Timeline for Central division and quaternion algebras
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 12, 2011 at 16:35 | vote | accept | Louis | ||
| Aug 12, 2011 at 13:47 | answer | added | naf | timeline score: 6 | |
| Aug 12, 2011 at 7:50 | comment | added | Louis | Excuse me M. Hansen but i don't quite understand. Does the central simple algebra you get with this process satisfy the assumption 1 of the question? If i understand well then, it suffice to look over the field k(x_1,..,x_6), to take D_1 as a square root of the quaternion (x_1,x_2), D_2 the quaternion algebra (x_3,x_4), and D_3 the quaternion algebra (x_5,x_6) ? | |
| Aug 11, 2011 at 22:45 | comment | added | David Hansen | If $D$ is a quaternion algebra, then its local Hasse invariant is $1/2$ at a finite even number of places and zero elsewhere, so you can get a square root of the Brauer class by changing the nonzero invts to $1/4$ at half the places and $3/4$ at the other half. | |
| Aug 11, 2011 at 14:02 | history | edited | Louis | CC BY-SA 3.0 |
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| Aug 11, 2011 at 14:01 | comment | added | Louis | exactly, i also correct this. Thank you. | |
| Aug 11, 2011 at 13:59 | comment | added | André Henriques | Could you explain what you mean by $D_1^2D_2$? Are you taking the tensor product of $D_1$, $D_1$, and $D_2$? | |
| Aug 11, 2011 at 13:51 | history | edited | Louis | CC BY-SA 3.0 |
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| Aug 11, 2011 at 13:50 | comment | added | Louis | oh my god sorry, i meant D_2 but wrote D_4... I corrected the typo, thank you. | |
| Aug 11, 2011 at 8:41 | comment | added | S. Carnahan♦ | There aren't any conditions on $D_2$, and $D_4$ seems to appear out of nowhere. | |
| Aug 10, 2011 at 20:10 | history | edited | Louis | CC BY-SA 3.0 |
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| Aug 10, 2011 at 9:20 | history | asked | Louis | CC BY-SA 3.0 |