Timeline for A profinite group which is not its own profinite completion?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21 at 11:45 | comment | added | Wojowu | @SpicetheBird You can see that it's not topologically finitely generated quite directly - if it was, there'd only be finitely many continuous maps $G_{\mathbb Q}\to\mathbb Z/2$, that is finitely many open index $2$ subgroups, that is finitely many degree $2$ extensions of $\mathbb Q$. But adjoining square roots of different primes gives you infinitely many quadratic extensions. | |
| Sep 7, 2012 at 0:22 | comment | added | Daniel Litt | (cont.) ... finitely many points, whence the (at least prime to p) part of the Galois group of the curve still remains manifestly finitely generated. | |
| Sep 7, 2012 at 0:21 | comment | added | Daniel Litt | @Spice the Bird, Moosbrugger: You're probably aware of this, but here's an analogy which might explain why both of these observations are to be expected. In the dictionary between number fields and function fields, one thinks as Galois groups as follows--take the etale fundamental group of your curve minus some points, and take the limit as one removes all the points. Removing a point gives a factor of \hat Z; the topological generator is analogous to a Frobenius lift on the number field side. On the other hand, if you only allow finite amounts of ramification, this is like removing... | |
| Nov 30, 2011 at 2:09 | comment | added | Moosbrugger | @Spice the Bird: On the bright side, Shaferevich conjectured that the the Galois group ramified at only finitely many primes is topologically finitely generated. I don't know if it's proved or not. | |
| Nov 30, 2011 at 0:53 | comment | added | Spice the Bird | This means that $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ is not finitely generated as a topological group. What a monster. | |
| Nov 29, 2011 at 16:52 | comment | added | Giuseppe | I now wonder: What is the profinite completion of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$? I've never seen anybody refer to this object. | |
| Nov 29, 2011 at 13:55 | history | answered | Keenan Kidwell | CC BY-SA 3.0 |