# If $\sigma: L \rightarrow L$ is a $K$-automorphism, where $K$ is dense in $L$ w.r.t some fitting topology, is $\sigma$ the identity?

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! :)

• 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 $K$. Commented Aug 1 at 13:33
• @NaïmFavier but you need to prove that the automorphisms are continous to use that fact. Commented Aug 1 at 13:49
• Any continuous $K$-automorphism of $L$ must be the identity. There could be other discontinuous $K$-automorphisms of $L$, however. Commented Aug 1 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. Commented Aug 1 at 14:10
• 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. Commented Aug 2 at 4:06

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.