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 xZx\in\mathbb{Z}, x0x\ne 0 is pnp^{-n}, where n=max{kN:x/pkZ}n=\max\{k\in\mathbb{N}: x/p^k\in\mathbb{Z}\}. In other words, the p-adic norm of x is 1/pn1/p^n where pnp^n is the maximal powerof p, which divides x. It is denoted as xp|x|_p . 0p=0|0|_p=0 by definition.

Properties:

  1. If x=±p1α1p2α2pmαmx=\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 xpk=pkαk|x|_{p_k}=p_k^{-\alpha_k}
  2. This norm is multiplicative: if x=±p1α1p2α2pmαmx=\pm p_1^{\alpha_1}p_2^{\alpha_2}\dots p_m^{\alpha_m} and y=±p1β1p2β2pmβmy=\pm p_1^{\beta_1}p_2^{\beta_2}\dots p_m^{\beta_m} then xy=±p1α1+β1p2α2+β2pmαm+βmxy=\pm p_1^{\alpha_1+\beta_1}p_2^{\alpha_2+\beta_2}\dots p_m^{\alpha_m+\beta_m} and xypk=pk(αk+βk)=xpkypk|xy|_{p_k}=p_k^{-(\alpha_k+\beta_k)}=|x|_{p_k}|y|_{p_k}
  3. x+ypmax{xp,yp}|x+y|_p\leq \max\{|x|_p, |y|_p\} . Indeed, if pnp^n divides xx and pmp^m divides y, then at least pmin{n,m}p^{\min\{n,m\}} divides x+y. So |x+y| is at most pmin{n,m}=max{pn,pm}p^{-\min\{n, m\}}=\max\{p^{-n},p^{-m}\} and x+ypmax{xp,yp}|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=k=makpkQpx=\sum\limits_{k=-m}a_kp^k\in\mathbb{Q}_p is pnp^{-n} where n is the least integer such that ak0a_k\ne0.

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


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