P-adic Norm is Norm/Proof 2

Proof
Recall that the $p$-adic norm is defined as:


 * $\forall q \in \Q: \norm r_p := \begin{cases}

0 & : r = 0 \\ p^{- k} & : r \ne 0 \end{cases}$

where:


 * $\displaystyle r = p^k \frac m n$

and:


 * $k, n \in \Z, m \in \Z_{\ne 0} : p \nmid m, n$

where $\nmid$ stands for "does not divide".

We must show the following hold for all $r_1$, $r_2 \in \Q$:

Norm Axiom $(\text N 1)$
Let $r \in \Q : r \ne 0$.

Let $k, m\in \Z, n \in \Z_{\ne 0} : p \nmid m, n$.

Suppose $r = 0$.

By definition:


 * $\norm {r}_p = 0$

Suppose $\displaystyle r = p^k \frac m n \ne 0$

By definition:


 * $\displaystyle \norm {r}_p = \frac 1 {p^k} > 0$

Suppose $\norm r_p = 0$.

By definition:


 * $r = 0$

Norm Axiom $(\text N 2)$
Suppose $r_1 = 0$ or $r_2 = 0$.

From axiom $(\text N 1)$, $\norm {r_1}_p = 0$ or $\norm {r_2}_p = 0$.

Suppose $r_1 \ne 0 \ne r_2$.

Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid n_1, n_2, m_1, m_2$

Let $\displaystyle r_1 = p^{k_1} \frac {m_1} {n_1}, r_2 = p^{k_2} \frac {m_2} {n_2}$

Then:


 * $\displaystyle r_1 r_2 = p^{k_1 + k_2} \frac {m_1 m_2}{n_1 n_2}$

We have that $p \nmid m_1$, $p \nmid m_2$.

Since $p$ is prime:


 * $p \nmid m_1 m_2$.

Similarly:


 * $p \nmid n_1 n_2$.

Therefore:

Norm Axiom $(\text N 3)$
Suppose one of the following is true:


 * $r_1 = 0$


 * $r_2 = 0$


 * $r_1 + r_2 = 0$

Then the result is straightforward.

Suppose $r_1 \ne 0$, $r_2 \ne 0$, $r_1 + r_2 \ne 0$.

Let $\displaystyle r_1 = p^{k_1} \frac {m_1}{n_1}, r_2 = p^{k_2} \frac{m_2}{n_2}$ where:


 * $\displaystyle k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$

Then:

By Fundamental Theorem of Arithmetic:


 * $\exists ! \tilde k \in \Z_{\ge 0} : \exists m \in \Z : p \nmid m : \tilde m = p^{\tilde k} m$

Obviously, $p \nmid n_1 n_2$

Hence:

All norm axioms are seen to be satisfied.

Hence the result.