Are the real numbers a nontrivial simple extension of another field?

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.

Solution 1:

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.