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Assume that the Continuum Hypothesis holds. If F is an uncountable field of real numbers, does F always necessesarily contain a proper uncountable sub-field? Are there many specific uncountable fields of real numbers whose existence can be proved without assuming the Axiom of Choice?

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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. –  Simon Thomas Jun 7 '10 at 15:13
    
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? –  Tran Chieu Minh Jun 7 '10 at 15:40
    
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 $K=\mathbb{R}$. –  Robin Chapman Jun 7 '10 at 16:10
    
Thank you, I see, so that does not work. –  Tran Chieu Minh Jun 7 '10 at 16:19
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2 Answers 2

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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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? –  Pete L. Clark Jun 7 '10 at 15:10
    
Note that the field $\mathbb{Q}(K)$ is actually Borel. –  François G. Dorais Jun 7 '10 at 16:17
    
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. –  Garabed Gulbenkian Jun 12 '10 at 17:53
    
@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. –  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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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). –  Pete L. Clark Jun 7 '10 at 15:12
    
What is the definition of a "polynomially independent subset"? –  Garabed Gulbenkian Jun 10 '10 at 14:41
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