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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?

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  • $\begingroup$ 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
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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) 1121-1129
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Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.

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  • $\begingroup$ 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
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    $\begingroup$ @Garabed: This is not correct. The hypothesis is zero Hausdorff dimension (of all Cartesian powers), not merely zero Lebesgue measure. For example, the standard middle-thirds 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
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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.

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    $\begingroup$ 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

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