Timeline for How to picture $\mathbb{C}_p$?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Feb 11, 2022 at 23:44 | comment | added | Z. M | @PeteL.Clark It seems that you are correct: math.stackexchange.com/questions/803524/… | |
| Jan 14, 2011 at 11:28 | comment | added | Pete L. Clark | In fact, now that I think about it, I strongly suspect that no choice is needed whatsoever... | |
| Jan 14, 2011 at 11:27 | comment | added | Pete L. Clark | @KConrad: sure, it's clear that one can get away with only a weak form of choice. But my question is whether you need any choice at all. | |
| Jan 14, 2011 at 7:52 | comment | added | KConrad | Pete: Since all the irreducible polynomials in Q_p[x] of a fixed degree split in some finite extension of Q_p (that's how I will say Q_p has only finitely many extensions of each degree in an algebraic closure without mentioning the term "algebraic closure), one should be able to construct an algebraic closure of Q_p without using AC in its most general form. | |
| Jan 13, 2011 at 16:24 | comment | added | Pete L. Clark | @Neil: AC isn't needed to talk about algebraic closures: it's needed to be sure that every field has an algebraic closure. For instance, certainly AC is not needed (or used) to show that $\mathbb{R}$ has an algebraic closure. I would be interested to know whether it is actually required for $\mathbb{Q}_p$. | |
| Jan 13, 2011 at 16:14 | history | edited | Neil Strickland | CC BY-SA 2.5 |
[Corrected as per Johannes Hahn's comment]
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| Jan 13, 2011 at 16:13 | comment | added | Neil Strickland | @Ketil: yes, but AC is already needed to construct algebraic closures, so we can't begin to talk about $\mathbb{C}_p$ without it. | |
| Jan 13, 2011 at 15:25 | comment | added | Ketil Tveiten | Doesn't the same-cardinalty-isomorphism depend on AC? In which case, you can't get a picture of anything... | |
| Jan 13, 2011 at 11:05 | comment | added | Johannes Hahn | Not all alg.closed fields of the same cardinality are isomophic. Example: $\overline{\mathbb{Q}}$ and $\overline{\mathbb{Q}(\pi)}$ are not isomorphic. Your result only holds for uncountable fields. | |
| Jan 13, 2011 at 10:59 | history | answered | Neil Strickland | CC BY-SA 2.5 |