Assume that the continuum hypothesis holds. If $F$ is an uncountable field of real numbers, does $F$ always contain a proper uncountable subfield? Are there many specific uncountable fields of real numbers whose existence can be proved without assuming the axiom of choice?

$\begingroup$ Perhaps you would get a more interesting question by asking whether there is an uncountable subfield $F$ of $\mathbb{R}$ for which the existence of a proper uncountable subfield cannot be proved without using some form of the the Axiom of Choice. $\endgroup$ – Simon Thomas Jun 7 '10 at 15:13

$\begingroup$ Correct me if I am wrong. I believe we can define a field $K$ by adjoining to $\mathbb{Q}$ all elements in $\mathbb{R}$ transcendental over $\mathbb{Q}$, without axiom of choice. The extension $\mathbb{R}/K$ must then be algebraic and hence $K$ or any extension of $K$ must be uncountable. I think this answer the second part of the question. It is not obvious to me how to define a uncountable subfield of $K$ without $AC$. Suppose, I want to prove that it is impossible to prove the existence of a uncountable proper subfield of $K$ is there any obvious path to proceed? $\endgroup$ – abcdxyz Jun 7 '10 at 15:40

$\begingroup$ If $a$ is transcendental and $b$ is algebraic and nonzero over $\mathbb{Q}$ then $ab$ is transcendental. If $K$ is obtained by adjoining all real transcendentals to $\mathbb{Q}$ then $K$ contains $a$ and $ab$ and so also $b$. Thus $K=\mathbb{R}$. $\endgroup$ – Robin Chapman Jun 7 '10 at 16:10
Take a compact Cantor set $K \subseteq \mathbb{R}$ of Hausdorff dimension zero. Actually we need all cartesian powers $K^n$ of dimension zero as well. The field $\mathbb{Q}(K)$ generated by it is uncountable, but still of Hausdorff dimension zero, so it is a proper subfield.
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That field consists of the values of rational functions $w(x_1,\dots,x_n)$ of many variables with rational coefficients, where the variables range over $K$. There are countably many such things, so you just have to show any one of them has dimension zero. The domain of any such $w$ (that is, the set where the denominator does not vanish) consists of an increasing countable union $\bigcup_k A_k$ of sets where the gradient is bounded, so that $w$ is Lipschitz continuous on each $A_k$. So the image of $w$ on $K^n$ is again a countable union of sets of dimension zero.
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G. A. Edgar & Chris Miller, Borel subrings of the reals. Proc. Amer. Math. Soc. 131 (2003) 11211129
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Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.

$\begingroup$ Interesting! This answer mixes field theory and metric geometry in a way I have not seen. Could you give some more details and/or references as to why $\mathbb{Q}(K)$ has Hausdorff dimension zero? $\endgroup$ – Pete L. Clark Jun 7 '10 at 15:10

$\begingroup$ Note that the field $\mathbb{Q}(K)$ is actually Borel. $\endgroup$ – François G. Dorais♦ Jun 7 '10 at 16:17

$\begingroup$ Thanks alot for your answer. Your argument seems to show that if K is any set of real numbers having zero Lebesgue measure, then the field Q(K) also has zero Lebesgue measure. This gives us a method of obtaining a very large number of uncountable real fields all different from R, without using the Axiom of Choice. $\endgroup$ – Garabed Gulbenkian Jun 12 '10 at 17:53

1$\begingroup$ @Garabed: This is not correct. The hypothesis is zero Hausdorff dimension (of all Cartesian powers), not merely zero Lebesgue measure. For example, the standard middlethirds Cantor set has Lebesgue measure zero, but the additive group it generates is the whole line. $\endgroup$ – Gerald Edgar Jun 13 '10 at 1:13
I think the following argument ought to answer your first question, but I haven't checked the details. An uncountable subfield F of R will contain an uncountable polynomially independent subset (by Zorn's lemma). And any proper subset of that polynomially independent subset will generate a proper subfield of F.

1$\begingroup$ Yes, this definitely works. It shows that any uncountable field has uncountably many uncountable subfields. It does not use CH, but it does use AC (for the existence of transcendence bases). $\endgroup$ – Pete L. Clark Jun 7 '10 at 15:12

$\begingroup$ What is the definition of a "polynomially independent subset"? $\endgroup$ – Garabed Gulbenkian Jun 10 '10 at 14:41