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Show that $\frac{2}{p-1}$ is a $p$-adic integer and find its p-adic expansion. P-adic numbers really make little sense to me so any help explaining what to do and why would be really appreciated. Thanks

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Hint: $\frac{1}{1-x}=1+x+x^2+x^3+x^4+\ldots$. –  Chris Eagle Apr 25 '12 at 11:25
    
If they make little sense to you, get the book "p-Adic Analysis Compared with Real" by Svetlana Katok. That is, if you have had real analysis and that made sense, then this book will help you understand p-adic numbers. It's at an advanced undergraduate level. I've read it twice. It's only about 150 pages total and it's only $23 on Amazon (in the U.S.). I have heard "p-adic Numbers: An Introduction" by Gouvea is good too but I have never read it. –  Graphth Apr 25 '12 at 12:58

1 Answer 1

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If $p=2$ then your expression is just equal to $2\in\Bbb Z$.

If $p$ is odd then numerator and denominator are prime to $p$, thus $p$-adic units. The quotient of $p$-adic units is a $p$-adic unit. Recall that a $p$-adic unit is an invertible element in the ring $\Bbb Z_p$ and can be recognized as those $p$-adic integers having a non-zero "costant" term in their $p$-adic expansion.

In order to get the $p$-adic expansion of $\frac2{p-1}$ just rewrite it as $-2\frac1{1-p}$ and apply Chris Eagle's hint.

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Right thanks for that everyone. After a bit of playing around I get that $a_0=(p-2)$ $ a_i=(p-3)$ for $i=1,2,...$ so $(p-2)+\sum_{i=1}^{\infty}(p-3)p^i$ is the expansion. Does that sound right? –  Mike Davies Apr 26 '12 at 18:49

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