Field reductions

In fact there are no nontrivial "field reductions" of $\mathbb{R}$: if $\mathbb{R} = F(a)$, then $a \in F$ and $F = \mathbb{R}$.

Case 1: $a$ is algebraic over $F$, hence $F(a) = F[a]$, so $d = [F[a]:F]$ is finite. Then $[\mathbb{C}:F] = 2d$, i.e., "the" algebraic closure of $F$ has finite degree over $F$.

Theorem: Let $K$ be a field and $\overline{K}$ any algebraic closure. If $[\overline{K}:K] = d$ is finite, then $d = 1$ or $d = 2$.

Proof: This is essentially the Grand Artin-Schreier Theorem (see e.g. Section 12.5 of http://math.uga.edu/~pete/FieldTheory.pdf for a proof of that.) Namely, Artin-Schreier says that if $\overline{K}/K$ is a finite Galois extension, then the degree is either $1$ or $2$. Certainly $\overline{K}/K$ is normal. And for the purposes of this question we are in characteristic zero, so everything is separable. Therefore $\overline{K}/K$ is Galois. (But here is a proof of the separability in the general case: if $\overline{K}/K$ is not separable, then there is a nontrivial subextension $L$ such that $[\overline{K}:L] = p^n$ and $\overline{K} = L(a^{p^{-n}})$. But this is impossible: the polynomial $t^p - a$ is irreducible over $L$ iff all of the polynomials $t^{p^n} - a$ are irreducible over $L$. So $\overline{K}/K$ is separable.)

Thus we must have $d = 1$, i.e., $F = \mathbb{R}$.

Case 2: $a$ is transcendental over $F$. But then the field $F(a)$ is isomorphic to the rational function field $F(t)$. Such a field cannot be isomorphic to $\mathbb{R}$, because it admits finite extensions of degree $n$ for all $n \in \mathbb{Z}^+$, e.g. $F(t^{\frac{1}{n}})$.