Why is it important that the $p$-adic absolute value satisfy multiplicativity?

Question

I suppose my question is really "why do we require norms in general to satisfy multiplicativity?". I ask this because for the usual absolute value on $\mathbb R$, I never feel like multiplicativity plays any "key role"; compare this to for example the omnipresent triangle inequality for the usual absolute value on $\mathbb R$, and in general in any metric space.

Suppose we are learning about $p$-adic numbers for the first time, say via this nice presentation: https://math.uchicago.edu/~tghyde/Hyde%20--%20Introduction%20to%20p-adics.pdf. I get that we would like some sort of norm $|a|_n$ for $a\in \mathbb Z_n$ that is smaller for values of $a$ that have a larger number $N$ of rightmost zeroes (in base $n$), and I agree something like $c^{-N}$ is a natural choice (really given the limited menagerie of "standard elementary functions" that's pretty much our only choice, other than perhaps an inverse power function like $\frac 1N$ or $\frac 1{N^k}$), but why do we emphasize that such a norm MUST be multiplicative?

EDIT: basically, I’m looking for the easiest example of the $p$-adic norm being used in some application to prove something “interesting” (outside the abstract theory of absolute values), that requires multiplicativity; in particular this rules out Ostrowski’s theorem.

Sort-of related is Why does the p-adic norm use base p? since it is sort of related to this "motivating the $p$-adic norm" business I have going on here.

Answer 1

The $p$-adic norm on $\mathbb{Q}$ is a standard example of an algebraic absolute value. In particular, we are viewing $\mathbb{Q}$ as a ring (more specifically an integral domain) here, and so the only notions of absolute values that are of interest are the ones that respect the ring operations. None of the other axioms refer to ring multiplication in any way, so the requirement that $\lvert \cdot \rvert$ be multiplicative can be thought of as the simplest way to make the absolute value respect the ring multiplication.

Answer 2

I apologize in advance that my answer is somehow nonmathematical, but I believe it's one aspect of the situation.

Other comments and answers have provided evidence that multiplicativity is an essential assumption in many results concerning $p$-adic norms. But if you think about how mathematical research is done, then the answer might be just the other way around.

Why do we require the $p$-adic norm to satisfy multiplicativity? Because researches in this direction turned out to produce many interesting and useful results, such as Ostrowski's theorem, while not much could be done without this requirement.

As a comparison, consider the related question of why $p$-adic norm uses prime number $p$. It turns out that in earlier literatures, there exists "$g$-adic (pseudo-)norm" with not necessarily prime $g$ (see e.g. Introduction to $p$-adic Numbers and their Functions, by Kurt Mahler). But not many important results have been established for these pseudonorms, hence you don't see them very often now.

Answer 3

Well, if you loosen the restriction of multiplicativity to submultiplicativity, you get a far more intricate theory which seems to be not fully developed yet, but which (as @asahay points out in his answer and comments) connects naturally to nonarchimedean functional analysis (books by Bosch/Güntzer/Remmert, and Schneider), to Berkovich Spaces, and to Scholze's Perfectoid Spaces. See Kedlaya's paper On Commutative Nonarchimedean Banach Fields (arXiv link) for a good introduction.

As a first teaser, note that for any finite family (and many infinite families) $(\Vert \Vert_i :i \in I)$ of submultiplicative ring norms, their supremum is again a submultiplicative ring norm. I have an inkling that on $\mathbb Q$, all submultiplicative norms arise in this way from the multiplicative absolute values well-known via Ostrowski, but to be honest I'm not even in possession of a proof for that.