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I currently have to cope with field automorphisms. I already understood that any field automorphism of $\mathbb{C}$ must fix all elements in $\mathbb{Q}$.

My question is the following: Assume a number $x$ is fixed by every field automorphism of $\mathbb{C}$. Does that imply that $x \in \mathbb{Q}$?

My strong guess is that this implication is true but I could not come up with an idea for a proof. Any hint into the right direction would be appreciated.

Mortymus
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  • I think that this is not correct. I don't have a proof on hand, but I do know that there is no nontrivial automorphism of $\mathbb{R}$. This is a counterexample to your intuition. – Alex G. Aug 18 '14 at 16:43
  • Dear @AdamHughes : it does not appear the poster is assuming continuity... something would have to be said about that to make the comment follow? – rschwieb Aug 18 '14 at 16:45
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    But there are algebraic automorphism of the complex numbers that are non-continuous, which is what the question is about. @AdamHughes – Thomas Andrews Aug 18 '14 at 16:48

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It turns out to be true. Since I don't think your question is a duplicate, but it is nicely addressed at this solution by Andres Caicedo, I'm giving you a community wiki answer to point you to it.

If you take a look at the discussion right before the second to last paragraph, you can learn why exactly the rationals are fixed.

Community
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rschwieb
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