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Thread: P-adic valuation

  1. #1
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    P-adic valuation

    Hello,

    I've been reading about p-adics and I think I understand most of the basics (Thanks to David A. Madore's "A First Introduction to p-adic Numbers" found at http://www.madore.org/~david/math/padics.pdf).

    I need some clarification (maybe an example would help) on how to get the inverse of a p-adic integer.

    The method used (in the paper) is for getting the inverse of $\displaystyle 1-\alpha$. So, if you have a p-adic integer, say $\displaystyle \gamma$, which ends in $\displaystyle 1$, you have to let $\displaystyle \gamma = 1 - \alpha$, and solve for $\displaystyle \alpha$, which will end with a $\displaystyle 0$. Then the inverse of $\displaystyle \gamma = 1 + \alpha + \alpha^2 + ...$. Is this the correct way of getting the inverse of $\displaystyle \gamma$?

    Now, if $\displaystyle \gamma$ ends in something else other than zero or one, what we are supposed to do is look for a digit $\displaystyle f$ so that $\displaystyle f\gamma$ will end in a one, and thus can be inverted by the above method (if it's correct ). Since we have the inverse of $\displaystyle f\gamma$, to get the inverse of $\displaystyle \gamma$, we multiply the inverse of $\displaystyle f\gamma$ by $\displaystyle f$ again. Is this correct?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    If $\displaystyle \alpha $ ends in $\displaystyle d $, pick the unique $\displaystyle f $ such that $\displaystyle 0<f<p $ and $\displaystyle df\equiv1\bmod{p} $.

    Now $\displaystyle \beta=f\alpha $ has a last digit of $\displaystyle 1 $. We then can find the inverse of $\displaystyle \beta $.

    So $\displaystyle \beta\beta^{-1}=1\implies(\alpha)\cdot\left(f\beta^{-1}\right)=1\implies (\alpha)\cdot\left(f(f\alpha)^{-1}\right)=1\implies\alpha^{-1}=f(f\alpha)^{-1} $.
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  3. #3
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    Thanks for the link!
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