Answer to Question #188979 in Functional Analysis for Saji

Question

What is the p-adic absolute value or p-adic norm?

Answer 1

p-adic norm of an integer $x\in\mathbb{Z}$, $x\ne 0$ is $p^{-n}$, where $n=\max\{k\in\mathbb{N}: x/p^k\in\mathbb{Z}\}$. In other words, the p-adic norm of x is $1/p^n$ where $p^n$ is the maximal powerof p, which divides x. It is denoted as $|x|_p$ . $|0|_p=0$ by definition.

Properties:

1. If $x=\pm p_1^{\alpha_1}p_2^{\alpha_2}\dots p_m^{\alpha_m}$ is a decomposition of x into a product of primes, then $|x|_{p_k}=p_k^{-\alpha_k}$
2. This norm is multiplicative: if $x=\pm p_1^{\alpha_1}p_2^{\alpha_2}\dots p_m^{\alpha_m}$ and $y=\pm p_1^{\beta_1}p_2^{\beta_2}\dots p_m^{\beta_m}$ then $xy=\pm p_1^{\alpha_1+\beta_1}p_2^{\alpha_2+\beta_2}\dots p_m^{\alpha_m+\beta_m}$ and $|xy|_{p_k}=p_k^{-(\alpha_k+\beta_k)}=|x|_{p_k}|y|_{p_k}$
3. $|x+y|_p\leq \max\{|x|_p, |y|_p\}$ . Indeed, if $p^n$ divides $x$ and $p^m$ divides y, then at least $p^{\min\{n,m\}}$ divides x+y. So |x+y| is at most $p^{-\min\{n, m\}}=\max\{p^{-n},p^{-m}\}$ and $|x+y|_p\leq \max\{|x|_p, |y|_p\}$.

By the multiplicativity rule p-adic norm can be expanded to the set of rational numbers.

The p-adic norm of a non-zero p-adic series $x=\sum\limits_{k=-m}a_kp^k\in\mathbb{Q}_p$ is $p^{-n}$ where n is the least integer such that $a_k\ne0$.

This norm is also multiplicative ($|xy|_{p}=|x|_{p}|y|_{p}$ ) and Archimedean ($|x+y|_p\leq \max\{|x|_p, |y|_p\}$ ).