Is there a field $K \subset \mathbb{R}$ such that $1 < [\mathbb{R} : K] < \infty$? i.e a proper subfield of $\mathbb{R}$ such that the field extension $\mathbb{R}/K$ is finite.
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Note that $K$ must be uncountable. – Pedro Dec 24 '14 at 19:22
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1And every element of $\mathbb R\setminus K$ must be algebraic. – Thomas Andrews Dec 24 '14 at 19:23
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Looks like you'd need to adjoin a set of positive measure to $\mathbb{Q}$ as a basis – David P Dec 24 '14 at 19:23
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2Is there any elemantary proof for nonexistence ? – mesel Dec 24 '14 at 19:30
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1A very natural question, but as @Timbuc’s answer points out, Artin-Schreier forbids such a field. Do you know the proof that $\mathbb R$ has no nontrivial automorphisms? It’s much more basic than A-S, and you can prove it yourself. It demonstrates the weaker proposition that there are no fields $K$ over which $\mathbb R$ is normal. – Lubin Dec 24 '14 at 20:47
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@Lubin that is a nice remark. Do you think this approach might be extended to obtain a proof of the full proposition? – Marco Flores Dec 24 '14 at 21:54
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@MarcoFlores, no, I think the two facts are independent, even if of a similar flavor. You can prove, by a philosophically similar method, that the $p$-adic fields $\mathbb Q_p$ also have no nontrivial automorphisms. – Lubin Dec 25 '14 at 18:08
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Related (but different): http://math.stackexchange.com/questions/1140259 – Watson Jan 31 '17 at 22:44
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The Artin-Schreier theorem implies that $\;[\Bbb C:K]\le2\;$ , and from here that the answer is no .
