# Does Con(ZF) imply Con(ZF + Aut C = Z/2Z)?

How many field automorphisms does $\mathbf{C}$ have? If you assume the axiom of choice, there are tons of them -- $2^{2^{\aleph_0}}$ I believe. And what if you don't -- how essential is the axiom of choice to constructing "wild" automorphisms of $\mathbf{C}$? Specifically, if you assume that ZF admits a model, does that imply that ZF admits a model where $\mathbf{C}$ has no wild automorphisms: $\mathop{Aut}\mathbf{C}=\mathbf{Z}/2\mathbf{Z}$?

I suppose if that's true, then the next logical question is to construct models of ZF where $\mathop{Aut}\mathbf{C}$ has cardinality strictly between 2 and $2^{2^{\aleph_0}}$--pretty disturbing if you ask me. Which finite groups can you hit?

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The use of inaccessible cardinals is not necessary here, the Baire property works just as well as Lebesgue measure. Shelah (Can you take Solovay's inaccessible away, Isr. J. Math. 48, 1984, 1-47) shows that ZF + DC + "every subset of R has the Baire property" is relatively consistent with ZF. (This is also the paper where Shelah also shows that the inaccessible cardinal is necessary for Solovay's result.)

The connection is an old theorem of Banach and Pettis which says that any Baire measurable homomorphism between Polish groups is automatically continuous. This result is provable in ZF + DC. Since C is a Polish group under addition, it follows that every additive endomorphism of C is continuous in Shelah's model. Since the continuous additive endomorphisms of C are precisely the R-vector space endomorphisms, it follows that the only field automorphisms of C in Shelah's model are the identity and conjugation.

As pointed out by Pete Clark in the comments, the Artin-Schreier Theorem goes through using only the Boolean Prime Ideal Theorem (PIT), which is significantly weaker than full AC. This shows that AC is not completely necessary to show that there is a unique conjugacy class of elements of order 2 in Aut(C) and that these correspond precisely to the finite subgroups of Aut(C).

Looking at Pete Clark's Field Theory Notes, specifically at Steps 4 and 5 of his proof of the Grand Artin-Schreier Theorem on pages 62-63, I think that it is a theorem of ZF that the only possible order for a nontrivial finite subgroup of Aut(C) is 2.

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+1: very nice! Also, the fact that orderability of formally real fields is equivalent to BPIT is the answer to a previous question of mine: mathoverflow.net/questions/8560/… With that fact in mind, the usual proof of Artin-Schreier can be seen to depend only on BPIT. –  Pete L. Clark Feb 28 '10 at 15:21
Thanks for the link Pete, the Berr, Delon, Schmid paper is exactly what I was thinking about. –  François G. Dorais Feb 28 '10 at 15:34
Great, François! This avoids the inaccessible cardinal completely and provides a complete, affirmative answer to the question. This is the right answer to the question. –  Joel David Hamkins Feb 28 '10 at 19:13
So many prompt and interesting responses: This is why I love MO. I agree that Pete's notes give a proof that in ZF, Aut C only has Z/2Z as a nontrivial finite subgroup. But I still don't see any reason why Aut C couldn't be something terrifying like a semidirect product of Z by Z/2Z. Any comments on this? –  Jared Weinstein Feb 28 '10 at 20:40
I'm not sure. I think Aut(C) has topological structure that prevents it from being a countably infinite group, but the topological structure can depend on AC in all sorts of ways. I'm pretty sure Aut(C) can't be countably infinite assuming PIT. –  François G. Dorais Feb 28 '10 at 21:08

Dear Jared,

First things first: assuming AC, it is indeed true that for any algebraically closed field $F$, $\# \operatorname{Aut}(F) = 2^{\# F}$. The main idea for this is that we can choose a transcendence basis and then every permutation of the transcendence basis extends to an automorphism of $F$. See e.g. Theorem 80 on p. 49 of

http://math.uga.edu/~pete/FieldTheory.pdf

for more details.

Second, yes, it is consistent with ZF that $\operatorname{Aut}(\mathbb{C})$ is just the identity and complex conjugation, at least if you believe in inaccesible cardinals. There are a lot of results of the form "A field automorphism of $\mathbb{C}$ is continuous (i.e., is the identity or complex conjugation) if...." One of these sufficient conditions is measurability, e.g.

Kestelman, H. Automorphisms of the field of complex numbers. Proc. London Math. Soc. (2) 53, (1951). 1--12.

http://www.math.uga.edu/~pete/Kestelman51.pdf

And it is well-known I believe I have heard that there are models of ZF in which every subset of $\mathbb{C}$ is measurable. [Addendum: As Prof. Edgar mentions in his comment, there is the Solovay model, whose construction relies on the existence of an inaccessible cardinal. So I'm not sure whether it is known unconditionally whether "All subsets of $\mathbb{C}$ are Lebesgue measurable" is consistent with ZF. But it seems that this is believed to be true, at least.]

As for your third question -- in conventional mathematics we have the Artin-Schreier theorem, which implies that for any algebraically closed field $F$, $\operatorname{Aut}(F)$ has no finite subgroups of order greater than $2$. (See e.g. loc. cit., Theorem 98 on p. 61.) But the proof of this uses AC. Without AC, I certainly don't know. I suspect you'll need an actual set theorist (such people exist on MO!) for that.

Addendum: As established in a previous MO question -- How much choice is needed to show that formally real fields can be ordered? -- the orderability of all formally real fields is equivalent to the Boolean Prime Ideal Theorem. It follows that (as Francois G. Dorais suggested) BPIT implies the Grand Artin-Schreier Theorem. Perhaps this could be helpful in answering Jared's last question (though I don't immediately see how).

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Does Solovay's model (where every set is measurable) require a large cardinal axiom as well as ZF? And it that what you refer to here? –  Gerald Edgar Feb 28 '10 at 12:07
Gerald -- if anything, I think I was referring to Dependent Choice. But to be honest, I know very little about this and am mostly just repeating what some analysts have told me. Because of your question, I looked in Schechter's Analysis and Foundations and found that, if LM = {all subsets of R are Lebesgue measurable}, it seems to be open whether ZF + DC + LM is consistent. So I had better weaken the language in my answer a bit and wait for someone else to weigh in. –  Pete L. Clark Feb 28 '10 at 12:20
Pete, the theory ZF+DC+LM is exactly equiconsistent to ZFC+"there is an inaccessible cardinal", by results of Solovay and Shelah, as I explain in my answer. Isn't it interesting how large cardinals come into this question? –  Joel David Hamkins Feb 28 '10 at 13:34
Pete, I briefly checked your notes. Since C has characteristic 0, you seem to be able to rule out all possible finite orders of subgroups of Aut(C) except 2 using Kummer Theory. Do you think this use of Kummer Theory implicitly uses the Axiom of Choice? –  François G. Dorais Feb 28 '10 at 16:05

Pete's answer is informative. But there is a subtle point that actually turns the answer somewhat upside down. It turns out that the answer is related to large cardinals! [Edit: François's answer shows how to avoid the inaccessible cardinal.]

Pete mentions the Kestelman article, which explains

Every function which defines a non-trivial automorphism of the complex numbers transforms every bounded set (in the Argand plane) into a set of Lebesgue measure zero or else into a non-measurable set.

By considering larger and larger bounded sets, this means that the existence of a nontrivial automorphism implies the existence of a nonmeasurable set. I believe that this part of the Kestelman article does not use AC, although I suppose that one must have Dependent Choices (DC) to have a decent theory of Lebesgue measure.

Pete mentions that there are known to be models of ZF in which every set of reals is measurable. These models, however, as Gerald mentions in his comments, are constructed from a ground model of ZFC having an inaccessible cardinal (Solovay's model). Shelah has proved that this large cardinal hypothesis cannot be omitted. Thus, the consistency of ZF + DC + "Every set is Lebesgue measurable" is equivalent to the theory "ZFC + there is an inaccessible cardinal". One way to explain what this means is that we should be exactly as confident in the consistency of inaccessible cardinals as we are that there is no analogue of the Vitali construction of a non-Lebesgue measurable set not using AC.

Since the Kestelman result shows that the existence of a nontrivial automorphism of C (in the presence of DC) implies the existence of a nonmeasurable set, this establishes:

• Con(ZFC + there is an inaccessible cardinal) implies Con(ZF + DC + there is no nontrivial automorphism of C).

This is the actual result that Pete's argument provides. The hypothesis here is strictly stronger than Con(ZF), if ZF is consistent. [Edit: François shows that by using the Baire property instead of measure, one avoids the need for inaccessible cardinals, so he has the optimal argument.]

Having DC in the conclusion seems what should be desired, when considering functions on R and C, since even to know that the epsilon-delta and convergent sequence characterizations of continuity are equivalent uses DC.

I'm not sure what happens if one drops DC in the conclusion. For example, it is known to be consistent with ZF that the reals are a countable union of countable sets, and this model does not have DC or even countable choice. Perhaps this is a good candidate model?

Finally, the question about realizing other groups is extremely interesting.

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+1: thanks for clarifying. My expertise on this is certainly limited to the structure of automorphism groups of fields under the "standard" set-theoretic assumptions (AC). I mentioned the set theoretic stuff at all only knowing that JDH would be here to catch me if I fell. –  Pete L. Clark Feb 28 '10 at 14:43
(JDH and/or FGD, to be more precise.) –  Pete L. Clark Feb 28 '10 at 15:47