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Is there a (proper) subfield $K$ of $\mathbb{R}$ such that $\mathbb{R}$ is an algebraic extension of $K$?

From this question, Is there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite?, it is clear that for such an $K$, $[\mathbb{R}:K]=\infty$.

This question seems to be non-trivial, and I suspect any existence result will be non-constructive. Any references will be appreciated.

Community
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  • What do you mean by $F$? – Cameron Buie Feb 09 '15 at 05:48
  • sorry, meant $K$ –  Feb 09 '15 at 05:55
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    Can't one just do the following? Let $S$ be a real transcendence basis for $\Bbb C$ over $\bar{\mathbb Q}$ and take $\mathbb Q(S)$ – PVAL-inactive Feb 09 '15 at 06:06
  • Just to be clear, you are asking whether or not there is a constructive or axiom of choice-free construction that would give that, or is one involving choice fine? If it's the latter, you can do as PVAL suggests in much the same way as you find a basis for ${\bf R}$ over ${\bf Q}$. – tomasz Feb 09 '15 at 06:28
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    If $K$ is a maximal subfield of $\mathbb R$ not containing $\sqrt2$ (such as exists by Zorn's lemma), isn't every element of $\mathbb R$ algebraic over $K$? – bof Feb 09 '15 at 06:56
  • Related: https://math.stackexchange.com/questions/426842/are-the-real-numbers-a-nontrivial-simple-extension-of-another-field – Watson Jun 08 '18 at 13:37

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Just to have at least one actual answer to the question, I'll convert the comment by PVAL to an answer.

Using Zorn's lemma there exists a transcendence basis $B$ of $\Bbb R$ over $\Bbb Q$: a maximal algebraically independent (over $\Bbb Q$) set of real numbers. Then $\Bbb Q(B)$ is a subfield of $\Bbb R$, and by maximality every element of $\Bbb R$ is algebraic over $\Bbb Q(B)$. So $\Bbb R/\Bbb Q(B)$ is an algebraic extension.

Marc van Leeuwen
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    Why is $\mathbb{Q}(B)$ a proper subfield of $\mathbb{R}$? –  Feb 09 '15 at 06:43
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    @Amudhan: Because $\Bbb Q(B)$ is a purely transcendental extension of $\Bbb Q$, so it contains no irrational algebraic (over$~\Bbb Q)$ elements such as $\sqrt{13}$. – Marc van Leeuwen Feb 09 '15 at 07:49