Timeline for index of smooth varieties
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 9, 2018 at 7:16 | comment | added | user43198 | Thank you very much for the answer and the subsequent comments. | |
| May 9, 2018 at 7:15 | vote | accept | user43198 | ||
| May 7, 2018 at 20:46 | history | edited | Jason Starr | CC BY-SA 4.0 |
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| May 7, 2018 at 20:22 | history | edited | Jason Starr | CC BY-SA 4.0 |
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| May 7, 2018 at 15:46 | comment | added | Jason Starr | @user43198 "I thought that any projective variety after base change by a finite field extension contains a rational point." That is certainly true, but there is no guarantee that the finite field extension is "unramified", i.e., is the fraction field of a finite, \'etale extension of DVRs. | |
| May 7, 2018 at 15:44 | comment | added | user43198 | To be clear you are saying that the generic fiber of $\pi$ satisfies the property that the base change of the generic fiber to any finite unramified extension of the fraction field of $R$, has index $d$. This is a little counter-intuitive for me since I thought that any projective variety after base change by a finite field extension contains a rational point. | |
| May 7, 2018 at 9:15 | comment | added | Jason Starr | @user43198 The construction does not require that the residue field of $R$ is algebraically closed. | |
| May 7, 2018 at 9:04 | comment | added | user43198 | Thank you. One last question. Your construction does not seem to require the residue field is algebraically closed. Is this correct i.e., will your construction also work if the residue field of $R$ is not algebraically closed? | |
| May 7, 2018 at 8:54 | comment | added | Jason Starr | @user43198 "Is this correct?" Yes, that is correct. | |
| May 7, 2018 at 8:23 | comment | added | user43198 | One further question. Redoing your arguments for a finite, unramified extension $L$ of $K$ (with residue degree $0$) and $R_L$ the integral closure of $R$ in $L$, we will be able to conclude that the generic fiber of the corresponding $\pi_L:\mathcal{X}_{R_L}^s \to \mbox{Spec}(R_L)$ is also of index $d$. The generic fiber of $\pi_L$ should be the base change to $\mbox{Spec}(L)$ of the generic fiber of $\pi$ (unramified extension will ensure that the uniformizer of $R$ and $R_L$ can be the same). Is this correct? | |
| May 7, 2018 at 0:16 | comment | added | Jason Starr | @user43198. "Is it obvious that the generic fiber of $\pi$ is smooth?" I do not know what is obvious to you. However, that follows from the Jacobian criterion, since the Jacobian ideal of the Fermat polynomial equals $\langle d t_0^{d-1},\dots, dt_{d-1}^{d-1} \rangle$. | |
| May 7, 2018 at 0:10 | comment | added | user43198 | Is it obvious that the generic fiber of $\pi$ is smooth? | |
| May 6, 2018 at 23:07 | history | edited | Jason Starr | CC BY-SA 4.0 |
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| S May 6, 2018 at 23:01 | history | answered | Jason Starr | CC BY-SA 4.0 | |
| S May 6, 2018 at 23:01 | history | made wiki | Post Made Community Wiki by Jason Starr |