Intereting Posts

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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?

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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 little fermat’s theorem it is easy to see $|x-a|_p<1$.

It is easy to prove $(x_n) $ is 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 $p|x-a$. Since $x^{p-1}=x$ and $x\in\mathbb{Q},$ implies that $x=1$. But $x$ cannot be 1. Suppose $x=1$. Then $p|(1-a)$. 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 $(x_n)$ not converges to $x$ in $\mathbb{Q}$ with $p$-adic norm.

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