P-adic Norm is Norm

= Theorem =

The p-adic measure of distance forms a norm on $\Q_p$.

= Proof =

We prove in succession each of the three properties of norms.

Property 1
From the definition of the p-adic function $|*|_p$, we have the first property of a norm.

Property 2
For either $x=0$ or $y=0$ property 2 reduces to property 1, so suppose $x,y \neq 0$ are p-adic numbers. Then $ord_p (xy) = ord_p (x) + ord_p (y)$, so $|xy|_p = \tfrac{1}{p^{ord_p(x)+ord_p(y)}} = \tfrac{1}{p^{ord_p(x)}} \tfrac{1}{p^{ord_p(y)}} = |x|_p |y|_p$.

Property 3
Finally, for either $x=0$, $y=0$, or $x+y=0$, property 3 is true by definition, so $x,y, x+y$ are all non-zero. First, assume $x,y \in \Q$. Let $x=a/b, y=c/d$ be in lowest terms. Then $x+y = \tfrac{ad+bc}{bd}$ and $ord_p(x+y)=ord_p(ad+bc)-ord_p(b)-ord_p(d)$. The highest power of $p$ dividing the sum of two numbers is at least the minimum of the highest power dividing the first and the highest power dividing the second. Therefore,

$ord_p(x+y) \geq min(ord_p(ad), ord_p(bc))-ord_p(b)-ord_p(d) \ $

$= min(ord_p(a)+ord_p(d),ord_p(b)+ord_p(c))-ord_p(b)-ord_p(d) \ $

$= min(ord_p(a) - ord_p(b), ord_p(c)-ord_p(d)) \ $

$= min(ord_p(x),ord_p(y)) \ $

Therefore,

$|x+y|_p = p^{-ord_p(x+y)} \leq max(p^{-ord_p(x)},p^{-ord_p(y)}) = max(|x|_p,|y|_p) \leq |x|_p+|y|_p \ $