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.


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.)

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.

  • $\begingroup$ Is there any straight-forward example of a Cauchy sequence that does not converge? $\endgroup$ – fretty Mar 24 '12 at 13:45
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    $\begingroup$ 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. $\endgroup$ – MJD Mar 24 '12 at 20:10

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.

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    $\begingroup$ This result is not due to Teichmüller, but to Kürschak (1913). See page 5 of www.rzuser.uni-heidelberg.de/~ci3/hist_val.pdf. $\endgroup$ – KCd Mar 24 '12 at 22:32
  • $\begingroup$ @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. $\endgroup$ – MichalisN Mar 26 '12 at 12:36

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