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A Question about p-adic numbers

  1. Jan 11, 2012 #1
    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
     
  2. jcsd
  3. Jan 11, 2012 #2

    micromass

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    The p-adic numbers can be written in the form

    [tex]\sum_{i=0}^{+\infty}{a_ip^i}[/tex]

    for [itex]0\leq a_i\leq p-1[/itex].

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

    [tex]\sum_{i=1}^{+\infty}{a_ip^i}[/tex]

    Now, we say that [itex]x=y~(mod~p\mathbb{Z}_p)[/itex] if [itex]x-y\in p\mathbb{Z}_p[/itex].
     
  4. Jan 11, 2012 #3
    Ok, thanks micromass. So put another way it means that x and y are congruent modulo p in Zp.

    Cheers!
     
  5. Jan 12, 2012 #4
    have another question , if [tex] p \rightarrow infty [/tex] , how can you prove that the infinite prime [tex] p= \infty [/tex] is just the hole of the Real numbers ??
     
  6. Jan 13, 2012 #5

    morphism

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    As phrased, this question makes no sense.

    I suspect what you're asking is, "Why do people say that [itex]\mathbb Q_p = \mathbb R[/itex] when [itex]p=\infty[/itex]?" 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".
     
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