Timeline for Galois orbit of a $k_{s}$ - torus
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2019 at 2:24 | comment | added | vutuanhien | Thanks a lot. So now there can only be finitely many different $\sigma(U)$, right? | |
| Sep 28, 2019 at 17:22 | comment | added | reuns | They have coefficients in $k_s\subset k^{alg}$ and finitely many algebraic numbers generate a finite extension. $Gal(k_s/k)$ corresponds to infinitely many isomorphisms $A_s\to \sigma(A_s)$ but there are only finitely many $\sigma(A_s)$ | |
| Sep 28, 2019 at 16:41 | comment | added | vutuanhien | @reuns I'm sorry if this question is silly but why those equations have coefficients in a finite extension of $k$? Is it related to the fact that every torus splits over a finite Galois extension or something else? | |
| Sep 28, 2019 at 14:55 | comment | added | reuns | I guess you can take an affine neighborhood $U$ of $1$ in your $k_s$-torus, affine means it is $Spec(A_s)$, locally finite type means $U$ can be chosen such that $A_s$ is a finitely generated $k_s$ algebra thus defined by finitely many algebraic equations whose coefficients are in a finite extension of $k$ so that $Gal(k_s/k)$ sends $A_s$ to finitely many different copies | |
| Sep 28, 2019 at 13:53 | history | edited | András Bátkai | CC BY-SA 4.0 |
added journal citation
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| Sep 28, 2019 at 13:10 | review | First posts | |||
| Sep 28, 2019 at 13:53 | |||||
| Sep 28, 2019 at 13:08 | history | asked | vutuanhien | CC BY-SA 4.0 |