Timeline for Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?
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| Apr 3, 2021 at 22:04 | comment | added | Uri Bader | I am convinced now that $F$ is an example of a field $\simeq\mathbb{C}$ which is endowed with a completely metrizable topology, but does not admit a compatible absolute value. This is the first example of this sort that I am aware of - thanks for giving it! However, at the moment I am not convinced that it has a closed subfield which is both separable and algebraically closed. | |
| Apr 3, 2021 at 21:40 | comment | added | Gerald Edgar | A countable algebraically closed field could be used to start with instead of $\mathbb C$. But I still have to go to a subfield go make it separable. The field of Hahn series is always complete (in the uniform sense compatible with group translation). If it is also metrizable, then use an invariant metric, and metric completeness follows from general completeness. BUT If $F_3$ algebraically closed is known only for value group contained in $\mathbb R$, then I do have a problem. I may have to extend $F_1 \subset F_2 \subset F_3 \subset \cdots$ countably many steps. | |
| Apr 3, 2021 at 20:54 | comment | added | Uri Bader | Thanks for the answer! I understand that $F$ is metrizable. Could you please elaborate on why is it completely so? Why do you insist on having $\mathbb{C}$ as your residue field rather than taking a countable algebraically closed field instead? Do you need this for completeness? I doubt it. If you insist on using $\mathbb{C}$, I find your argument for having $F_3$ algebraically closed insufficient. Note that Krasner's Lemma assumes the existence of an absolute value. Could you please elaborate here as well? Thanks. | |
| Apr 3, 2021 at 17:17 | history | answered | Gerald Edgar | CC BY-SA 4.0 |