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I'm going to take Galois theory in the upcoming semester and I'm working on some basic problems right now. I was trying to figure out what $\mathrm{Aut}(\mathbb{C})$ should be. If $f: \mathbb{C} \to \mathbb{C}$ is an automorphism, then it's straightforward to see that $f(i) = \pm i$. Obviously, if I can prove that $f \mid_{\mathbb{R}}=\mathrm{id}_{\mathbb{R}}$, I'm done and $\mathrm{Aut}(\mathbb{C}) \cong \mathbb{Z}_2$.

However, no matter how much I struggled with the problem, I failed to prove that an automorphism of $\mathbb{C}$ must send real numbers to real numbers. In fact, now I even doubt that it's true. Is it possible for an automorphism of complex numbers to send a real number to a complex number? If yes, how exotic $\mathrm{Aut}(\mathbb{C})$ is? Is it uncountable, for example? Can we construct any of these automorphisms without relying on the axiom of choice?

stressed out
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  • There are infinite automorphisms of $;\Bbb C;$ . Most of them though doesn't look as nice as the two ones we get in Galois Theory. In fact, they can be pretty creepy, and if you want continuous ones then the identity and conjugation are the only ones there are. – DonAntonio Jan 15 '19 at 08:31
  • Even if your automorphism of $\mathbb C$ sends real numbers to real numbers, there could be nontrivial automorphisms of $\mathbb R$. – bof Jan 15 '19 at 08:32
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    There are no nontrivial automorphisms of $\mathbb{R}$. A real number is a square if and only if it's nonnegative, so any field automorphism of $\mathbb{R}$ also preserves order. That plus the fact that it preserves $\mathbb{Q}$ are enough to ensure it preserves everything. – jmerry Jan 15 '19 at 08:43
  • Why the downvote? – stressed out Jan 15 '19 at 13:07

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