TITLE: Is $\mathbb{R}$ a finite field extension?
QUESTION [5 upvotes]: Is there a field $K \subseteq \mathbb{R}$ with $[\mathbb{R} : K] < \infty$?
My intuition tells me no: I would imagine that $K(\alpha)$ would be missing some $n$th root of $\alpha$ for all $x \in \mathbb{R} \setminus K$. But I'm not sure how to approach this: it's certainly not true of all finite field extensions (e.g. $\mathbb{R} \subseteq \mathbb{C}$), and there are irrationals $\beta$ s.t. $\beta$ has arbitrarily (but finitely) many roots in $\mathbb{Q}(\beta)$: e.g. $n$th powers of numbers satisfying $x^n - x - 1$.

REPLY [2 votes]: If such a $K$ exists, $[C:K]$ is finite where $C$ is the field of complex numbers, you can apply the Artin-Shreir theorem to show that $K=R$.
https://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index/8759#8759