Timeline for Absolute Galois group, number theory and the Axiom of Choice
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2022 at 7:42 | comment | added | Maxime Ramzi | And because any two countable algebraic closures are isomorphic, this method produces an algebraic closure which is isomorphic to the one obtained by "picking a subfield of $\mathbb C$" ! | |
| Jun 10, 2022 at 7:40 | comment | added | Maxime Ramzi | a first root, then do the same procedure in the new field to figure out which new irreducible factor I should split and so on. For any polynomial $P$, I only need to do this at most $deg(P)$ times so this terminates and I can move on to the next one, and I still have a chosen ordering of $K$. I think this works | |
| Jun 10, 2022 at 7:39 | comment | added | Maxime Ramzi | Ok I think I get it : at every stage, you have canonical orderings. You can choose, once and for all, a recipe that, given a bijection of your field $K$ with $\mathbb N$, produces a bijection of $K[X]$ with $\mathbb N$, and given a polynomial $P$, one of $K[X]/(P)$. So you order the polynomials of $\mathbb Q$ once and for all, and when you add roots to, say, $P$, you add them in the following way : $K$ has been created at a given stage so by induction I have an explicit bijection with $\mathbb N$, so I can order the irreducible factors of $P$ in $K$, and for each irreducible factor I can add | |
| Jun 10, 2022 at 7:04 | comment | added | Maxime Ramzi | @TimothyChow : Yes I absolutely agree with the beginning of your comment, which is why I don't understand how your construction allows you to do this and guarantee countability. Let's say you have a polynomial $P$, and you add one root. In your new field, $P$ has some new irreducible factors (which are no longer part of your ordering of polynomials) : which one do you now choose ? | |
| Jun 9, 2022 at 21:56 | comment | added | Timothy Chow | @MaximeRamzi First of all, I don't think you can add just one root per polynomial, can you? Given $x^3 - 2$, don't you want to adjoin all the roots? If you adjoin just one of them then you don't get the others, so when are the other roots going to get adjoined if not now? Secondly, the argument I have in mind does not begin by first constructing $\mathbb{C}$ somehow and then picking out a subfield. We start with $\mathbb{Q}$ and we build larger and larger extensions and take the union. | |
| Jun 9, 2022 at 21:49 | comment | added | Maxime Ramzi | @TimothyChow : I'm a but confused as to how this allows you to enumerate all roots - a priori when you do this you add one root per polynomial. What the lexicographic order argument tells you is that the subfield of $\mathbb C$ of algebraic elements is countable - nothing about other ones or different embeddings | |
| Jun 9, 2022 at 21:43 | comment | added | Timothy Chow | @MaximeRamzi Oh, I see now what you were concerned about. I'm slightly worried that if you try to rely on $\mathbb{C}$ then you run into subtleties about how to pick your embedding into $\mathbb{C}$. The way I would go about it is, when handed an irreducible polynomial such as $x^3 - 2$, to construct the splitting field directly from the polynomial, as outlined for example on Wikipedia (which, by the way, does caution about circular reasoning when proving the existence of an algebraic closure). | |
| Jun 9, 2022 at 16:05 | comment | added | Maxime Ramzi | @Timothy Chow : sure but you still need to order the ones you've added and that you have left. For this you can use the lexicographic order in $\mathbb C$ | |
| Jun 9, 2022 at 15:10 | comment | added | Timothy Chow | @MaximeRamzi A countable union of finite sets is not necessarily countable without AC, but as you're enumerating the polynomials, you can discard any roots that you have encountered already. | |
| Jun 9, 2022 at 13:35 | comment | added | Maxime Ramzi | Ah nevermind, I am silly: you can order the roots, say lexicographically in $\mathbb C$. | |
| Jun 9, 2022 at 13:34 | comment | added | Maxime Ramzi | I'm a bit confused, how can one prove that the algebraic closure is countable ? The "order the polynomials canonically" only seems to give that the algebraic closure is a countable union of finite sets. Must that be countable without AC ? (once you have a well-order on $\overline {\mathbb Q}$ clearly there is no issue) | |
| Jun 2, 2022 at 16:03 | comment | added | Simon Henry | Interesting. So there is only one countable algebraic closure. Thanks ! | |
| Jun 2, 2022 at 14:30 | comment | added | Emil Jeřábek | @SimonHenry Every automorphism of a finite extension of $\mathbb Q$ lifts to an automorphism of the “usual” algebraic closure, as the latter is countable. | |
| Jun 2, 2022 at 14:29 | comment | added | David E Speyer | @SimonHenry Is it that the usual algebraic closure can have few automorphisms? Specifically, using $\overline{\mathbb{Q}}$ for the algebraic closure of $\mathbb{Q} \subset \mathbb{C}$, is it possible that there is a Galois extension $K/\mathbb{Q}$ for which $\text{Aut}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{Gal}(K/\mathbb{Q})$ is not surjective? I used to think this could happen, but Timothy Chow's comment on the original question suggests not, and now I think I can prove that it can't happen. | |
| Jun 2, 2022 at 14:17 | comment | added | Simon Henry | I think it is more correct to say that when people talk about "algebraic number theory" they only talk about finite extention of $\mathbb{Q}$. The "usual" algebraic closure can have very few automorphisms. | |
| Jun 2, 2022 at 13:43 | history | answered | Timothy Chow | CC BY-SA 4.0 |