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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?

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    $\begingroup$ You can pick any base for the norm, but then you don't get the relation $\prod |x|=1$ (where the product is over one representative of each possible absolute value) which is nice for adelic stuff later on. Not sure what the original motivation was though, other than prime-power sizes appear as cardinality of quotient stuff maybe? $\endgroup$
    – runway44
    Dec 20, 2019 at 5:11
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    $\begingroup$ For any $c_p \in (0,1)$ you can set $|n|_p = c_p^{v_p(n)}$ for integers $n$ to define the $p$-adic absolute value. Once you want to use all the different absolute values on $\mathbf Q$ together (height functions on projective space over $\mathbf Q$ in terms of local height functions, Haar measure on adeles of $\mathbf Q$ in terms of local Haar measures) it's important to have the product formula $\prod_v |x|_v = 1$ for all nonzero $x \in \mathbf Q$. Setting $x = p$, a prime number, the product formula forces you to have $|p|_\infty c_p = 1$, so you need to use $c_p = 1/|p|_\infty = 1/p$. $\endgroup$
    – KCd
    Dec 20, 2019 at 7:24
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    $\begingroup$ Here is a more intrinsic reason, in terms of $\mathbf Q_p$ alone, to have $|p|_p = 1/p$. If $\mu$ is a Haar measure on $\mathbf Q_p$ then for each $\alpha \in \mathbf Q_p$ there is a unique number $c(\alpha) > 0$ such that $\mu(\alpha U) = c(\alpha)\mu(U)$ for all open subsets $U$ of $\mathbf Q_p$. The number $c(\alpha)$ is independent of the choice of Haar measure on $\mathbf Q_p$. It turns out that $c \colon \mathbf Q_p \rightarrow [0,\infty)$ is an absolute value on $\mathbf Q_p$, and in fact it is the standard $p$-adic absolute value where $|p| = 1/p$. $\endgroup$
    – KCd
    Dec 20, 2019 at 7:29
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    $\begingroup$ @Kcd Otherwise the motivation for choosing $|p|_p=p^{−1}$ is that for any $a\in \Bbb{Q}_p$ then $a|a|_p\in \Bbb{Z}_p$ $\endgroup$
    – reuns
    Dec 20, 2019 at 15:05
  • $\begingroup$ @reuns absolute values are usually viewed as real numbers, and products of real and $p$-adic numbers are not usually defined. Therefore a product $a|a|$ for general $a \in \mathbf Q_p$ and some unknown absolute value $|\cdot|$ on $\mathbf Q_p$ a priori does not make sense. $\endgroup$
    – KCd
    Dec 20, 2019 at 17:14

1 Answer 1

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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

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  • $\begingroup$ Can you please define $\mathfrak{P}$ and provide some intuition for them? I don't know what the rational primes are. I also don't see where you're using $|~|_\infty$ in the answer? $\endgroup$
    – Siddharth Bhat
    Dec 20, 2019 at 5:18
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    $\begingroup$ I think it should be over all primes (not "rational primes"). So it is the product of the usual absolute value of $r$ (denoted $|r|_\infty$ here) with all the $p$-adic absolute values (norms as you say) of $r$. This formula stems from the prime decomposition of $r$. $\endgroup$
    – Stefan Lafon
    Dec 20, 2019 at 5:20
  • $\begingroup$ @StefanLafon your latex is broken, and it's hard to see what you're trying to say in the space of a comment box. Could you please create a new answer, along with the definition of $\mathfrak P$? Thanks! $\endgroup$
    – Siddharth Bhat
    Dec 20, 2019 at 5:25
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    $\begingroup$ Please take a look at en.wikipedia.org/wiki/P-adic_order#p-adic_absolute_value $\endgroup$
    – Stefan Lafon
    Dec 20, 2019 at 5:27
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    $\begingroup$ @SiddharthBhat There's nothing technically wrong with the given definition of $\mathfrak{P}$. You can easily google "rational prime" - it just means your standard run-of-the-mill prime number, as opposed to prime elements or prime ideals in more general, abstract contexts. (I do find the choice of letter confusing though, since the letter $\mathfrak{P}$ is usually reserved for factoring prime ideals in extensions.) And $|\cdot|_{\infty}$ is used since the product has $|\cdot|_p$ where $p=\infty$ is allowed. $\endgroup$
    – runway44
    Dec 20, 2019 at 5:42

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