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The following basic result has been discussed numerous times on this forum.

The set of subfields of $\mathbb{C}$ not containing $\sqrt{2}$ has an inclusion-maximal element.

Proof. If $K_1\subset K_2\subset\ldots$ is a chain of such subfields, then their union is still a subfield of $\mathbb{C}$ that does not contain $\sqrt{2}$. Therefore, the result follows from Zorn's lemma.

Let me recall that Zorn's lemma is an equivalent reformulation of the Axiom of Choice. Therefore, this proof does not give any explicit construction of such a field. Can we do without Choice?

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heptagon
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  • Aside: I think you could prove the existence of a maximal algebraic extension of $\mathbb{Q}$ with that property. A maximal subfield of $\mathbb{C}$ with that property is another question. –  Jan 01 '17 at 15:49
  • You may find http://math.stackexchange.com/questions/411705/do-maximal-proper-subfields-of-the-real-numbers-exist helpful. – Gerry Myerson Jan 01 '17 at 15:54
  • @Hurkyl The algebraic version of the question is still interesting to me. Can you please give a hint? – heptagon Jan 01 '17 at 16:04
  • @GerryMyerson I think this is a different question. The question you mean seems to ask if there is a field $K$ such that $K(\sqrt{2})=\mathbb{R}$ . – heptagon Jan 01 '17 at 16:08
  • It is a different question – but it still might be helpful. – Gerry Myerson Jan 01 '17 at 16:10
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    @Heptagon: The key point is that $\mathbb{Q}$ is, like any countable set, well-orderable. I believe there's a theorem that any well-orderable field has a unique (up to isomorphism) well-orderable algebraic closure, and you should be able to use a well-order on that to do whatever transfinite iteration you need. (note that, I think, that without AoC it's possible that it may have other algebraic closures as well, but they won't be well-orderable) –  Jan 01 '17 at 16:28
  • @Hurkyl: You are correct about your last comment. – Asaf Karagila Jan 01 '17 at 18:47
  • I *think* that @Gerry is correct. If $F$ is a maximal subfield of $\Bbb C$ such that $F(\sqrt 2)=\Bbb R$, then $K=F\cap\Bbb R$ is a subfield of $\Bbb R$ such that $K(\sqrt 2)[\sqrt{-1}]=\Bbb C$. In which case, the two questions---the one here, and the one linked by Gerry---are equivalent. – Asaf Karagila Jan 01 '17 at 18:49
  • @Hurkyl Thanks for the explanation! Maybe I should create a sister question for algebraic numbers instead of $\mathbb{C}$ in order to keep your answer recorded. I don't know if I am allowed to do so by the rules here though. – heptagon Jan 01 '17 at 18:59
  • @AsafKaragila Sorry, but I do not understand why is my question equivalent to those that you discuss. It seems easy to me to get that a field $K$ can satisfy $K(\sqrt{2})=\mathbb{R}$ or $K(\sqrt{2})=\mathbb{C}$ only if $K$ is $\mathbb{R}$ or $\mathbb{C}$, but I do not see how to get the answer to my question from there. – heptagon Jan 01 '17 at 19:07
  • You can create as many questions as you like, heptagon, but if they are related then please make sure you link each to the others. – Gerry Myerson Jan 01 '17 at 19:39
  • Thanks for explaining, @GerryMyerson! – heptagon Jan 01 '17 at 22:06

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