Timeline for uncountable algebraically closed field other than C ?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Apr 2, 2015 at 12:01 | comment | added | Andreas Blass | As pointed out by Wesley Calvert in a now-deleted answer, the algebraic closure of $\mathbb C(X)$ would work only in the weak sense that it's not identical with $\mathbb C$. It is isomorphic to $\mathbb C$ (provided the axiom of choice holds). | |
| May 20, 2010 at 15:42 | comment | added | BCnrd | @Gerald: another pseudo-example, not as widely known, lies at the bottom of $p$-adic Hodge theory: we begin with the valuation ring $O$ of $C = \mathbf{C}_ p$ and form the inverse limit $R$ under $p$-power maps of copies of $O/pO$. Then incredibly $R$ turns out to be a valuation ring (in particular, domain) whose fraction field is algebraically closed. OK, so as input we have to use the field $C$ which is algebraically closed. But the remarkably part is that it is in no way evident that the fraction field of $R$ should be alg. closed, since the alg. closedness of $C$ was "in char. 0". | |
| May 20, 2010 at 15:13 | comment | added | BCnrd | @Gerald: The field of Puiseux series in 1 variable over C. :) | |
| May 20, 2010 at 12:49 | comment | added | Gerald Edgar | How about a more interesting question: Are there any examples (besides C) of algebraically closed fields, where "algebraic closure" is not part of the construction? Maybe a borderline answer is certain real-closed fields F, then F(i) is algebraically closed. | |
| May 20, 2010 at 11:51 | vote | accept | Laurent | ||
| May 20, 2010 at 11:41 | answer | added | coudy | timeline score: 7 | |
| May 20, 2010 at 11:30 | vote | accept | Laurent | ||
| May 20, 2010 at 11:51 | |||||
| May 20, 2010 at 11:24 | answer | added | Pete L. Clark | timeline score: 24 | |
| May 20, 2010 at 11:21 | comment | added | Wadim Zudilin | Of course, the algebraic closure of $\mathbb C(x)$ is meaningful. The usual question is whether a given power series is algebraic over $\mathbb C(x)$ or not. For example, $e^x$ does not satisfy any polynomials equation with coefficients from $\mathbb C(x)$. | |
| May 20, 2010 at 11:13 | history | asked | Laurent | CC BY-SA 2.5 |