7
$\begingroup$

It is widely known that the algebraic closure of the $p$-adic completion $\mathbb{Q}_p$ of $\mathbb{Q}$ isn't complete anymore. It's completion is complete and known as $\mathbb{C}_p$.

I have read in a book about non-archimedean analysis that in this case the process ends, which means that $\mathbb{C}_p$ is also algebraically closed.

My question is: is there an example of a field K, in which the algebraic closure $K^{alg}$ isn't complete, and the completion of $K^{alg}$ isn't algebraically closed ? And how do I construct such an example.

Improve this question
$\endgroup$
2
  • $\begingroup$ Notice you can assume that $K$ is algebraically closed to begin with, and then you are looking for an algebraically closed field whose completion for an absolute value is not algebraically closed. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 21, 2010 at 7:20
  • $\begingroup$ The question is a bit unclear: if $K$ is an arbitrary normed field, then its algebraic closure does not inherit canonically a norm (I don't know if it always does). If $K$ is complete then the norm has a unique extension. Hence I see the question making sense assuming $K$ complete (in which case one cannot assume $K$ algebraically closed as suggested by @MarianoSuárez-Álvarez). Or assuming that $K$ is an algebraically closed normed field (not assuming that it's algebraic over a complete subfield), which is more general. If I understand correctly, the answer is positive anyway. $\endgroup$
    – YCor
    Commented Jan 31, 2020 at 16:28

2 Answers 2

Reset to default
9
$\begingroup$

No, there is not.

If the valuation is archimedean, by Ostrowski the field is isomorphic to the real or complex numbers, so the algebraic closure will already be complete.

If the valuation is non-archimedean, the completion of the algebraic closure will always be algebraically closed. See for example here: http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Just a comment on this nice handout: it turns on the fact (so important in many applications of local fields) that if you have a polynomial with coefficients in the completion, you can approximate it coefficient-by-coefficient arbitrarily well by elements of the uncompleted field, and then a basic result says that if two polynomials over a complete field are sufficiently close in coefficients and one is irreducible, then so is the other and their splitting fields are isomorphic. What Brian does not say is that this has a name: <b>Krasner's Lemma</b>. $\endgroup$
    – Pete L. Clark
    Commented Jan 21, 2010 at 8:37
  • $\begingroup$ ...I should mention that in exchange for not saying "Krasner", Brian's handout pays close attention to what happens in positive characteristic. This needs a little more argument, since (and I didn't say this above) Krasner's Lemma is only true for separable polynomials. $\endgroup$
    – Pete L. Clark
    Commented Jan 21, 2010 at 8:40
3
$\begingroup$

There is a theorem of Kurschák which asserts that the completion of a valued algebraically closed fied is algebraically closed. This is proved in Paulo Ribenboim's The theory of classical valuations.

Improve this answer
$\endgroup$
2
  • $\begingroup$ This is a special case of Zariski's lemma, which says that any finite degree extension F/K is algebraic, from which the nullstellensatz follows immediately. $\endgroup$
    – Harry Gindi
    Commented Jan 21, 2010 at 8:47
  • $\begingroup$ And of course, Zariski's lemma is a very specialized form of Zariski's Main Theorem. $\endgroup$
    – Harry Gindi
    Commented Jan 21, 2010 at 8:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .