Field reductions If there is a field $F$ that is a field reduction of the real numbers, that is $F(a)=\mathbb{R}$ for some $a$, let's also denote this $F=\mathbb{R}(\setminus a)$, then given $x \in \mathbb{R}$ is there a general method to determine whether $x$ is in $F$ or $x$ is in $\mathbb{R}\setminus F$ ?
 A: 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://alpha.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.
$F(t^{\frac{1}{n}})$.
