Is it possible for an incompletely valued field to have a finite-degree extension which is complete?

Asked Jun 13 '23 at 15:48

Active Jun 30 '23 at 01:01

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Is it possible to have a completely valued field $(L,|\cdot|)$ and a subfield $K$ of $L$ such that

$\bullet$ $L/K$ is a finite extension

and that

$\bullet$ $(K,|\cdot| \, |_K)$ is not complete?

Note that this is possible if $L/K$ is infinite, for example we can have $L = \mathbb{R}$ (see here for an example). Thank you for any help in advance for $L/K$ being finite.

This link deals with the case where $L/K$ is separable. I would be happy to receive some arguments about the inseparable case.

edited Jun 30 '23 at 01:01

asked Jun 13 '23 at 15:48

Jianing Song

1,707

Does this answer your question? Subfield of a complete valuation field also complete – Torsten Schoeneberg Jun 23 '23 at 20:02
@TorstenSchoeneberg No, that question assumes that $L/K$ is separable and that $|\cdot|$ is discrete, which are much too strong. – Jianing Song Jun 29 '23 at 11:50
If you read the answer to the end, you see that the condition of the valuation to be discrete can be replaced by the much weaker one that the bigger field is not algebraically closed. This narrows down the search for possible examples significantly. – Torsten Schoeneberg Jun 29 '23 at 15:24
@TorstenSchoeneberg Ok thanks, and what about separability? – Jianing Song Jun 29 '23 at 15:29
I guess that is the one case left open. If I'm not mistaken, that should boil down to excluding the case $L= K(\sqrt[p]{a})$ in characteristic $p$. It seems likely that this case could be done by hand (the elements and the valuations of $L$ and $K$ are quite explicitly linked to each other), in any case you could at least refine your question to that. – Torsten Schoeneberg Jun 29 '23 at 17:27
@TorstenSchoeneberg Thank you for the suggestion! – Jianing Song Jun 30 '23 at 01:01

Is it possible for an incompletely valued field to have a finite-degree extension which is complete?

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