In the paper Algebraische Konsequenzen des Determiniertheits-Axioms (U. Felgner – K. Schulz, Arch. Math. (Basel) 42 (1984), pp. 557–563), the authors show that in models of Zermelo-Fraenkel set theory without the Axiom of Choice, but with the Axiom of Determinacy, one can show that the field of complex numbers $\mathbb{C}$ only has one nontrivial automorphism (being complex conjugation).

Can the same be said about any algebraically closed field in characteristic $0$ (and if so, why, or where can I found a clear reference)?

EDIT: in view of the answer and comments below, I want to add two more nuanced questions.

  • Suppose we work in Zermelo-Fraenkel without Choice. Is it consistent to say that for every field $k$ in characteristic $0$, there exists and algebraic closure $\overline{k}$ of $k$ for which the automorphism group has size at most $2$?

  • Same setting. Is it consistent to say that every uncountable algebraically closed field has at most $2$ automorphisms?


1 Answer 1


Läuchli constructed an algebraic closure of the rationals which does not have any non-identity automorphisms.

Läuchli, H., Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, 1-18 (1962). ZBL0108.01002.

This field was studied in more details by Hodges in the paper

Hodges, Wilfrid, Lauchli’s algebraic closure of Q, Math. Proc. Camb. Philos. Soc. 79, 289-297 (1976). ZBL0324.02058.

  • 3
    $\begingroup$ Yes, but the standard construction of algebraic closure of Q gives, in ZF, a well-ordered (countable) algebraically closed field, which has all the usual $2^\omega$ automorphisms. Hence irrespective of any further examples like this, the answer to the original question is NO. $\endgroup$ Mar 9 at 14:38
  • $\begingroup$ I don't have access to those articles, but is the issue that representing a complex number in the algebraic closure as 'a+bi' with a, b real requires choice? $\endgroup$
    – Will Boney
    Mar 9 at 15:12
  • $\begingroup$ @asafkaragila: and what if I change my question into: "at most one nontrivial automorphism" ? $\endgroup$
    – THC
    Mar 9 at 15:13
  • 3
    $\begingroup$ @THC: No, Emil is correct. AC is only needed to prove that all algebraic closures of the rationals are isomorphic. But there is always a countable one. Indeed, the one you get inside $\Bbb C$ itself is countable. $\endgroup$
    – Asaf Karagila
    Mar 9 at 15:29
  • 2
    $\begingroup$ @THC: The problem here is that the failure of choice does not preclude very large well-orderable objects, as Emil points out. You can get some kind of approximation to this fact in the following sense. Every field (of any characteristics!) has an extension which has an algebraic closure with no automorphisms. This can be found in my paper "Iterated failures of choice". $\endgroup$
    – Asaf Karagila
    Mar 9 at 17:51

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