Timeline for Is every field extension of an ultrafield an ultrafield?
Current License: CC BY-SA 4.0
14 events
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| Mar 26, 2019 at 18:44 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals in title, romanized lim
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| Mar 26, 2019 at 10:24 | answer | added | YCor | timeline score: 5 | |
| Mar 26, 2019 at 10:24 | history | edited | YCor |
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| Mar 19, 2011 at 22:19 | vote | accept | user12940 | ||
| Mar 19, 2011 at 21:57 | comment | added | user12940 | @Pete: You're right. I had in mind infinite extensions, regardless of their trascendence degree. | |
| Mar 19, 2011 at 19:35 | comment | added | Chris Eagle | @Pete: Yes, $\mathbb{C}$ is an ultrafield: there's only one characteristic-0 algebraically closed field of cardinality continuum, so $\mathbb{C}$ is isomorphic to any nonprincipal ultrapower of the algebraic numbers. | |
| Mar 19, 2011 at 17:45 | answer | added | Laurent Moret-Bailly | timeline score: 9 | |
| Mar 19, 2011 at 5:46 | history | edited | Pete L. Clark |
edited tags; edited tags
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| Mar 19, 2011 at 5:45 | comment | added | Pete L. Clark | Your question seems to skip over the case of finite transcendence degree, which it seems to me may already be a source of counterexamples. For instance, $\mathbb{C}$ is an ultrafield (right?), but is $\mathbb{C}(t)$ an ultrafield? | |
| Mar 19, 2011 at 2:08 | answer | added | user6976 | timeline score: 8 | |
| Mar 19, 2011 at 0:19 | comment | added | user1437 | For those unfamiliar with the terminology in the book, an ultrafield is a ultraproduct of an infinite collection of fields over a nonprincipal ultrafilter, which in this case of the ultraproduct is a field. | |
| Mar 18, 2011 at 23:43 | comment | added | user12940 | I'm following the terminology of Schouten's "The Use of Ultraproducts in Commutative Algebra": an ultrafield is simply an ultraproduct of fields. | |
| Mar 18, 2011 at 23:29 | comment | added | Qiaochu Yuan | What is an ultrafield? | |
| Mar 18, 2011 at 23:23 | history | asked | user12940 | CC BY-SA 2.5 |