Timeline for Classify all the fields with abelian absolute Galois group
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 4, 2020 at 9:57 | vote | accept | S. Li | ||
| Feb 17, 2020 at 20:47 | answer | added | Arno Fehm | timeline score: 19 | |
| Jun 15, 2017 at 18:38 | comment | added | user05811 | @LeonidPositselski $\mathbb{Z}_p\times\mathbb{Z}/2$ cannot be the absolute Galois group of a field, for any prime $p$. More generally, the normalizer of an involution in an absolute Galois group of a field is the group of order 2 generated by this involution. See Proposition 19.4.3 in: I. Efrat, Valuations, Orderings, and Milnor K-Theory, AMS 2006. | |
| Feb 14, 2017 at 0:20 | comment | added | Leonid Positselski | My argument shows that, denoting the $p$-adic integers by $\mathbb Z_p$, the group $\mathbb Z_2\times \mathbb Z/2$ is not an absolute Galois group. Still I don't know whether $\mathbb Z_p\times \mathbb Z/2$ can be an absolute Galois group for odd $p$. | |
| Feb 13, 2017 at 23:22 | comment | added | S. Li | @LeonidPositselski Thanks to your comment, so procyclicity is also wrong. Good to know, thanks again. | |
| Feb 13, 2017 at 21:08 | comment | added | Leonid Positselski | On the other hand, $\widehat{\mathbb Z}\times \mathbb Z/2$ is not an absolute Galois group, I believe (because the modified supercommutativity rule has to hold in the Galois cohomology with constant coefficients $\mathbb Z/2$). | |
| Feb 13, 2017 at 20:52 | comment | added | Leonid Positselski | Consider the iterated formal power series field $K=k((T_1))((T_2))\cdots((T_n))$ with $k$ an algebraically closed field of characteristic zero, and you'll get a field $K$ with $Gal(\overline{K}/K)=\widehat{\mathbb Z}^n$. | |
| Feb 13, 2017 at 17:59 | comment | added | S. Li | Ah, thanks to all of your comments. So it seems to me that all the examples listed above have absolute galois group either $\hat{\mathbb{Z}}$ or $\mathbb{Z}/2$? | |
| Feb 13, 2017 at 16:33 | comment | added | David Lampert | Exercise 3, Chapter 8, Lang's "Algebra": $K$ = a maximal subfield of $\overline{\mathbb Q}$ not containing $\sqrt{2}$. | |
| Feb 13, 2017 at 15:44 | comment | added | Watson | The finite fields can be replaced by quasi-finite fields. | |
| Feb 13, 2017 at 15:40 | comment | added | KConrad | The real numbers can be replaced by real closed fields. Finite fields can be replaced by algebraic extensions of finite fields. Do you know all the closed subgroups of the profinite integers $\widehat{\mathbf Z}$? | |
| Feb 13, 2017 at 15:30 | history | asked | S. Li | CC BY-SA 3.0 |