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Let $K$ be some field extension of $\mathbb{Q}$ containing some complex number $c=a+bi$ with $a,b\in \mathbb{R}$ and $b\neq 0$. Is it possible that $K\cong \mathbb{R}$ as fields?

I tried to disprove this. We can define an order on $K$ by defining $a\geq b$ if and only if $\sigma(a)\geq \sigma(b)$. In this way, $(K,\leq)$ is a totally ordered field. Hence either $c>0$ or $c<0$. In both cases we have that $(a^2-b^2)+2abi=c^2>0$. This could lead to a contradiction if I could assume $a=0$ and $b\in \mathbb{Q}$, however I don't even manage to get a purely complex number in $K$ (real part zero).

Another approach is to look at polynomials. We have that $c$ is a solution of the polynomial $x^2+2ax+a^2-b^2\in \mathbb{R}[X]$, however we cannot necessarily view this as a polynomial over $K$.

I tried to construct an example of such a field $K$ (which I don't believe exits), this leads to nowhere either. I feel there should be some easy argument to solve this problem, but I can't find it.

Mathematician 42
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    Such fields exist, but none of them is explicit. They come from the wild automorphisms of $\mathbb C$, see e.g. http://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers and https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/PaulBYale.pdf – PseudoNeo May 03 '16 at 09:52
  • Right, so I can take such a wild automorphism and simply restrict it to $\mathbb{R}$, if the automorphism is wild enough, the image will contain some purely complex numbers. I knew these wild automorphisms existed, but they are way beyond the scope of the course notes I was reading. I guess the exercise is misplaced. – Mathematician 42 May 03 '16 at 10:03
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    Actually, an automorphism of $\mathbb C$ sending $\mathbb R$ to itself can only be the identity or the complex conjugation (that's not hard to prove), so in your sentence, "wild enough" simply means $\neq {\mathrm{id},\bar\cdot}$. With this construction, you get a lot of fields $F \subset \mathbb C$ isomorphic to $\mathbb R$ (actually you get all of those fields such that $\mathbb C/F$ is a finite extension). But there is another source of examples, still not explicit: not only $\mathbb C$ has nontrivial automorphisms, it also has nontrivial selfembeddings $\mathbb C \to \mathbb C$… – PseudoNeo May 03 '16 at 10:18
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    … for example because the algebraic closure of $\mathbb C(X)$ is (non explicitly) isomorphic to $\mathbb C$. You get this way other subfields $F \subset \mathbb C$ isomorphic to $\mathbb R$, but such that $\mathbb C/F$ is an infinite extension. – PseudoNeo May 03 '16 at 10:19

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