Subfield of $\mathbb R$ with algebraic closure as $\mathbb C$.

Question

Does there exist a proper subfield of $\mathbb{R}$ whose algebraic closure is $\mathbb{C}$ ?

A weaker question: Does there exist a proper subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is algebraic over $F$?

Answer 1

The two conditions you cited are equivalent, since ${\bf C}$ is the algebraic closure of ${\bf R}$

Yes, there is such a field. Just take a transcendental basis $B$ of $\bf R$ over ${\bf Q}$, and then ${\bf Q}(b)_{b\in B}\subseteq{\bf R}$ is algebraic and $\bf C$ is the algebraic closure of ${\bf Q}(b)_{b\in B}$.

I'm not sure if this is true without axiom of choice, so it might be hard to give a “concrete” example.