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:

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:

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

Axiom $(\text V 2)$
By definition of the restricted $p$-adic valuation:


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

Axiom $(\text V 3)$
Let $m, n \in \Z$.

let:
 * $p^\alpha \divides n$
 * $p^\beta \divides m$

where $\alpha \ge \beta$.

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

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

Hence by the definition of $\nu_p^\Z$:
 * $\map {\nu_p^\Z} {m + n} \ge \min \set {\map {\nu_p^\Z} m, \map {\nu_p^\Z} n}$