## 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 $\nu_p$:

$(1): \quad \nu_p$ is a mapping into $\Z \cup \set {+\infty}$
$(2): \quad \nu_p$ satisfies the valuation axioms $\text V 1$, $\text V 2$ and $\text V 3$
$(3): \quad \nu_p$ 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 $x \ne 0$.

$\exists v \in \Z: \norm x_p = p^{-v}$

Hence:

 $\ds \map {\nu_p} x$ $=$ $\ds -\log_p \norm x_p$ Since $x \ne 0$ $\ds$ $=$ $\ds -\log_p p^{-v}$ Definition of $v$ $\ds$ $=$ $\ds -\paren {-v}$ Definition of Real General Logarithm $\ds$ $=$ $\ds v$ $\ds$ $\in$ $\ds \Z$ Definition of $v$

$\Box$

### $\nu_p$ satisfies $(\text V 1)$

If $x = 0$ then:

 $\ds \map {\nu_p} {0 \cdot y}$ $=$ $\ds \map {\nu_p} 0$ $\ds$ $=$ $\ds +\infty$ Definition of $\nu_p$ $\ds$ $=$ $\ds +\infty \cdot \map {\nu_p} y$ Definition of Extended Real Multiplication $\ds$ $=$ $\ds \map {\nu_p} 0 \cdot \map {\nu_p} y$ Definition of $\nu_p$

Similarly, if $y = 0$ then:

 $\ds \map {\nu_p} {x \cdot 0}$ $=$ $\ds \map {\nu_p} x \cdot \map {\nu_p} 0$

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

 $\ds \map {\nu_p} {x y}$ $=$ $\ds -\log \norm {x y}_p$ $x y \ne 0$ $\ds$ $=$ $\ds -\log \norm x_p \norm y_p$ Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity $\ds$ $=$ $\ds -\paren {\log \norm x_p + \log \norm y_p}$ Sum of General Logarithms $\ds$ $=$ $\ds \paren {-\log \norm x_p} + \paren {-\log \norm y_p}$ $\ds$ $=$ $\ds \map {\nu_p} x + \map {\nu_p} y$ Definition of $\nu_p$

$\Box$

### $\nu_p$ satisfies $(\text V 2)$

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:

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

$\Box$

### $\nu_p$ satisfies $(\text V 3)$

Suppose $x = 0$.

Then:

 $\ds \map {\nu_p} {0 + y}$ $=$ $\ds \map {\nu_p} y$ $\ds$ $\ge$ $\ds \min \set {\map {\nu_p} 0, \map {\nu_p} y}$ Definition of Min Operation

Similarly, if $y = 0$ then:

 $\ds \map {\nu_p} {x + 0}$ $\ge$ $\ds \min \set {\map {\nu_p} x, \map {\nu_p} 0}$

Suppose $x + y = 0$.

Then:

 $\ds \map {\nu_p} {x + y}$ $=$ $\ds \map {\nu_p} 0$ $\ds$ $=$ $\ds +\infty$ Definition of $\nu_p$ $\ds$ $\ge$ $\ds \map {\nu_p} x$ Definition of Extended Real Number Line $\ds$ $\ge$ $\ds \min \set {\map {\nu_p} x, \map {\nu_p} y}$ Definition of Min Operation

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

Then:

 $\ds \norm {x + y}$ $\le$ $\ds \max \set {\norm x_p, \norm y_p}$ Non-Archimedean Norm Axiom $\text N 4$: Ultrametric Inequality $\ds$  $\ds$ $\ds \leadsto \ \$ $\ds \log \norm {x + y}$ $\le$ $\ds \log \max \set {\norm x_p, \norm y_p}$ Logarithm is Strictly Increasing $\ds$ $=$ $\ds \max \set {\log \norm x_p, \log \norm y_p}$ Logarithm is Strictly Increasing $\ds$  $\ds$ $\ds \leadsto \ \$ $\ds -\log \norm {x + y}$ $\ge$ $\ds -\max \set {\log \norm x_p, \log \norm y_p}$ Inversion Mapping Reverses Ordering in Ordered Group $\ds$ $=$ $\ds \min \set {-\log \norm x_p, -\log \norm y_p}$ $\ds$  $\ds$ $\ds \leadsto \ \$ $\ds \map {\nu_p} {x + y}$ $\ge$ $\ds \min \set {\map {\nu_p} x, \map {\nu_p} y}$ Definition of $\nu_p$

$\Box$

### $\nu_p$ 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:

 $\ds \map {\nu_p} x$ $=$ $\ds -\log \norm x_p$ Definition of $\nu_p$ $\ds$ $=$ $\ds \map {\nu_p^\Q} x$ Definition of $p$-adic norm $\norm {\,\cdot\,}_p$ on rational numbers $\Q$

$\blacksquare$