8
$\begingroup$

If one accepts the Axiom of Choice (AC), then the automorphism group of $\mathbb{C}$ is a huge and wild group, very poorly understood.

But apparently if one does not accept the Axiom of Choice, then the automorphism group of $\mathbb{C}$ has size $2$, only consisting of the identity and complex conjugation. I should be more precise here (see the comments below): it is consistent with this assumption that the automorphism group of $\mathbb{C}$ has size $2$ (i. e., we can find models of ZF without AC in which this group has size $2$).

What are other interesting (classes of) fields where similar things can be said about the automorphism group ? (That is, upon not accepting the Axiom of Choice, one ends up with consistency of an "easy" or even trivial automorphism group.)

$\endgroup$
8
  • 2
    $\begingroup$ How are you going to prove that $\mathbb C$ has no more than two automorphisms? $\endgroup$ – მამუკა ჯიბლაძე Oct 23 '20 at 12:34
  • 4
    $\begingroup$ Not accepting AC is not nearly enough. You're saying "Here is a finite set; I do not accept the claim it is empty. Therefore, it has exactly 4 elements". $\endgroup$ – Asaf Karagila Oct 23 '20 at 13:03
  • 9
    $\begingroup$ @მამუკაჯიბლაძე It is consistent with ZF that every automorphism of a Polish group is continuous, in that case only the identity and conjugations are automorphisms of $\Bbb C$. $\endgroup$ – Asaf Karagila Oct 23 '20 at 13:04
  • 6
    $\begingroup$ @მამუკაჯიბლაძე I seem to vaguely recall that there was a question whether it is consistent with ZF that Aut($\mathbb C$) has size strictly between $2$ and $2^{2^\omega}$, with inconclusive answers (but I can’t find anything). I believe that at least the Artin–Schreier theorem can be made choiceless enough to show that if there are more than 2 automorphisms, there are infinitely many. However, none of this should concern $\mathrm{Gal}(\tilde{\mathbb Q}/\mathbb Q)$, which should be as wild as usual even in ZF, as $\tilde{\mathbb Q}$ is a countable (and therefore well ordered) field. $\endgroup$ – Emil Jeřábek Oct 23 '20 at 14:49
  • 3
    $\begingroup$ @LSpice Indeed, they need not. $\endgroup$ – Emil Jeřábek Oct 30 '20 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.