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In this question, Is the algebraic closure of a $p$-adic field complete

GEdgar's answer,

Any $x$ which is algebraic of degree $n$ over $\mathbb Q_2$ has a unique series expansion $$ x = \sum 2^{u_j} $$ where $u_j \to \infty$ (unless it is a finite sum) and all $u_j$ are rationals with denominator that divides $n!$

But why this claim holds?

I asked in the comment form, but this answer was posted $9$ years ago, so it is difficult to get reaction.

Thank you in advance.

GEdgar
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    As I commented under your previous question https://math.stackexchange.com/questions/4273717/any-x-which-is-algebraic-of-degree-n-over-mathbb-q-2-has-a-unique-series : I find it strange, since it implies that every finite extension of $\mathbb Q_2$ is totally ramified, which is not true (or I'm confused somewhere) – user8268 Oct 11 '21 at 21:22
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    To make it more concrete and elementary, let $x$ be a solution of $x^2 = x + 1$. Then all the $u_j$'s should be non-negative, and either all are actually positive, and so $v_2(x)>0$ and $v_2(1+x)=0$, or the lowest one is $1$, and so $v_2(x+1)>0$ and $v_2(x)=0$. Both of these are in contradiction with $v_2(x+1)=v_2(x^2)=2v_2(x)$. I wonder what's the right formulation of @GEdgar 's claim. – user8268 Oct 12 '21 at 09:52
  • sorry, should be "...or the lowest one is $0$..." in my comment above – user8268 Oct 12 '21 at 12:00
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    If there is a right formulation, then I think it should hold only for totally ramified extensions, because the proposed series has only coefficients $\in {0,1}$ i.e. the residue field seems to still be $\mathbb F_2$ (accordingly, @user8268 exhibits the problem for the smallest unramified extension). But even when restricting to those, the issue remains that for most $k$, something like "$2^{1/k}$" is not even a well-defined symbol to begin with, cf. https://math.stackexchange.com/q/4256962/96384. – Torsten Schoeneberg Oct 12 '21 at 15:20

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This seems completely false, even if you make the restriction that @TorstenSchoeneberg suggested in a comment.

For instance, I don’t believe that $\sqrt6$, generator of the totally ramified quadratic extension $\Bbb Q_2(\sqrt6\,)$, can be written in the form quoted by @GEdgar. If I’m wrong here, then I beg Gerry to show us the appropriate expansion.

Most serious of all is the observation of @user8268 that all Edgar’s numbers seem clearly to live only in ramified extensions of $\Bbb Q_2$, while the roots of unity of odd order are not there at all.

Lubin
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