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Let $a\in \mathbb{Z}$ be relatively prime to $p$ prime. Then show that the seqeunce $\{a^{p^{n}}\}$ converges in the $p$-adic numbers.

This to me seems very counter intuitive. Since $(a,p)=1$ the norm will always be $1$. I really have no idea what to do with $|a^{p^{n}}-a^{p^{m}}|$ factoring gets me nothing and we can't use the nonarchimedian property because they have the same norm. Any hints or ideas would be great. Thanks.

Jyrki Lahtonen
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TheNumber23
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1 Answers1

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Recall that the Euler totient function has values $\phi(p^n)=p^{n-1}(p-1)=p^n-p^{n-1}$ for all $n$. This means that for all $a$ coprime to $p$ we have the congruence $$ a^{p^n}\equiv a^{p^{n-1}}\pmod{p^n}. $$ By raising that congruence to power $p^{m-n}$ a straightforward induction on $m$ proves that $$ a^{p^m}\equiv a^{p^{n-1}}\pmod{p^n} $$ for all $m\ge n$. This holds for all $n$ implying that the sequence is Cauchy, and thus convergent w.r.t. the $p$-adic metric.

Jyrki Lahtonen
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  • Very good hint. It also explained my incorrect intuition. – TheNumber23 Nov 06 '14 at 20:09
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    And if you think of what this means in the $p$-adic integers $\mathbb Z_p$, you see that it’s saying that the limit of that sequence is the $(p-1)$-th root of unity that’s congruent to $a$ modulo $p$. – Lubin Nov 10 '14 at 04:08