Equivalence of Definitions of P-adic Norms/Lemma 1

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Theorem

Let $p \in \N$ be a prime number.

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


Then:

$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$


Proof

Let $x \in \Z_{\ne 0}$.

By definition of the $p$-adic valuation:

$\map {\nu_p} x = \sup \set {v \in \N: p^v \divides x}$

Let $\map {\nu_p} x = k$.

Then:

$p^k \nmid x$

By definition of a divisor:

$\exists y \in Z : x = p^k y$


Aiming for a contradiction, suppose:

$p \divides y$

By definition of a divisor:

$\exists y' \in Z : y = p y'$

Hence:

\(\ds x\) \(=\) \(\ds p^k \paren{p y'}\)
\(\ds \) \(=\) \(\ds p^{k + 1} y'\)

This contradicts:

$k = \sup \set {v \in \N: p^v \divides x}$

Hence: $p \nmid y$


Since the choice of $x \in \Z_{\ne 0}$ was arbitrary then:

$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$

$\blacksquare$