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I am looking at some notes that give an example of a Cauchy sequence that doesn't converge in $\mathbb{Q}$ with respect to the $p$-adic absolute value. Their example is to let $1 < a< p-1$ and consider the sequence $(x_n) = (a^{p^n}).$ They claim that the sequence doesn't converge in $\mathbb{Q}$. Their argument is that if the sequence had a limit $x \in \mathbb{Q}$, then we would have $x^p = x$ (which I understand) and $|x-a|_p < 1$ (which I also understand). But then they claim that therefore $a$ is a non-trivial $(p-1)$-th root of unity. I do not understand that. Could someone please clarify?

IAmNoOne
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Mehta
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  • Take $x^p = x$ and divide by $x$ to get $x^{p=1}=1$. – lhf Sep 06 '14 at 02:28
  • That would imply that, since $x$ is either 1 or -1. But the notes say that $a$ must be a root of unity, and I don't see where the contradiction is. – Mehta Sep 06 '14 at 02:31
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    I think that there is a typo: Therefore $x$ (and not $a$) is a non trivial...We see easily that $|a-1|_p=|a+1|_p=1$, so $x\not =\pm 1$. – Kelenner Sep 06 '14 at 05:09

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Suppose $(x_n)=(a^{p^n})$ converges to $x$ in $\mathbb{Q}$ with respect $p$-adic norm. Then using strong triangle inequality and Fermat's Little Theorem it is easy to see $|x-a|_p<1$. It is easy to prove $(x_n) $ is a Cauchy sequence. Also $x=\lim_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}x_{n+1}=\lim_{n\rightarrow\infty}x_n^p=(\lim_{n\rightarrow\infty}x_n)^p=x^p.$ We proved $|x-a|_p<1$ and $x^{p-1}=1$. Since $|x-a|_p<1, $ implies that $p\mid x-a$. Since $x^{p-1}=x$ and $x\in\mathbb{Q},$ implies that $x=1$. But $x$ cannot be 1. Suppose $x=1$. Then $p\mid 1-a$. That is $1-a=pk$ for some $k\in\mathbb{Z}$, that is $a=-pk+1$. If $k<0$ then $a>p$ and if $k>0$ then $a<1$, which is a contradiction, since $1<a<p-1.$ Hence $(x_n)$ does not converge to $x$ in $\mathbb{Q}$ with $p$-adic norm.

user26857
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S. Pitchai Murugan
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    Please, can you help me understand why is $|x-a|<1? I don't understand how to use the triangle inequality. Thanks! – robbis Apr 26 '19 at 17:43
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    Could there be a mistake in this answer? Specifically the part about $|x-a|_p <1$. Using Fermat's little theorem we could argue that $a^{p^{n-1}}-1 \equiv 0 \mod p^{n-1}$ and so that its norm is less than 1, but we don't have exactly that. Instead we have (not $x$ but for $x_n$): $|a^{p^n}-a|_p = |a|_p|a^{p^n-1}-1|_p$. But as far as I can see $p^n-1$ isn't a useful value of Euler's $\phi$ function that we can use Fermat's theorem with. Or should this be approached differently? – typedrums Nov 20 '19 at 18:14