15
$\begingroup$

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.

$\endgroup$
19
$\begingroup$

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 $\overline{\mathbb Q_2}$.

added

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.

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

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.

$\endgroup$
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.