Timeline for Action of a profinite group
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S May 1, 2014 at 11:28 | history | bounty ended | Pablo | ||
| S May 1, 2014 at 11:28 | history | notice removed | Pablo | ||
| May 1, 2014 at 11:27 | vote | accept | Pablo | ||
| May 1, 2014 at 10:05 | answer | added | Jeremy Rickard | timeline score: 3 | |
| Apr 30, 2014 at 15:32 | vote | accept | Pablo | ||
| May 1, 2014 at 11:27 | |||||
| Apr 30, 2014 at 15:32 | answer | added | Pablo | timeline score: 0 | |
| S Apr 24, 2014 at 17:05 | history | bounty started | Pablo | ||
| S Apr 24, 2014 at 17:05 | history | notice added | Pablo | Improve details | |
| Apr 22, 2014 at 21:16 | answer | added | Julian Rosen | timeline score: 4 | |
| Apr 22, 2014 at 21:13 | comment | added | YCor | @Alex: I assume that $I$ is supposed infinite, otherwise you're right. | |
| Apr 22, 2014 at 20:38 | comment | added | Pablo | I think that if $G$ is finite, you may pick some nontrivial $v \in V$ and take $U = <gv : g \in G>$. This shows that the answer to 1 is yes. | |
| Apr 22, 2014 at 20:24 | comment | added | Alex B. | I suspect that I don't understand the question, but if you take $G$ finite of order not divisible by $p$, $M$ any non-trivial irreducible $\mathbb{Q}_p[G]$-module, and $V$ a $\mathbb{Z}_p[G]$-lattice in $M$, then isn't the answer to all your questions trivially (no pun intended) "no"? | |
| Apr 22, 2014 at 19:02 | history | edited | Pablo | CC BY-SA 3.0 |
added 66 characters in body
|
| Apr 22, 2014 at 17:24 | history | edited | Pablo | CC BY-SA 3.0 |
added 60 characters in body
|
| Apr 22, 2014 at 17:21 | comment | added | Pablo | Yes, I mean that the submodule will be nonntrivial, and generated as a group (forgetting the action of $G$) by finitely many elements. | |
| Apr 22, 2014 at 17:15 | comment | added | YCor | Ah OK you just mean $\mathbb{Z}_p^I$... Question 1 looks weird: $\{0\}$ answers positively the question. Also the closed submodule generated by any element is t.f.g. as a submodule... What do you mean? A nonzero submodule that is t.f.g. as topological group? | |
| Apr 22, 2014 at 17:12 | comment | added | Pablo | $p$ is a fixed prime not to be changed. $\mathbb{F}_p$ is a field with $p$ elements. The infinite product means that we take a power of disjoint copies of the same group, to say, vectors with coordinates in $\mathbb{Z}_p$ (only one group - not changing the prime). $G$ is assumed to be a finitely generated profinite group so compactness is not an issue. | |
| Apr 22, 2014 at 17:05 | comment | added | YCor | Does $\mathbb{Z}_p$ denotes the $p$-adics? Do you mean $p\in I$ instead of $i\in I$ (otherwise the product is meaningless). What is $\mathbb{F}_p$? Also "i.e. $V$ is a profinite $G$-module" is not just a restatement: there is something (not hard) to check, which fails when $G$ is not assumed compact. | |
| Apr 22, 2014 at 16:42 | history | asked | Pablo | CC BY-SA 3.0 |