7
$\begingroup$

Following this question:

Given a valued field $K$, denote with $\bar{K}$ its algebraic closure and with $\hat{K}$ the completion. Then both $\hat{\bar{K}}$ and $\hat{\bar{\hat{K}}}$ are complete and algebraically closed. If $K=\mathbb{Q}$, these constructions always give $\mathbb{C}$ or $\mathbb{C}_p$ for a prime $p$, i.e. both constructions give the same fields.

Is this true for any field $K$? Or is there a field $K$ with a valuation $v$ such that for no extensions of $v$ to $\bar{K}$ we have $\hat{\bar{K}}=\hat{\bar{\hat{K}}}$?

$\endgroup$
  • 3
    $\begingroup$ I don't understand what you mean by "these constructions" since $\mathbb{R}$ and $\mathbb{Q}_p$ are not algebraically closed. $\endgroup$ – Laurent Moret-Bailly Aug 16 '16 at 16:14
  • 2
    $\begingroup$ And yes, it should say “$\mathbb C$ or $\mathbb C_p$”. $\endgroup$ – Emil Jeřábek Aug 16 '16 at 16:17
  • $\begingroup$ you're right, I edited the question $\endgroup$ – Martin Aug 22 '16 at 10:07
14
$\begingroup$

First you have to observe that since all extensions of the valuation to $\bar{K}$ are conjugate, $\hat{\bar{K}}$ is well-defined up to (non-unique) isomorphism.
Now, since $\hat{\bar{K}}$ is complete and $K$ is dense in $\hat{K}$, the inclusion $K\subset \hat{\bar{K}}$ extends continuously to $K\subset \hat{K}\subset \hat{\bar{K}}$ (in fact you can identify $\hat{K}$ with the closure of $K$ in $\hat{\bar{K}}$). Since $\hat{\bar{K}}$ is an algebraically closed extension of $\hat{K}$, it contains a unique copy of $\bar{\hat{K}}$, namely the algebraic closure of $\hat{K}$ in it. Moreover this copy obviously contains $\bar{K}$ which is dense in $\hat{\bar{K}}$, hence it is also dense. Since $\hat{\bar{K}}$ is complete, it is therefore isomorphic to the completion of $\bar{\hat{K}}$.

$\endgroup$
3
$\begingroup$

The completion of the algebraic closure of a valued field is algebraically closed and complete.

So any further operation of closure or completion gives you a field isomorphic to $\hat{\bar{K}}$.

$\endgroup$
  • 8
    $\begingroup$ I don't think this answers the question. $\endgroup$ – Laurent Moret-Bailly Aug 16 '16 at 18:53

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.