Timeline for Trigonal curves of genus three: can their Galois closure be non-abelian
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2015 at 17:25 | comment | added | Noam D. Elkies | What has the Klein curve to do with this? For any linear and quartic forms $P_1(x,y)$ and $P_4(x,y)$, the curve $X: z^3 P_1 = P_4$ admits a cyclic degree-3 map $y/x$ to the projective line, and $X$ is rarely (is it ever?) isomorphic with the Klein quartic. | |
| Mar 25, 2013 at 8:48 | vote | accept | Tridib | ||
| Mar 23, 2013 at 13:15 | answer | added | Michael Stoll | timeline score: 10 | |
| Mar 23, 2013 at 12:54 | comment | added | Jason Starr | @Michael: I didn't see your comment before I added my comment. | |
| Mar 23, 2013 at 12:53 | comment | added | Jason Starr | Every group of order 6 that acts faithfully on a set of 3 elements is the full symmetric group on that set. So, if the Galois closure does have degree 6, then it automatically has Galois group isomorphic to the symmetric group on 3 elements. | |
| Mar 23, 2013 at 12:47 | comment | added | Michael Stoll | If a field extension of degree 3 is not Galois, then its Galois closure has Galois group $S_3$, which is non-abelian. So either your morphism $X \to {\mathbb P}^1$ is already Galois (and then abelian), or else $Y \to {\mathbb P}^1$ has non-abelian Galois group. | |
| Mar 23, 2013 at 12:28 | history | asked | Tridib | CC BY-SA 3.0 |