Timeline for Known cases of Tate conjecture for varieties which are smooth over a curve
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 17 at 14:13 | answer | added | Will Sawin | timeline score: 3 | |
| Nov 16 at 22:53 | comment | added | Vik78 | @JasonStarr I see my mistake— I intended the question to allow the base to be any smooth projective curve, which need not necessarily itself be smooth over $\mathbb{P}^1$. I apologize for the confusion and have edited the post | |
| Nov 16 at 22:50 | history | edited | Vik78 | CC BY-SA 4.0 |
deleted 33 characters in body
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| Nov 16 at 22:33 | comment | added | Jason Starr | @Vik78 A smooth projective morphism from a curve $B$ (the general case where your base curve $B$ has arbitrary genus) to $\mathbb{P}^1$ is a finite etale morphism. Since the etale fundamental group of $\mathbb{P}^1$ is trivial (over a separably closed field), this forces $B$ to be a genus $0$ curve. So there is no reduction in your question from the case of a base curve of arbitrary genus to a curve of genus $0$. | |
| Nov 16 at 21:16 | comment | added | Vik78 | @AriyanJavanpeykar Every curve admits a nonconstant morphism to $\mathbb{P}^1$, so that one can always compose a map to a curve with such a morphism. I see now that this alone does not guarantee smoothness in positive characteristic as the map of curves may be inseparable. Is that what goes wrong in your example? | |
| Nov 16 at 20:34 | comment | added | Ariyan Javanpeykar | @Vik78 Why do you write "equivalently to $\mathbb{P}^1$" in your question? There are surfaces of general type which do not admit a smooth morphism to $\mathbb{P}^1$, but which do admit a smooth morphism to some higher genus curve. These are sometimes called Kodaira surfaces. | |
| Nov 16 at 18:35 | history | edited | Vik78 | CC BY-SA 4.0 |
Edited to remove redundant flatness assumption
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| Nov 16 at 18:32 | comment | added | Vik78 | @willsawin thanks for the response. Could you please expand on how you conclude when the fibration splits after a finite cover? | |
| Nov 16 at 18:11 | comment | added | Will Sawin | Smooth implies flat so that condition isn't needed. Wlog the fiber is a curve. The key invariants are the genus of the base and the genus of the fiber. If either is 0 the statement is known for boring reasons. If the fibration splits as a product after a finite cover of the base then it's OK by the Tate conjecture for divisors. Otherwise you can maybe use progress on function field BSD. | |
| Nov 16 at 17:40 | history | asked | Vik78 | CC BY-SA 4.0 |