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I don't understand something about $\Bbb{Q}_p[\zeta_{p^\infty-1},p^{1/\infty}]$, especially trying to use its elements to approximate those of $\overline{\Bbb{Q}}_p\subset \Bbb{C}_p$.

For $a\in\overline{\Bbb{Q}}_p^*$ consider the map $a\mapsto U(a)$ where $U(a)$ is the unique unramified root of unity such that $v(a p^{-v(a)}-U(a)) > 0$. Set $U(0)=0$. Then construct the sequence $b_0=U(a)p^{v(a)},b_{k+1}=b_k + U(a-b_k)p^{v(a-b_k)}$ so that the sequence $v(a-b_k)$ is strictly increasing.

  • If the sequence $b_k$ doesn't converge then $\Bbb{Q}_p[p^{v(a-b_0)},p^{v(a-b_1)},\ldots]$ is an infinite extension of $\Bbb{Q}_p$ and the orbit of $a$ under the action of $Aut(\overline{\Bbb{Q}}_p/\Bbb{Q}_p)$ is infinite, impossible.

  • If the sequence $b_k$ converges, to $a$, then again $\Bbb{Q}_p[p^{v(a-b_0)},p^{v(a-b_1)},\ldots,U(a),U(a-b_0),U(a-b_1),\ldots]$ must be a finite extension of $\Bbb{Q}_p$ (as otherwise $a$ has infinitely many conjugates) so that any $a\in \overline{\Bbb{Q}}_p $ is in fact an element of $\Bbb{Q}_p[\zeta_{p^n-1},p^{1/n}]$ for some $n$, which doesn't make sense.

What am I missing?

ern
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  • I think I am overlooking that some $v(a-b_k)$ may have a power of $p$ in their denominator, so that the $Aut(\overline{\Bbb{Q}}p/\Bbb{Q}_p)$ conjugates have some additional $p^k$-th root of unity appearing, those are not in $\Bbb{Q}_p[\zeta{p^\infty-1},p^{1/\infty}]$ which allows $\Bbb{Q}_p[p^{v(a-b_0)},p^{v(a-b_1)},\ldots]$ to be an infinite extension of $\Bbb{Q}_p$. – ern Nov 05 '22 at 17:13
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    Already writing $p^{1/\infty}$ is too much abuse of notation for my taste. Do you mean that you are adjoining all $n$-th roots of $p$ for all $n$, or just one compatible sequence? (You are aware, I assume, that $p$ has $n$ different $n$-th roots in the algebraic closure, and in general they generate different field extensions.) – Torsten Schoeneberg Nov 06 '22 at 05:22
  • That being said, indeed I do not understand why said orbit should be infinite. – Torsten Schoeneberg Nov 06 '22 at 05:24
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    @TorstenSchoeneberg The idea is that $a$ has at least $[\Bbb{Q}p(b_k):\Bbb{Q}_p]$ distinct conjugates, because they are distinct in $\overline{\Bbb{Q}}_p/(a-b_k)\overline{\Bbb{Z}}_p$, but this holds only if there are no powers of $p$ in the denominators of the $v(a-b_k)$. Ie. things get more complicated when some powers of $p$ appear in the denominators of the $v(a-b_k)$, ie. when $a$ is not tame over $\Bbb{Q}_p$. So adding all the roots of $p$ (ie. adding the $\zeta{p^k}$) or just one compatible sequence is part of the confusion. – ern Nov 06 '22 at 08:44
  • I realized this after posting the question, so I guess my main concern now is if for $a$ non-tame, does the sequence $b_k$ have to converge or not? – ern Nov 06 '22 at 08:46
  • Related: https://mathoverflow.net/q/107481/27465. Your extension is bigger than the maximal tamely ramified extension precisely by also containing $p$-power roots of $p$ (some or all, depending on what the notation is supposed to mean). – Torsten Schoeneberg Nov 06 '22 at 17:47
  • Incidentally I do not understand how you conclude in your last paragraph, assuming convergence, that a) the extension you wrote down is finite and b) that then it is of the form you claim at the very end. – Torsten Schoeneberg Nov 06 '22 at 17:53
  • Not sure if that is the issue, but maybe there is a false conclusion from the terms of a sequence to its limit there, similar to one I pointed out in https://math.stackexchange.com/a/4274251/96384? – Torsten Schoeneberg Nov 06 '22 at 17:59

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