P-adic valuation

**Bingk** · July 7th 2010, 05:57 PM

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

Now, if $\gamma$ ends in something else other than zero or one, what we are supposed to do is look for a digit $f$ so that $f\gamma$ will end in a one, and thus can be inverted by the above method (if it's correct

). Since we have the inverse of $f\gamma$ , to get the inverse of $\gamma$ , we multiply the inverse of $f\gamma$ by $f$ again. Is this correct?

**chiph588@** · July 7th 2010, 06:15 PM

If $\alpha$ ends in $d$ , pick the unique $f$ such that $0<f<p$ and $df\equiv1\bmod{p}$ .

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

So $\beta\beta^{-1}=1\implies(\alpha)\cdot\left(f\beta^{-1}\right)=1\implies (\alpha)\cdot\left(f(f\alpha)^{-1}\right)=1\implies\alpha^{-1}=f(f\alpha)^{-1}$ .

**melese** · July 8th 2010, 12:17 AM

Thanks for the link!

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