Timeline for The variety induced by an extension of a field
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 6, 2020 at 10:10 | comment | added | Federico Fallucca | Can you help me also for this question, please? math.stackexchange.com/q/3815141 | |
| Sep 6, 2020 at 10:07 | vote | accept | Federico Fallucca | ||
| Sep 6, 2020 at 10:07 | comment | added | Federico Fallucca | Yes, ok, but this would mean simply that the elements of the type $(a,...,a)$ belongs to the stabilizer of each point, right? We are interested only of that actions that are faithful, so this mean we must consider the new faithful action $((Z/nZ)^k)/(\cap_x G_x)$, that is isomorphic exactly to (Z/nZ)^{k-1} via the isomorphism $(a_1,..,a_k)\to (a_2-a_1, \cdots , a_k-a_1)$ ? | |
| Sep 5, 2020 at 19:03 | comment | added | Francesco Polizzi | The group acting effectively is $(Z/nZ)^{k-1}$, because the element $(-1,\ldots,-1) \in (Z/nZ)^k$ acts trivially on the homogeneous coordinates. | |
| Sep 5, 2020 at 15:03 | vote | accept | Federico Fallucca | ||
| Sep 5, 2020 at 15:03 | |||||
| Sep 5, 2020 at 15:03 | comment | added | Federico Fallucca | The problem is that the group $(Z/nZ)^k$ acts on $X$ and not $(Z/nZ)^{k-1}$. What is the mistake? | |
| Sep 5, 2020 at 14:58 | vote | accept | Federico Fallucca | ||
| Sep 5, 2020 at 15:03 | |||||
| Aug 12, 2020 at 21:36 | history | answered | amateur | CC BY-SA 4.0 |