4

How to Prove that $\operatorname{Aut}(\mathbb C/\mathbb Q)$ is infinite?

I had already proved that $\operatorname{Aut}(\mathbb R/\mathbb Q)$ is trivial group containing identity use continuity argument.

But When I thought about $\operatorname{Aut}(\mathbb C/\mathbb Q)$ problem I could not even start.

I had just started course in field theory in which we have done Galois group definition and some example.

Please give me hint so that I can solve the above problem.

Another User
  • 5,048
Curious student
  • 6,583
  • 5
    You need some sort of axiom of choice to extend any automorphism of a finite Galois extension $K/\Bbb{Q}$ or of $\Bbb{Q(\pi)/Q}$ or ... to an automorphism of $\Bbb{C/Q}$. Morally $\Bbb{C}$ is the algebraic closure of some purely transcendental extension of $K(\pi)$, but the transcendental basis is uncountable and its existence relies on choice. – reuns Sep 01 '19 at 05:03
  • 1
    Does an automorphism have to be continuous? – Arthur Sep 01 '19 at 05:48
  • 1
    No But Here we can for actually that automorphism is continous – Curious student Sep 01 '19 at 06:25
  • 4
    This is very tricky, see the comment by reuns. IIRC the answer depends on the choice of set theory in the sense that there may exist a set theory such that $Aut(\Bbb{C}/\Bbb{Q})$ only has the two obvious automorphisms. It makes me squirm whenever I read about a problem in pure algebra such that the answer depends on what kind of objects are qualified to be sets. It does not really make sense to me, but that's how it goes :-). Caveat: I may be wrong about this particular case, but there definitely exist questions in algebra with answers depending on the set theory universe we work in. – Jyrki Lahtonen Sep 01 '19 at 06:55
  • 1
    Anyway, with the usual set theory in use, there will be $2^{2^{\aleph_0}}$ automorphisms, and $\Bbb{C}$ has infinitely many subfields isomorphic to $\Bbb{R}$. Don't ask me to exhibit one :-) – Jyrki Lahtonen Sep 01 '19 at 07:00
  • 1
    Oh, and only two of those automorphisms are continuous with respect to the usual topology. That is actually easy to prove. – Jyrki Lahtonen Sep 01 '19 at 07:01
  • 1
    Thank a Lot Sir. I had one doubt .Identity and Conjugation map are continous .But How to claim that there is no other possible continous automorphism ? – Curious student Sep 01 '19 at 07:03
  • 1
    @JyrkiLahtonen: would you mind providing a link(s) to examples of the questions you mentioned? I'm curious. Thanks. – Robert Lewis Sep 01 '19 at 07:09
  • 4
    SRJ, an automorphism must map each rational number to itself. $\Bbb{Q}$ is dense in $\Bbb{R}$, so a continuous automorphism must send each real number to itself. $\Bbb{C}/\Bbb{R}$ is algebraic of degree two, leaving us with two choices. – Jyrki Lahtonen Sep 01 '19 at 09:10
  • 1
    @RobertLewis I'm afraid I cannot. For then the recollections were not that vague. 1) The easiest to describe is that if we don't have the axiom of choice, then it is possible to have vector spaces that don't have any bases whatsoever. 2) The existence of non-obvious automorphisms of $\Bbb{C}$ similarly relies on Zorn's lemma. 3) In GTM series book on Homological Algebra by Hilton & Stammbach they mention a result about extensions of abelian groups. Of these I remember seeing (but not studying) a result that with a different set theory, the answer will change. – Jyrki Lahtonen Sep 01 '19 at 09:18
  • 1
    (cont'd) I am not positive about how the choice of set theory affects the answer to 2). I will ping Asaf here, he would know where the links are. – Jyrki Lahtonen Sep 01 '19 at 09:19
  • I don't know, if there are results where factors other than including the Axiom of Choice play a role. – Jyrki Lahtonen Sep 01 '19 at 09:26
  • Two comments up: "these" = items 1 and 3. Sorry about poor proof-reading. – Jyrki Lahtonen Sep 01 '19 at 09:35
  • 3
    First let me point out that every field automorphism will preserve the prime field. So every automorphism of $\Bbb C$ will fix the rationals pointwise. Now, as pointed by @Jyrki there are a lot of subfields of $\Bbb C$ which have $\Bbb C$ as their algebraic closure, but this result is using non-trivial facts from model theory and the compactness theorem of FOL. And indeed it is consistent that every automorphism of the additive group of $\Bbb C$ is continuous, and therefore the identity of conjugation. So indeed the choice of set theory matters. – Asaf Karagila Sep 01 '19 at 10:04
  • See https://math.stackexchange.com/a/2070297/86856 (I prove a different result there but the argument easily adapts to prove the automorphism group is infinite and in fact uncountable). – Eric Wofsey Sep 01 '19 at 13:33

1 Answers1

5

Fixing a transcendency basis $I$, we have $|I|=\mathfrak{c}=2^{\aleph_0}$. Then any permutation of $I$ extends to a (non-unique) automorphism. Hence the cardinal of this automorphism group is non only infinite, but equal to $2^{\mathfrak{c}}=2^{2^{\aleph_0}}$.

More generally, for any algebraically closed field $K$ of uncountable cardinal $\alpha$, the cardinal of $\mathrm{Aut}(K)$ is $2^\alpha$. (This is also true whenever $\alpha$ is countable, although the previous argument only applies when the transcendence degree is infinite.)

YCor
  • 17,878