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P-adic norm, valuation, and expansion

  1. Feb 25, 2015 #1
    In this book I'm working from (P-adic numbers: An introduction by Fernando Gouvea), he gives an explanation of the p-adic expansion of a rational number which I'm pretty sure I understand this, but a bit later he talks about the p-adic absolute value, which from what I understand is the same as the norm. He defines the valuation like this:
    Which I can understand that, my interpretation is that this is basically another way of decomposing an integer that emphasizes the prime factor that we care about. So ##v_p(n)## is just the exponent on p in the the prime factorization of n.

    Then he discusses the p-adic absolute value, and in short he defines it as
    $$
    |x|_p=\left\{
    \begin{array}{l}
    |x|_p=p^{-\alpha}\text{ if } p^{\alpha}|x\\
    |x|_p=1 \text{ if }p\not|\text{ } x
    \end{array}
    \right.
    $$

    I'm a little foggy as to how these 3 things are related. I'm pretty sure he defines the valuation the way he does for the proof that ##|x|_p## is a metric. But I don't see how the expansions are related.
     
  2. jcsd
  3. Mar 2, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Mar 4, 2015 #3
    I've resolved this question. When we find the p-adic expansion we can directly read off the value of the p-adic norm. Numbers whose p-adic norms are equal to 1 correspond to the numbers who p-adic expansions have a non-zero constant term.
     
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