P-adic Number is Power of p Times P-adic Unit

Theorem

Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Let $\Z_p^\times$ be the $p$-adic units.

Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ denote the $p$-adic valuaton on the $p$-adic numbers.

Let $a \in \Q_p$.

Then there exists $u \in \Z_p^\times$ such that:

$a = p^{\map {\nu_p} a} \cdot u$

Proof

From P-adic Number times P-adic Norm is P-adic Unit, there exists $n \in \Z$ such that:

$p^n a \in \Z_p^\times$

where

$p^n = \norm a_p$

We have:

\(\ds \map {\nu_p} a\)

\(=\)

\(\ds -\log_p \norm a_p\)

Definition of P-adic Valuation on P-adic Numbers

\(\ds \)

\(=\)

\(\ds -\log_p p^n\)

\(\ds \)

\(=\)

\(\ds -n\)

Let $u = p^n a$.

Then:

\(\ds a\)

\(=\)

\(\ds p^{-n} u\)

\(\ds \)

\(=\)

\(\ds p^{\map {\nu_p} a} \cdot u\)

$\blacksquare$