic root congruent to $2$ modulo $5$.

If $p$ is odd, then $p=p^1\cdot 1$ has odd valuation, so [L1] rules out its being a square in ${\mathbb{Q}}_p$.

If $p=2$, then $2=2^1\cdot 1$ also has odd valuation, so the $2$-adic criterion in [L1] rules it out as well.

Hence no prime has a square root in its own $p$-adic field.

The residue class $3$ satisfies $3^2-2=7\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}7)$, and $f^{\mathrm{\prime }}(3)=6\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}7)$. By [L1], there is a unique root $a\in {\mathbb{Z}}_7$ with $a\equiv 3(\mathrm{m}\mathrm{o}\mathrm{d}7)$.

The Newton step from $a_0=3$ is $a_1=3-\frac{7}{6},$ which is well defined in ${\mathbb{Q}}_7$ and already lies in the same residue class modulo $7$; iterating stays in that class and converges to the lifted root from step 1.1.

One has $17=2^0\cdot 17$ with $17\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}8)$, so [L1] shows that $17$ is a square in ${\mathbb{Q}}_2$.

One has $5=2^0\cdot 5$ with $5\equiv 5(\mathrm{m}\mathrm{o}\mathrm{d}8)\mathrm{\ne }1$, so [L1] shows that $5$ is not a square in ${\mathbb{Q}}_2$.

By [L1], $\mathrm{\mid }{p}^n{\mathrm{\mid }}_p=p^{-n}\to 0$, so under the named isometric embedding the sequence converges to $0$ in ${\mathbb{Q}}_p$.

In the usual absolute value, $\mathrm{\mid }{p}^n{\mathrm{\mid }}_{\mathrm{\infty }}=p^n\to \mathrm{\infty }$, so the same sequence runs away instead of converging to $0$.

The map $({\epsilon }_n{)}_{n\ge 0}⟼{\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n{p}^n$ from $\{0,1{\}}^{\mathbb{N}}$ to ${\mathbb{Z}}_p$ is injective by uniqueness of digit expansions in [L1]. Therefore ${\mathbb{Z}}_p$ is uncountable.

The subset of ${\mathbb{Q}}_p$ consisting of elements integral over $\mathbb{Z}$ is countable: there are only countably many monic polynomials with integer coefficients, and each has only finitely many roots in the field ${\mathbb{Q}}_p$.

Hence some element of ${\mathbb{Z}}_p$ is not integral over $\mathbb{Z}$. That element lies in ${\mathbb{Z}}_p\subseteq {\mathbb{Q}}_p$ by [L1], so ${\mathbb{Z}}_p$ cannot equal the integral closure of $\mathbb{Z}$ in ${\mathbb{Q}}_p$.