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So there are easy examples for algebraic closures that have index two and infinite index: $\mathbb{C}$ over $\mathbb{R}$ and the algebraic numbers over $\mathbb{Q}$. What about the other indices?

EDIT: Of course $\overline{\mathbb{Q}} \neq \mathbb{C}$. I don't know what I was thinking.

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    $\begingroup$ I bet you were not expecting the answer to be a theorem. It's one of the coolest little theorems in all of Galois theory. $\endgroup$ Dec 13, 2009 at 14:58
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    $\begingroup$ Incidentally, C is not an algebraic closure of Q, since it contains transcendental elements like e (or more generally, because it has uncountable cardinality). $\endgroup$
    – S. Carnahan
    Dec 13, 2009 at 16:16

2 Answers 2

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Theorem (Artin-Schreier, 1927): Let K be an algebraically closed field and F a proper subfield of K with $[K:F] < \infty$. Then F is real-closed and $K = F(\sqrt{-1})$.

See e.g. Jacobson, Basic Algebra II, Theorem 11.14.

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    $\begingroup$ Sure, you put LaTeX in your answer, but I answered first! =) $\endgroup$ Dec 13, 2009 at 14:53
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    $\begingroup$ @Harry: You beat him by three and a half minutes only. That is about the time it would take Pete to type in his answer plus look up the exact reference in Jacobson's book. A clear case of independent responses if I ever saw one. $\endgroup$ Dec 13, 2009 at 16:24
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    $\begingroup$ Original reference, for the record: Emil Artin und Otto Schreier: Eine Kennzeichnung der reell abgeschlossenen Körper, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5 #1 (1927) 225–231 (doi:10.1007/BF02952522). Quoth Zentralblatt: “In Ergänzung der Untersuchungen über die algebraische Konstruktion reeller Körper der Verf. (ibidem 5 (1926), 85-99; F. d.M. 52) wird bewiesen, dass die reell abgeschlossenen Körper identisch sind mit den Körpern, die durch endliche Erweiterung algebraisch abgeschlossen werden können, ohne selbst algebraisch abgeschlossen zu sein.’ $\endgroup$ Dec 13, 2009 at 18:04
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    $\begingroup$ Am I the only person who finds this kind of competition to be an unhealthy expenditure of effort? $\endgroup$
    – Boris Bukh
    Jan 7, 2010 at 15:50
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    $\begingroup$ @Boris: You are not. $\endgroup$ Mar 22, 2011 at 20:03
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The Artin-Schreier theorem says that every algebraic closure of finite index has index 2, and it's the algebraic closure of a real-closed field.

Page 299 of Algebra by Serge Lang. Google Books Link to the page.

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