Is the field of real numbers $\mathbb{R}$ a finite extension of some subfield $k\subset \mathbb{R}$?
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$\begingroup$ Although I agree this question is not really suitable for this site, it probably should have been migrated to a better site. To find the answer, google Artin-Schreier. $\endgroup$– Todd Trimble ♦Sep 10, 2013 at 13:47
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there is no such sub-field. [Edited] It is a theorem of Emil Artin that the only automorphism of the field of complex numbers of finite order is of order two. If such a sub-field $F$ existed, then $Aut ({\mathbb C}/F)$ would have only elements of order two and hence abelian. In particular, ${\mathbb R}/F$ would be abelian (and Galois). But $\mathbb R$ has no field automorphisms and hence $F=\mathbb R$.
I thank Peter clark for pointing out that I was attributing a wrong result to Emil Artin(!).
answered Jul 16, 2013 at 15:29
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2$\begingroup$ More directly: there is a theorem that the algebraic closure of $k$ is a finite extension of $k$ only if $k$ is algebraically closed or real-closed. $\endgroup$ Jul 16, 2013 at 15:45
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1$\begingroup$ In fact the theorem stated here is not true: if it were, then every real-closed field of continuum cardinality would be isomorphic to $\mathbb{R}$. But the real-closure of $\mathbb{R}((t))$ is a counterexample. It follows from another question that I asked here that in fact there are $2^{\# \mathbb{R}}$ conjugacy classes of order $2$ automorphisms on the complex numbers. $\endgroup$ Jul 27, 2013 at 7:24
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$\begingroup$ I think peter clark is right; what is true is that the only finite order automorphisms of $\mathbb C$ is of order two. If $F$ is a "finite index" subfield of the reals, then it is easy to see that there will be elements of order more than two fixing $F$ and that cannot be. This is the result that I am using. $\endgroup$ Jul 27, 2013 at 11:51