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Is there a proper subfield $K$ of the real numbers and a real number $\theta$ such that $\mathbb R = K(\theta)$?

I thought of this question earlier idly wondering about what the structure of the poset of all subfields of $\mathbb C$ looks like and I'm surprised that I have no idea how to answer it.

Mees de Vries
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    A perhaps surprising conclusion to this : if $(\alpha_i){i\in I}$ is a set of generators of $\mathbb{R}$ over $\mathbb{Q}$ then for every finite $S\subset I$ $(\alpha_i){i\in I\backslash S}$ is still a set of generators. – orangeskid Mar 30 '18 at 10:53

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This is impossible.

If $\theta$ is algebraic over $K$, then $\Bbb{R}$ has finite degree over $K$, and so $\Bbb{C}$ has finite degree over $K$. But this is impossible: by Artin-Schreier, if the algebraic closure of a field is a finite extension, its degree is $1$ or $2$.

If $\theta$ is transcendental over $K$, then $\Bbb{R}=K(\theta)$ does not contain a square root of $\theta$ or of $-\theta$. But of course this is impossible: every positive real has a real square root.

Chris Eagle
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