Completion and algebraic closure commutable

Question

The following corollary of Krasner´s Lemma says:

Let k be a global field and p a prime of k. Then $(\overline{k})_p=\overline{k_p}$. Im wondering if $(\overline{k})_p$ means the completion of $\overline{k}$ because i know that $\overline{\mathbb{Q}_p}$ is not complete. So i think it means

$\bigcup L_{ip}$ with the $L_{ip}$ ranging over all finite extensions of k. In particular $\bigcup L_{ip}$ doesn´t need to be complete. Am i correct with that? Thx for any help given!

You can find the corollary for example in Neukirch´s cohomology of number fields, 8.1.5.

I think the corollary says that the completion of an algebraic closure is an algebraic closure of the completion. For example, if you take $\overline{\mathbb{Q}}$ and you complete it with respect to an extension of the p-adic valuation you obtain an algebraic closure of $\mathbb{Q}_p$. Does make sense this? — user253407, Aug 06 '15 at 15:56

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

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The equality stated in that proposition is to be read in the following way: If you take the algebraic closure of $k_p$ you obtain the same field as if you take the union $\bigcup L_{iw}$ of all completions of finite extensions $L_i$ of $k$. This does not imply $\overline{k}_p$ to be complete. Actually it is false for $\overline{ \mathbb{Q}_p}$ as you mentioned. (Look here for a short proof of the last statement: link )

edited Apr 13 '17 at 12:21

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answered Aug 07 '15 at 20:48

cosinus

227

What completions are we considering of the $L_i$? Just ones lying above the $p$-adic valuation on $\mathbb{Q}$? – Dorebell Dec 10 '15 at 01:00
Yes, just those. – Lubin Dec 11 '15 at 05:11

Completion and algebraic closure commutable

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