4

By this question: Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice? We have that over ZF, algebraic closures of $\mathbb Q$ aren't unique. Are their Galois groups as extensions over $\mathbb Q$ still isomorphic? In general, what other obstructions to doing constructive Galois theory over $Gal(\bar{\mathbb Q}/\mathbb Q)$ might we run into?

Community
  • 1
fhyve
  • 2,495
  • It is unique if you assume it is well-orderable (i.e. countable); but otherwise what sort of obstructions do you expect to run into? (I mean by the first sentence that the non-uniqueness is because one of them is countable, and the other is not. So by always taking the countable one, you avoid this issue and can proceed.) – Asaf Karagila Mar 01 '15 at 10:09
  • Well, from an algebraic perspective, is there a reason to prefer the construction that gives you the countable closure over the uncountable one(s)? I don't exactly know how to articulate it, but it feels like if we just pick the countable one because it is countable, how do we know we are looking at the "real" absolute Galois group, and that we are not somehow losing information? Can we show, somehow, that across theories, the one constructed from the countable closure is the same as the unique one in ZFC? – fhyve Mar 01 '15 at 21:06
  • My intuition is that if I know what happens in the case without choice, and in what way choice is required (if it is required at all), then it will give me insight into why understanding the group is difficult. – fhyve Mar 01 '15 at 21:07
  • Well, countable models are just automatically nicer. Because they are countable. – Asaf Karagila Mar 01 '15 at 21:27
  • This is sort of a strawman, but it emphasizes my point: That's kind of like saying, "I want to study the behavious of algebraic closures of Q. There are a few of them, and one happens to be countable. So lets only study the countable one and forget about the others because the countable one is easier." It's a post hoc justification. Is there a reason to pick the construction that gives the countable closure over ones that give uncountable closures, without knowing beforehand that the first one is countable? Just that it is countable, and that countable is easier, isn't good enough for me. – fhyve Mar 03 '15 at 08:02
  • If a field is countable, then there is a canonical construction which will always produce a countable algebraically closed extension. Is there a reason to choose it over others? Yes, it provably exists; and has nice properties as a bonus. Since you're not looking for a first-order structure of this closure (because you want to investigate an external construction), you have the benefit of being able to choose from what you have. It would be nice, of course, if there was some model theoretic argument, that holds without choice, to transfer some statements about one Galois group to another. – Asaf Karagila Mar 03 '15 at 08:08
  • And by the way, the fact that there is so much emphasis on Noetherian rings, finite extensions of $\Bbb Q$ and so on, only shows you that this "strawman" is really a legitimate argument. We work with nice things because it's easier to prove statements about nice things. Countable things are *very* nice. – Asaf Karagila Mar 03 '15 at 08:10
  • Is the countable closure always unique for a countable field? I'm pretty sure in the other link someone said that it is for $\mathbb Q$, but is it true in general?

    As for Noetherian rings, I don't think that that is quite the same argument. In that case, you have a bunch of things that you care about that already satisfy nice size properties, and in the general case, restricting yourself to rings with those nice size properties is easier. In the case that I'm asking about however, we don't know (well, we might now know) whether the thing we care about is the countable one or not.

     – fhyve Mar 04 '15 at 02:10
  • Related: Lauchli's algebraic closure of Q BY WILFRID HODGES – Watson Oct 19 '17 at 21:01

0 Answers0