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> **Possible Duplicate:**  
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> [Examples of algebraic closures of finite index](https://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index) 
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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)



Is there a subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is a finite algebraic extension of $F$.
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](http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf).


