# Is there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite?

Is there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite? Here $[\mathbb R:K]$ means the dimension of $\mathbb R$ as a $K$-vector space.

What I have tried:

1. If we can find a finite subgroup $G\subset Gal (\mathbb C/\mathbb Q)$ such that $G$ contains the complex conjugation, it will be done by letting $K$ be the fixed field of $G$. But I don't know whether such a group exists. Maybe we can start with finding a suitable subgroup of $Gal(\bar{\mathbb Q}/\mathbb Q)$ and then lift it to $Gal(\mathbb C/\mathbb Q)$, where $\bar{\mathbb Q}$ denotes the algebraic closure of $\mathbb Q$.
2. By isomorphism extension theorem, we can find many automorphisms of $\mathbb C$, none of them carries $\mathbb R$ to itself except for the identity. This is because $Gal(\mathbb R/\mathbb Q)$ is the trivial group. For example, now suppose $\{x_\alpha\}\subset \mathbb R$ is a transcendence basis over $\mathbb Q$. Let $\sigma$ be a permutation of $\{x_\alpha\}$, then by the isomorphism extension theorem, $\sigma$ extends to an automorphism of $\mathbb C$, which we still denote by $\sigma$. Then $L=\sigma(\mathbb R)$ is a copy of $\mathbb R$ and $\mathbb R$ is algebraic over $K=\mathbb R\cap L$. How large can $K$ be? Is it possible that $[\mathbb R:K]$ is finite?

Does anyone has some ideas?

Thanks!

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According to Pete Clark's answer at http://mathoverflow.net/questions/13769/orders-of-field-automorphisms-of-algebraic-complex-numbers, if $L/K$ is a field extension with $L$ algebraically closed and $[L:K]\lt\infty$, then $[L:K]=1 {\rm\ or\ } 2$. Now, $\bf C$ is algebraically closed, and if $K$ is a subfield of $\bf R$, then $[{\bf C}:K]=2[{\bf R}:K]$, which pretty much settles it.

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That's awesome! Thanks very much! – Yang Zhou Oct 12 '12 at 12:10