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Is the field of real numbers $\mathbb{R}$ a finite extension of some subfield $k\subset \mathbb{R}$?

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closed as off-topic by Mark Sapir, Daniel Moskovich, Emil Jeřábek, Andrey Rekalo, Theo Johnson-Freyd Jul 16 '13 at 17:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Mark Sapir, Theo Johnson-Freyd
If this question can be reworded to fit the rules in the help center, please edit the question.

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. – Todd Trimble Sep 10 '13 at 13:47
up vote 4 down vote accepted

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(!).

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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. – Emil Jeřábek Jul 16 '13 at 15:45
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. – Pete L. Clark Jul 27 '13 at 7:24
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. – Venkataramana Jul 27 '13 at 11:51

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