Is there any "well-known" algebraically closed field that is uncountable other than $\mathbb{C}$ ? The algebraic closure of $\mathbb{C}(X)$ would work, but is it meaningful, is this field used in some topics ? Have you other examples ?
Thank you.
$\begingroup$
$\endgroup$
5
-
6$\begingroup$ 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 $\mathbb C(x)$. $\endgroup$– Wadim ZudilinCommented May 20, 2010 at 11:21
-
1$\begingroup$ 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. $\endgroup$– Gerald EdgarCommented May 20, 2010 at 12:49
-
7$\begingroup$ @Gerald: The field of Puiseux series in 1 variable over C. :) $\endgroup$– BCnrdCommented May 20, 2010 at 15:13
-
6$\begingroup$ @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". $\endgroup$– BCnrdCommented May 20, 2010 at 15:42
-
$\begingroup$ 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). $\endgroup$– Andreas BlassCommented Apr 2, 2015 at 12:01
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
9
The algebraic closure of $\mathbb{F}_p((t))$ is uncountable of characteristic $p$. It comes up naturally in number theory and algebraic geometry.
For every characteristic $p \geq 0$ and uncountable cardinal $\kappa$, there is up to isomorphism exactly one algebraically closed field of characteristic $p$ and cardinality $\kappa$. The examples of $\mathbb{C}$ and closures of Laurent series fields as above give you the ones of continuum cardinality and all characteristics. Indeed I do not know any specific reason to consider algebraically closed fields of larger than continuum cardinality.
answered May 20, 2010 at 11:24
-
$\begingroup$ I agree that I have never had to think about infinite sets of cardinalities other than countable and continuum. (Occasionally, I hear about crank mathematicians who want to claim that we should work in a version of "set theory" in which this is all there are.) But, not being a set theories, I have no understanding of why this might be. Do you have any stories you tell yourself for why there are no specific reasons to algebraically closed fields of larger than continuum cardinality? Or do you also only have the observation that there seems to be no interesting math in those examples? $\endgroup$ Commented Nov 29, 2011 at 7:54
-
$\begingroup$ @Theo: (Well, of course I'm not a set theorist either.) There are various ways to try to justify this. For instance, in Weil-style algebraic geometry it was recognized that an algebraically closed field of (not even necessarily uncountably) infinite absolute transcendence degree is roomy enough to encompass all needed geometric constructions. The language of the time was universal domain, but I think a better take on it is that such fields are precisely the saturated models of the theory of algebraically closed fields, so are in a precise sense sufficiently rich. $\endgroup$ Commented Nov 29, 2011 at 13:23
-
$\begingroup$ But that won't even get you to continuum cardinality. For that I think we have to talk about topological completion: you start with a countably infinite topological space and realize that in some sense you are missing some points, so you "complete" it (in any of various ways) and tend to get a space of continuum cardinality. For instance the completion of a countably infinite separable metric space without isolated points will necessarily be of continuum cardinality. $\endgroup$ Commented Nov 29, 2011 at 13:26
-
$\begingroup$ It seems that the only "familiar" operation which takes us beyond the continuum is formation of the power set...but taking unrestricted power sets is not something you need to do in field theory. Well, maybe I've just pushed the issue around the landscape a bit. I hope you found these comments at least somewhat helpful. $\endgroup$ Commented Nov 29, 2011 at 13:28
-
$\begingroup$ @Pete: Thanks for the comments --- they are very helpful. Indeed, being neither a set theorist nor an algebraic geometer, I wasn't aware of the "roomy enough" results (although I am not surprised by them). $\endgroup$ Commented Nov 29, 2011 at 22:03
$\begingroup$
$\endgroup$
1
The algebraic closure of the p-adic field $Q_p$ is also of interest. One may even want to consider the completion (with respect to the p-adic absolute value) of this algebraic closure. The resulting field is both complete and algebraically closed. It is denoted by $C_p$, and is considered as an p-adic analog of $C$.
-
13$\begingroup$ $\mathbb{C}_p$ is, as a field, isomorphic to $\mathbb{C}$. $\endgroup$ Commented May 20, 2010 at 12:04