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Let $K$ be a finite extension of $\mathbf{Q}_p$, i.e., a $p$-adic field. (Is this standard terminology?)

Why is (or why isn't) an algebraic closure $\overline{K}$ complete?

Maybe this holds more generally:

Let $K$ be a complete Hausdorff discrete valuation field. Then, why is $\overline{K}$ complete?

I think I can show that finite extensions of complete discrete valuation fields are complete.

Asaf Karagila
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Harry
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    $\sum_{n=0}^\infty 2^{1/n} p^n$ converges but it is not in $\overline{\mathbf{Q}_p}$, it is in only in $\mathbf{C}_p$. – reuns Jun 23 '19 at 03:52

2 Answers2

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The algebraic closure has countably infinite dimension over $\mathbb Q_p$, and therefore (by the Baire category theorem) is not metrically complete. (Except the case $\mathbb Q_\infty = \mathbb R$, where the algebraic closure has finite dimension, and is metrically complete.)

explanation

In $\mathbb Q_p$, let $x_n$ be a solution of $X^{n}=p$. Then $\{x_2,x_3,x_4,\dots\}$ is linearly independent over $\mathbb Q_p$. But we still need a proof that the algebraic closure does not have uncountable dimension.

Torsten Schoenberg provided the missing part:
Countable dimension follows from Krasner / Hensel and compactness of $\mathbb Z_p$ which shows for each $d \in \mathbb N$ , $\mathbb Q_p$ only has only countably many (actually finitely many) extension of degree $d$ in $\overline{\mathbb Q_p}$. I doubt there is a more elementary argument for that.

former explanation
We see in Incompleteness of algebraic closure of p-adic number field, from $9$ years ago question that it is incorrect...

How about an example? In $\mathbb Q_2$, the partial sums of the series $$ \sum_{n=1}^\infty 2^{n+1/n} $$ belong to $\overline{\mathbb Q_2}$, but the sum of the series does not. The partial sums form a Cauchy sequence with no limit in $\overline{\mathbb Q_2}$.

Why does this sum not exist in $\overline{\mathbb Q_2}$ ?

It is not trivial, but interesting: 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 $n!$. (Maybe divides $n$ in fact?) But the series expansion in this example has arbitrarily large denominators.

GEdgar
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  • Is there any straight-forward example of a Cauchy sequence that does not converge? – fretty Mar 24 '12 at 13:45
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    I am astounded that GEdgar was able to answer fretty's request for an example 50 seconds before the request was posted. Surely that ought to be worth at least 1,000 points. – MJD Mar 24 '12 at 20:10
  • Could you tell me the reason why the denominator of $j_j$ devides $n!$? –  Oct 11 '21 at 14:15
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    One can avoid the use of countable dimension by applying Baire directly to the countable family of subsets $H_n := {x \in \overline{\mathbb Q_p} : [\mathbb Q_p(x) : \mathbb Q_p] \le n }$ whose union clearly is $\overline{\mathbb Q_p}$ and which are easily shown to be closed, but none of which can contain an open subset of $\overline{\mathbb Q_p}$ (because otherwise one could easily show it would contain all of $\overline{\mathbb Q_p}$). This is how it is proved in Schikhof's Ultrametric Calculus, Theorem 16.6. – Torsten Schoeneberg Oct 22 '21 at 03:33
  • Alternatively, countable dimension follows from Krasner / Hensel and compactness of $\mathbb Z_p$ which shows for each $d \in \mathbb N$, $\mathbb Q_p$ only has only countably many (actually finitely many) extension of degree $d$ in $\overline{\mathbb Q_p}$. I doubt there is a more elementary argument for that. – Torsten Schoeneberg Oct 22 '21 at 03:36
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Small addition: One might try to repeat this process of algebraic closure / analytic completion. Kürschak proved that this process terminates after one more step, i.e. the completion of $\overline{\mathbb{Q}_p}$ is algebraically closed.

More generally, if you start with a non-archimedean valued field $K$ and denote $\overline{K}$ the algebraic closure and $\widetilde{K}$ the analytic completion, then $\widetilde{\overline{\widetilde{K}}}$ is algebraically closed.

MichalisN
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    This result is not due to Teichmüller, but to Kürschak (1913). See page 5 of www.rzuser.uni-heidelberg.de/~ci3/hist_val.pdf. – KCd Mar 24 '12 at 22:32
  • @KCd: That's interesting, I think the professor who taught me this explicitly mentioned Teichmüller at the time, but perhaps it was a related result to this. Anyway, I corrected my answer. – MichalisN Mar 26 '12 at 12:36