Timeline for Are all real-closed subfields of $\overline{\mathbb{Q}}$ conjugate?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 16 at 18:33 | comment | added | Jianing Song | For comparison, if $K$ is an algebraic closed field with characteristic $0$ and transcendence degree $\kappa>\omega$, then it has $2^\kappa$ non-isomorphic maximal real-closed subfields: see here. | |
| Sep 5, 2018 at 13:17 | comment | added | LSpice | Specifically to @JarekKuben's reference, if (like me) you can't read German fluently but can read French, then you can find a stronger form of this result as Exercise 32 on p. A VI.42 of Bourbaki - Algèbre. (Don't ask me about the page numbering.) | |
| Sep 4, 2018 at 21:55 | comment | added | Jarek Kuben | This fact is mentioned - with some references - at page 5 of this notes by Keith Conrad. | |
| Sep 4, 2018 at 21:39 | vote | accept | benblumsmith | ||
| Sep 4, 2018 at 21:38 | answer | added | Emil Jeřábek | timeline score: 13 | |
| Sep 4, 2018 at 21:34 | comment | added | benblumsmith | Perhaps the argument goes like this: pick any real-closed subfield of $\overline{\mathbb{Q}}$. Order-complete it to embed it as a subfield of $\mathbb{R}$. The embedding extends to an embedding of $\overline{\mathbb{Q}}$ in $\mathbb{C}$ by a choice of $\sqrt{-1}\mapsto \pm i$. Since all embeddings of $\overline{\mathbb{Q}}$ in $\mathbb{C}$ have the same image, this means we can automorph $\overline{\mathbb{Q}}$ in such a way that any given real-closed subfield is the intersection with $\mathbb{R}$. Does that work? | |
| Sep 4, 2018 at 21:26 | comment | added | benblumsmith | I don't see that the argument he gives would lead one to conclude they are all conjugate. Am I missing something obvious? | |
| Sep 4, 2018 at 21:22 | comment | added | David E Speyer | The question he is answering is "Is it true that any element of order 2 in G is conjugate to σ?" But I agree that his answer is not unambiguous on this point. | |
| Sep 4, 2018 at 21:20 | comment | added | benblumsmith | I think Matt E is saying that all the involutions are "complex conjugations" i.e. they are involutions fixing a real-closed field, but I don't believe he is asserting that they are conjugate in $G_\mathbb{Q}$... | |
| Sep 4, 2018 at 21:20 | comment | added | LSpice | Hurrah for sufficient weasel words that I wasn't technically wrong. :-) | |
| Sep 4, 2018 at 21:18 | comment | added | David E Speyer | This answer by Matt E on math.SE says that all involutions in the absolute Galois group are conjugate: math.stackexchange.com/a/622935/448 | |
| Sep 4, 2018 at 21:17 | comment | added | benblumsmith | Again, you are right.... | |
| Sep 4, 2018 at 21:17 | comment | added | LSpice | Oh, wait. An involution in a quotient of $G_{\mathbb Q}$ doesn't obviously lift to an involution in $G_{\mathbb Q}$, so maybe it's subtler than that. | |
| Sep 4, 2018 at 21:15 | comment | added | benblumsmith | Yes I think you're right! Or, $C_2\times C_2\times C_2$ is a Galois group over $\mathbb{Q}$. Wow, that was easy. Please feel free to make this an answer. | |
| Sep 4, 2018 at 21:15 | comment | added | LSpice | For example, $\mathrm S_5$ is a Galois group over $\mathbb Q$ (right?), and its involutions don't form a single conjugacy class. | |
| Sep 4, 2018 at 21:14 | comment | added | LSpice | $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ is probably beyond apprehension on its own, but we know a lot of its quotients. If this is false, maybe there's some easy quotient that witnesses its falsity? | |
| Sep 4, 2018 at 21:02 | history | edited | benblumsmith | CC BY-SA 4.0 |
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| Sep 4, 2018 at 20:56 | history | asked | benblumsmith | CC BY-SA 4.0 |