Answer to Question #188979 in Functional Analysis for Saji

Question #188979

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


1
Expert's answer
2021-05-07T12:32:00-0400

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\\}" ).


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