Timeline for "Forms" of quadrics
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2014 at 19:42 | comment | added | Daniel Loughran | @Mattias: Its ok, I have thought about it a bit as well, and I could not find an "elementary" way to describe the obstruction class in my case either. | |
| Sep 24, 2014 at 18:48 | comment | added | Matthias Wendt | @DanielLoughran: true, both comments. Sorry for bringing up the Milnor conjecture, but the easier ways of proving this (Hilbert 90 etc.) are also explained in Gille-Szamuely. I removed the remark with the cup product; but I still wonder how to interpret the construction. The conics correspond to quaternion algebras, hence also to elements in $H^2(k,\mathbb{Z}/2)\cong {}_2Br(k)$. The only operation I see would be the additive structure of ${}_2Br(k)$... Sorry, I did not have the time yet to sit down and actually compute the obstruction class of your example. | |
| Sep 24, 2014 at 18:43 | history | edited | Matthias Wendt | CC BY-SA 3.0 |
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| Sep 23, 2014 at 21:25 | comment | added | Daniel Loughran | Also I'm not sure your interpretation of my construction is correct; conics over $k$ are classified by $H^1(k,\mbox{PGL}_2)$, not $H^1(k,\mathbb{Z}/2)$. | |
| Sep 23, 2014 at 21:08 | comment | added | Daniel Loughran | Thanks for working this out. Note that you don't need to invoke any deep theorems in K-theory to show that $H^2(k, \mathbb{Z}/2) \cong {}_2\mbox{Br}(k)$; this follows easily from Kummer theory and Hilbert's Theorem 90. | |
| Sep 23, 2014 at 19:22 | history | edited | Matthias Wendt | CC BY-SA 3.0 |
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| Sep 23, 2014 at 19:14 | history | answered | Matthias Wendt | CC BY-SA 3.0 |