So I found this paper "Uncountable fields have proper uncountable subfields" by Butcher et al but couldn't access it. However just the title implies that there are uncountable proper subfields of the reals. I was wondering if there is a general construction to get many (if not all) such subfields. At first I thought taking an uncountable subset $S$ of a Hamel basis $H$ with uncountable complement $H - S$ might work, but the problem is that even though this would guarantee that not all Hamel basis elements are in the set of rational linear combinations of elements of $S$, it does not guarantee that we won't get all of $H$ when we start iteratively including multiplying and taking multiplicative inverses along with taking rational linear combinations. So, is there some somewhat concrete (e.g., starting with a Hamel basis) way to construct uncountable subfields of ${\mathbb R}$?
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Do you want non-isomorphic subfields or simply different subfields? If it is the latter, I believe that every subset of the primes yields a different subfield by adjoining $\mathbb{Q}$ with the roots of these primes. But I don't have a proof for this. – Dominik Sep 18 '15 at 15:51
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@Dominik That's a very interesting hypothesis. I'm ok with the subfields found being isomorphic, although obviously getting infinitely many that are not isomorphic would be better. But currently I don't even know how to construct just one uncountable proper subfield. – user2566092 Sep 18 '15 at 15:53
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1@Dominik Also in your example it seems like the added generating elements to ${\mathbb Q}$ would be countable, so wouldn't the subfield be countable? – user2566092 Sep 18 '15 at 15:55
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I misread the question, I thought you wanted uncountably many subfields, not an uncountable subfield. – Dominik Sep 18 '15 at 15:55
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Starting from a transcendental basis of $\mathbb{R}$ over $\mathbb{Q}$, you can create an uncountable subfield like here. – Dominik Sep 18 '15 at 15:59
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@Dominik Thanks, the piece I was missing was that I was trying to use Hamel bases instead of "transcendental bases". Never heard of them before. If you post as an answer I'll upvote and accept. – user2566092 Sep 18 '15 at 16:04
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See Barbara Osofsky's answer (and my comments to her answer) to Are $\mathbb{R}$ and $\mathbb{Q}$ the only nontrivial subfields of $\mathbb{R}$?. – Dave L. Renfro Sep 18 '15 at 17:38
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As already written in the comments, the key is to use a transcendental basis instead of a Hamel basis. The corresponding construction can be found in this Thread.
