Prove some properties of the $p$-adic norm

Question

I need to prove that the $p$-adic norm is an absolute value in the rational numbers, by an absolute value in a field $K$ I mean a function $|\cdot|:K \to \mathbb{R}_{\ge 0}$ such that:

$$\begin{align} \text{I)}&~~~~ |x|=0 \implies x=0\\ \text{II)}&~~~~|x\cdot y|=|x|\cdot|y|\\ \text{III)}&~~~~\text{The triangle inequality.} \end{align} $$

I do not know how to prove II) and III) for the $p$-adic norm.

I have for II) that:

$|x|_{p}·|y|_{p}=p^{-r}·p^{-a}=p^{-r-a}$ and $|x·y|_{p}=p^{-m}$.

How can $p^{-m}=p^{-r}·q^{-a}$?

Answer 1

For (II), write

$q_1=p^mr, q_2=p^ns$ where $r,s\in\Bbb Q$ has numerator and denominator prime to $p$ and $m,n\in\Bbb Z$

Then

$$|q_1q_2|_p=p^{-(n+m)}=p^{-n}p^{-m}=|q_1|_p\cdot|q_2|_p$$

For (III) assume, that $n> m$, then

$$|q_1+q_2|_p=p^{-m}|p^{n-m}s+r|_p=p^{-m}\le p^{-m}+p^{-n}$$

because

$$p|(p^{n-m}s+r)\implies p\bigg|\big\lbrace (p^{n-m}s+r)-p^{n-m}s\big\rbrace =r$$

a contradiction, proving the triangle inequality there.

If $n=m$ write

$$|q_1+q_2|_p=p^{-n}|r+s|_p$$

Then since the common denominator of $r,s$ is still prime to $p$, we can reduce to the case $r,s\in\Bbb Z$.

But then it's easy, we just need to show, for integers, $r,s$ that

$$|r+s|_p\le |r|_p+|s|_p.$$

For this we note that if $r=p^nr', \; s=p^ns'$ with $\gcd(p,r')=\gcd(p,s')=1$ then $r+s=p^n(r'+s')$ and if $p|(r'+s')$ then

$$|r+s|_p\le p^{-n-1}\le 2p^{-n}=|r|_p+|s|_p$$

if $p\not |(r'+s')$ then

$$|r+s|_p=|p^n|_p\cdot |r'+s'|= p^{-n}\le 2p^{-n}=|r|_p+|s|_p.$$

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Note: By breaking things into cases like this we have an excellent observation: Almost always we have that $|q_1+q_2|_p=\max\{|q_1|_p, |q_2|_p\}$. And in the case that equality doesn't hold we have that $|q_1|_p=|q_2|_p$ which means all triangles in the $p$-adic world are isosceles, when we measure their lengths with this absolute value!