What Does a = b (mod pZp) Mean in p-adic Numbers?

  • Thread starter Thread starter pablis79
  • Start date Start date
  • Tags Tags
    Numbers
pablis79
Messages
6
Reaction score
0
Hi all,

If Zp is the ring of p-adic integers, what does the notation a = b (mod pZp) mean ? I understand congruence in Zp, i.e., a = b (mod p) implies a = b +zp, where z is in Zp (and a, b in Zp). However, I don't get what is meant by (mod pZp) ... does this mean a = b (mod p^k) for all k >= 1 ?

Thanks,
P
 
Physics news on Phys.org
The p-adic numbers can be written in the form

\sum_{i=0}^{+\infty}{a_ip^i}

for 0\leq a_i\leq p-1.

The ideal p\mathbb{Z}_p is the ideal generated by p. It contains elements like

\sum_{i=1}^{+\infty}{a_ip^i}

Now, we say that x=y~(mod~p\mathbb{Z}_p) if x-y\in p\mathbb{Z}_p.
 
Ok, thanks micromass. So put another way it means that x and y are congruent modulo p in Zp.

Cheers!
 
have another question , if p \rightarrow infty , how can you prove that the infinite prime p= \infty is just the hole of the Real numbers ??
 
zetafunction said:
have another question , if p \rightarrow infty , how can you prove that the infinite prime p= \infty is just the hole of the Real numbers ??
As phrased, this question makes no sense.

I suspect what you're asking is, "Why do people say that \mathbb Q_p = \mathbb R when p=\infty?" This is more a matter of convention (and convenience) than anything. There is no "let p -> infinity" going on. What is going on is that Q has several absolute values: up to equivalence, these are the p-adic absolute values (|.|_p) and the usual absolute value (|.|). One then completes Q at these absolute values to obtain the fields Q_p and R, respectively. One says "Q_p is the completion at of Q at p". Then there are good reasons to think of the usual absolute value as coming from an "infinite" prime, and to say that "R is the completion of Q at the infinite prime".
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top