P-adic Valuation Extends to P-adic Numbers

Theorem
Let $p$ be a prime number.

Let $\nu_p^\Q: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on the set of rational numbers.

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

Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ be defined by:
 * $\forall x \in \Q_p : \map {\nu_p} x = \begin{cases}

-\log_p \norm {x}_p : x \ne 0 \\ +\infty : x = 0 \end{cases}$

Then $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ is a valuation that extends $\nu_p^\Q$ from $\Q$ to $\Q_p$.

Proof
It needs to be shown that $\nu_p$:
 * $\quad \nu_p$ is a mapping into $\Z \cup \set {+\infty}$
 * $\quad \nu_p$ satisfies the valuation axioms $V(1)$, $V(2)$ and $V(3)$
 * $\quad \nu_p$ extends $\nu_p^\Q$.

Let $x, y \in \Q_p$.

$\nu_p$ is a mapping into $\Z \cup \set {+\infty}$
If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition.

Let $x \ne 0$.

By P-adic Norm of p-adic Number is Power of p then:
 * $\exists v \in \Z: \norm x_p = p^{-v}$

Hence:

$\nu_p$ satisfies $(V1)$
If $x = 0$ then:

Similarly, if $y = 0$ then:

If $x \ne 0, y \ne 0$ then:

$\nu_p$ satisfies $(V2)$
If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition.

If $x \ne 0$ then $\map {\nu_p} x \in \Z$ by the above.

Hence:
 * $\map {\nu_p} x = +\infty \iff x = 0$

$\nu_p$ satisfies $(V3)$
Suppose $x = 0$.

Then:

Similarly, if $y = 0$ then:

Suppose $x + y = 0$.

Then:

Suppose $x \ne 0, y \ne 0, x + y \ne 0$.

Then:

$\nu_p$ extends $\nu_p^\Q$
Let $x \in \Q$.

If $x = 0$ then $\map {\nu_p} 0 = +\infty = \map {\nu_p^\Q} 0$.

Let $x \ne 0$.

The $p$-adic norm $\norm{\,\cdot\,}_p$ on $p$-adic numbers $\Q_p$ is an extension of the $p$-adic norm $\norm{\,\cdot\,}_p$ on rational numbers $\Q$ by definition.

Hence: