Timeline for Why do twists of an algebraic group over k correspond to k-torsors over G
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 18, 2012 at 2:28 | comment | added | anon | Look at Hom from G to the twist (as a functor). This is a G-torsor. | |
| Mar 17, 2012 at 21:25 | comment | added | Harry | Just a minor nitpick: the bijection between the set of $k$-twists and $H^1$ does not follow from the references he gives between parentheses...(That's what had me going.) You need the analogues of those "statements for varieties" in the setting of "torsors" which he sort of explains in Remark 4.5.6. | |
| Mar 17, 2012 at 21:20 | comment | added | Harry | Ow I see. Well then it's clear. | |
| Mar 17, 2012 at 21:18 | comment | added | user18237 | He means twists of $\mathbf{G}$ as a $G$-torsor, not twists of $G$ as a variety. | |
| Mar 17, 2012 at 21:11 | comment | added | Harry | On page 98 of B. Poonen's notes www-math.mit.edu/~poonen/papers/Qpoints.pdf it clearly states that the set of $k$-torsors is equal to the set of twists of $\mathbf{G}$, where $\mathbf{G}$ is the algebraic group $G$ endowed with its right action of $G$ given by translation. Aren't the twists of $\mathbf{G}$ the same as the twists of $G$? | |
| Mar 17, 2012 at 20:47 | comment | added | user18237 | This isn't true. For example a quadratic twist of an elliptic curve is a twist in your sense, but not a torsor under the original elliptic curve. Even worse a twist of $G$ in your sense need not be a torsor for any group whatsoever. Here is a silly example: $G=\mathbf{Z}/5$, $X=\text{spec}(k\times k\times k')$ where $k'/k$ is a cubic separable extension. | |
| Mar 17, 2012 at 19:28 | history | asked | Harry | CC BY-SA 3.0 |