# Restricted P-adic Valuation is Valuation

## Theorem

Let $\nu_p^\Z: \Z \to \Z \cup \set {+\infty}$ be the $p$-adic valuation restricted to the integers.

Then $\nu_p^\Z$ is a valuation.

## Proof

To prove that $\nu_p^\Z$ is a valuation it is necessary to demonstrate:

 $(\text V 1)$ $:$ $\ds \forall m, n \in \Z:$ $\ds \map {\nu_p^\Z} {m n}$ $\ds =$ $\ds \map {\nu_p^\Z} m + \map {\nu_p^\Z} n$ $(\text V 2)$ $:$ $\ds \forall n \in \Z:$ $\ds \map {\nu_p^\Z} n = +\infty$ $\ds \iff$ $\ds n = 0$ $(\text V 3)$ $:$ $\ds \forall m, n \in \Z:$ $\ds \map {\nu_p^\Z} {m + n}$ $\ds \ge$ $\ds \min \set {\map {\nu_p^\Z} m, \map {\nu_p^\Z} n}$

### Axiom $(\text V 1)$

Let $m, n \in \Z$.

Let $m = 0$ or $n = 0$.

Then:

$\map {\nu_p^\Z} m = +\infty$

or:

$\map {\nu_p^\Z} n = +\infty$

Also:

$n m = 0$

and hence:

 $\ds \map {\nu_p^\Z} {n m}$ $=$ $\ds +\infty$ $\ds$ $=$ $\ds \map {\nu_p^\Z} n + \map {\nu_p^\Z} m$

Let $n m \ne 0$.

Then by definition of the restricted $p$-adic valuation:

$p^{\map {\nu_p^\Z} n} \divides n$
$p^{\map {\nu_p^\Z} n + 1} \nmid n$

Also:

$p^{\map {\nu_p^\Z} m} \divides m$
$p^{\map {\nu_p^\Z} m + 1} \nmid m$

Hence:

$p^{\map {\nu_p^\Z} n + \map {\nu_p^\Z} m} \divides n m$
$p^{\map {\nu_p^\Z} n + \map {\nu_p^\Z} m + 1} \nmid n m$

So:

$\map {\nu_p^\Z} {n m} = \map {\nu_p^\Z} n + \map {\nu_p^\Z} m$

$\Box$

### Axiom $(\text V 2)$

By definition of the restricted $p$-adic valuation:

$\forall n \in \Z: \map {\nu_p^\Z} n = +\infty \iff n = 0$

$\Box$

### Axiom $(\text V 3)$

Let $m, n \in \Z$.

$p^\alpha \divides n$
$p^\beta \divides m$

where $\alpha \ge \beta$.

Then $\exists t \in \Z, k \in \Z$ such that:

 $\ds n + m$ $=$ $\ds p^\alpha k + p^\beta t$ $\ds$ $=$ $\ds p^\beta \paren {p^{\alpha - \beta} k + t}$

Thus:

$p^\beta \divides \paren {m + n}$

$\Box$

Hence by the definition of $\nu_p^\Z$:

$\map {\nu_p^\Z} {m + n} \ge \min \set {\map {\nu_p^\Z} m, \map {\nu_p^\Z} n}$

$\blacksquare$