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I am currently studying some Galois theory, and after solving the following problem

Show that if $\sigma: \Bbb R \rightarrow \Bbb R$ is a field automorphism, then $\sigma = \text{id}_\Bbb R$.

I have been wondering if this is limited to $\Bbb R$, or if there is a more general notion that extends this. Notably, proving this theorem involved the fact that $\mathbb Q$ is dense in $\mathbb R$ (which gets fixed by any automorphism on a field containing $\mathbb Q$) and that $\sigma$ is continuous.

Question. Is it true that given a field $L$, a fitting topology on $L$ (intuitively feels like Hausdorff would make sense) and a subfield $K$ which is dense in $L$, any $K$-automorphism on $L$ is the identity?

I would appreciate any thoughts about this question, and how/why the statement could be weakened. Furthermore, this is a question out of pure interest, so I don't need a rigorous proof; intuition based explanations are totally fine. Thanks! :)

clorx
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  • This is true more generally for topological spaces: if $L$ is a Hausdorff space and $K \subseteq L$ is dense, then any endomorphism $L \to L$ is determined by its restriction to $K$. – Naïm Camille Favier Aug 01 '24 at 13:33
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    @NaïmFavier but you need to prove that the automorphisms are continous to use that fact. – mrod1605 Aug 01 '24 at 13:49
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    Any continuous $K$-automorphism of $L$ must be the identity. There could be other discontinuous $K$-automorphisms of $L$, however. – Geoffrey Trang Aug 01 '24 at 14:05
  • @GeoffreyTrang it feels like an interesting project to study those discontinous automorphism, but we should have a suitable topology for the field K. – mrod1605 Aug 01 '24 at 14:10
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    A similar argument works for the $p$-adic fields $\Bbb{Q}_p$ as opposed to $\Bbb{R}$. IIRC the existence of wild automorphisms of $\Bbb{C}$ referred to in Matsmir's answer relies on the axiom of choice. At least their existence proof relies on Zorn's lemma at several points, and I have a vague recollection that id we drop Choice, they need not exist. But that's beyond my payscale. – Jyrki Lahtonen Aug 02 '24 at 04:06
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    An easier type of example is eg $K = \Bbb Q$, and $L = \Bbb Q(\sqrt 2)$, thought of as a subfield of $\Bbb R$, along with the subspace topology. This is a nice topology on $L$ making it into a topological field, and $K$ is dense, but there is a nontrivial $K$-automorphism that sends $\sqrt 2$ to $-\sqrt 2$. The "that $\sigma$ is continuous" part is the remarkable thing for $\Bbb R$! – Izaak van Dongen Aug 22 '24 at 18:48

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This may not be a complete answer, since one can add a lot of assumptions on fields and topologies.

However, consider the field $F = \mathbb Q + i \mathbb Q$ of complex numbers with rational real and imaginary parts. Clearly, $F$ is dense in $\mathbb C$ in standard topology, but there exist nontrivial automorphisms of $\mathbb C$ that are identity on $F$. Clearly an automorphism $\phi: \mathbb C \rightarrow \mathbb C$ is identity on $\mathbb Q$ and either $\phi(i) = i$, or $\phi(i) = -i$. Thus, one of the automorphisms $x \rightarrow \phi(x)$ and $x \rightarrow \overline{\phi(x)}$ is an identity on $F$. Finally, I refer to the classical result, that there are infinitely many automorphisms of $\mathbb C$ besides identity and conjugation Wild automorphisms of the complex numbers. Taking any one of them as $\phi$ (and applying conjugation if needed) we get a counterexample.

As for the result for real numbers, it seems to me that the essence of the proof is that an automorphism of $\mathbb R$ preserves order, and only because of that it is continuous. The counterexamples for $F$ and $\mathbb C$ obviously are discontinuous.

Matsmir
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