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Is there an uncountable proper subfield of $\mathbf{R}$?

Richard
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2 Answers2

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Sure. Note that ${\bf R}$ itself is uncountable, so it has an uncountable transcendental basis $\{x_\alpha\mid \alpha\in {\mathfrak c}\}$ over ${\bf Q}$. Then ${\bf Q}(x_\alpha\mid \alpha\in {\mathfrak c})$ is a proper subfield of ${\bf R}$ (because neither of $x_\alpha,-x_\alpha$ has a square root, for example, which is not true in ${\bf R}$).

Alternatively, ${\bf Q}(x_\alpha\mid 0<\alpha\in {\mathfrak c})$ is even more clearly distinct from ${\bf R}$, as it does not contain $x_0$.

user26857
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tomasz
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  • Thank you very much! But I'm not familiar with abstract-algebra. Can you explain to me what means transcendental basis? – Richard Dec 23 '13 at 05:09
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    @Richard: A finite set ${x_1,\ldots,x_n}$ is algebraically independent over a field $K$ if there is no nonzero polynomial $p\in K[X_1,\ldots,X_n]$ (of $n$ variables with coefficients in $K$) such that $p(x_1,\ldots,x_n)=0$ An infinite set is algebraically independent if all finite subsets are algebraically independent (this is the same as linear independence in many ways). A transcendental basis of a field $L\supseteq K$ is a maximal algebraically independent set. – tomasz Dec 23 '13 at 13:53
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    @Richard: The cardinality of transcendental basis of $L$ over $K$ (the cardinality is always unique, though the basis certainly isn't; the proof is the same as for linear basis) is called transcendence degree, and denoted by $\operatorname{trdeg}_KL$, and it's not hard to show that $\lvert L\rvert=\lvert K\rvert+\operatorname{trdeg}_KL+\aleph_0$, unless the transcendence degree is zero (then we say that $L$ is an algebraic extension of $K$). – tomasz Dec 23 '13 at 14:01
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    @Richard: I forgot one thing: If the transcendence degree is zero, then $\lvert K\rvert\leq \lvert L\rvert+\aleph_0$, that's why you can conclude that ${\bf R}$ has a basis of cardinality $\mathfrak c$ over ${\bf Q}$. – tomasz Dec 23 '13 at 14:12
  • [This is wrong; I am stupid :(] Simple example: when $x_0 = \sqrt{2}$, the field becomes $(\mathbb{R} \setminus \sqrt{2} \mathbb{Q}) \cup {0}$. – JiminP Dec 23 '13 at 14:31
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    @JiminP: $\sqrt 2$ is not transcendental over $\bf Q$ and $(\mathbb{R} \setminus \sqrt{2} \mathbb{Q}) \cup {0}$ is not even closed under addition. – tomasz Dec 23 '13 at 14:32
  • Oops... I thought too simply... :( – JiminP Dec 23 '13 at 14:34
  • I wonder, does this construction require the axiom of choice, probably for the existence of the uncountable trancendental basis? Or can we get there by transfinite induction only (still, I am no expert here and I am not sure that transfinite induction up to some ordinal does not need the axiom of choice)? – M. Winter Nov 13 '19 at 08:21
  • It does. Existence of transcendental basis for arbitrary fields should be equivalent to AC. Moreover, a transcendental basis for reals can be easily used to construct a linear basis over rationals. – tomasz Nov 14 '19 at 14:02
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For an "explicit" answer you might take the field generated by a perfect set of Hausdorff dimension $0$. See e.g. Measure 0 subfield of R.

user26857
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Robert Israel
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