Is there a subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is a finite algebraic extension of $F$. [duplicate]

Question

Possible Duplicate:
Examples of algebraic closures of finite index

The question is in the title. I can prove that if such field $F$ exist then the extension $\mathbb{R}/F$ cannot be of degree $2$ (essentially because there are no automorphisms of $\mathbb{R}$ and an extension of degree 2 would create one)

Answer 1

If the degree of $\mathbf R$ over $F$ were $d$, then the degree of $\mathbf C$ over $F$ would be $2d$, contradicting the Artin-Schreier theorem.