To recapitulate some definitions, A norm on a field $K$ is a function $|\cdot|: K \rightarrow \mathbb R$ such that:
- $|x|_p = 0 \iff x = 0$
- $|xy| = |x||y|$
- $|x + y| \leq |x| + |y|$
To define the $p$-adic norm on $\mathbb Q$, we first create a valuation function $v_p: \mathbb Z \rightarrow \mathbb R$, such that $v_p(n)$ is the unique number such that ${n = p^{v_p(n)} m,~p \nmid m}$. That is, $v_p(n)$ is the highest power of $p$ occuring in the prime factorization of $n$. I set $v_p(0) = \infty$, since $0$ can be divided by $p$ infinitely many times.
Next, we extend the valuation to the rationals by setting $v_p : \mathbb Q^\times \rightarrow \mathbb R$, $v_p(a/b) \equiv v_p(a) – v_p(b)$.
We now note that the valuation is additive while the norm is multiplicative. Also, as the power $i$ in $p^i$ grows larger, so too does $v_p(p^i)$. However, we want large powers of $p$ to become smaller (so that infinite series in $p^i$ converge). We solve both of these by creating the p-adic norm as:
$|n|_p \equiv p^{-v_p(n)}$
However, the choice of base $p$ is arbitrary here. As far as I can tell, we could have just as well chose $|n|_p \equiv e^{-v_p(n)}$, and all the properties of a norm would have worked out.
So, why do we pick base $p$? Is there something particular nice that happens? Why can't we pick any base $c > 1$ for the norm?
Best Answer
the nice thing is that for any positive $r \in \mathbb{Q}$ you get:
$$ \prod_{p \in \mathcal{P \cup \{\infty\}}} |r|_p = 1 $$
where $\mathcal{P}$ is the set of (rational) primes and $|\,|_\infty$ is the usual archimedean absolute value