Timeline for Applications of the Ax Kochen Ershov (AKE) princicple
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2013 at 14:01 | comment | added | Emil Jeřábek | @Immi Halupczok: Finitely ramified fields always have characteristic $0$. In your example, the value group of $\mathbb F_p((t))$ has infinitely many elements between $0$ and $v(p)=\infty$. | |
| Aug 14, 2013 at 13:56 | comment | added | Emil Jeřábek | I’m afraid the AKE principle does not hold the way it is stated here. For example, $\mathbb Q_p$ and $\mathbb Q_p(\sqrt p)$ are finitely ramified, henselian, and have the same value groups and residue fields, but are not elementarily equivalent. See encyclopediaofmath.org/index.php/Model_theory_of_valued_fields for some variants that do hold. | |
| Mar 15, 2013 at 16:11 | answer | added | Immi Halupczok | timeline score: 7 | |
| Mar 15, 2013 at 15:17 | comment | added | Immi Halupczok | I think the version of the transfer principle stated above is not entirely correct: one has to require that the valued fields themselves have characteristic 0; otherwise $\mathbb{Q}_p$ vs. $\mathbb{F}_p((t))$ is a counter-example. | |
| Oct 6, 2012 at 15:11 | answer | added | Lilach Leibovich | timeline score: 2 | |
| Oct 6, 2012 at 13:53 | history | edited | elvis | CC BY-SA 3.0 |
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| Oct 6, 2012 at 13:50 | comment | added | elvis | Thank you very much ! But for unramified field of characteristic 0 ? What prevent them from being algebraically closed ? | |
| Oct 6, 2012 at 13:47 | history | edited | elvis | CC BY-SA 3.0 |
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| Oct 6, 2012 at 0:25 | comment | added | Felipe Voloch | I don't think a finitely ramified Henselian field can ever be algebraically closed, as it won't contain the $n$-th root of a prime element for $n$ large. | |
| Oct 5, 2012 at 19:28 | history | asked | elvis | CC BY-SA 3.0 |