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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 $\mathbb{R}\setminus F$ ?

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The answer probably depends on how F is given... Can you describe F explicitly? I do not know a single example! –  Mariano Suárez-Alvarez Nov 27 '10 at 17:33
All you know is that F(a)=R, and anything you can deduce from that. –  Mouswi Nov 27 '10 at 17:37

1 Answer 1

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. $F(t^{\frac{1}{n}})$.

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Thanks Pete. What then is the largest subfield of R that doesn't contain the square root of 2 for example ? –  Mouswi Nov 27 '10 at 17:46
@Mouswi: such fields exist by Zorn's Lemma, but they are far from unique: for instance, I can choose one such field which contains $\sqrt{-1}$ and another such field which contains $\sqrt{-2}$, but there is no field which contains both of these elements. –  Pete L. Clark Nov 27 '10 at 18:00
@Pete They are not subfields of $\mathbb{R}$, but one gets the idea :-) To address Mouswi's implicit question: such an extension will certainly not be simple. Actually, it is not immediately clear to me whether this extension can be chosen to be algebraic. –  Alex B. Nov 27 '10 at 18:35
@Alex: Good point! Take for instance $\sqrt{3}$ and $\sqrt{6}$ instead. And by the way, please feel free to suggest a rephrasing / followup question. Do you mean the following: if $K$ is a subfield of $\mathbb{R}$ maximal with respect to the exclusion of $\sqrt{2}$, is $\mathbb{R}/K$ algebraic? I think it certainly can be: first adjoin to $\mathbb{Q}$ the elements of a transcendence basis... –  Pete L. Clark Nov 27 '10 at 18:45
@Mouswi: they are maximal elements in the partially ordered set of subfields of $\mathbb{R}$ which do not contain $\sqrt{2}$. In other words, such a field $K$ is a subfield of $\mathbb{R}$ which does not contain $\sqrt{2}$ but any strictly larger subfield of $\mathbb{R}$ must contain $\sqrt{2}$. Anyway, perhaps it is time for you to tell us a little about the context of your question. Where did you get the idea of "field reduction"? –  Pete L. Clark Nov 27 '10 at 19:06

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