Timeline for An fppf cover with trivial Picard group
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2017 at 6:16 | vote | accept | user2831784 | ||
| Oct 2, 2017 at 9:43 | answer | added | Jason Starr | timeline score: 12 | |
| Oct 2, 2017 at 8:25 | comment | added | Jason Starr | The morphism does not have to be etale, but I am assuming that it is finite. However, that is irrelevant as an argument. I will write more details in an answer below. | |
| Oct 2, 2017 at 1:51 | comment | added | user2831784 | @JasonStarr Thanks. If I understand correctly, in your comment the fppf cover $f$ is assumed to be an etale morphism. Can there be no such $f$ which is finite flat but ramified, or smooth of relative dimension $\ge 1$? | |
| Oct 2, 2017 at 1:08 | comment | added | Jason Starr | If $X$ has genus $\geq 1$, then for every fppf cover, every connected component $Y_i$ of $Y$ is a smooth, quasi-projective curve of genus $\geq 1$. In other words, it is a dense open subscheme of a smooth projective curve $\overline{Y}_i$ of genus $\geq 1$. The Picard group of such a curve is infinitely generated; in fact, the kernel of the degree homomorphism is infinitely generated. The complement of $Y_i$ in $\overline{Y}_i$ is generated by finitely many points. Thus, $\text{Pic}(Y_i)$ is the quotient of an infinitely generated group by a finitely generated group. So it is nonzero. | |
| Oct 2, 2017 at 0:44 | comment | added | user2831784 | @JasonStarr Could you give some details of the argument that no such fppf cover $f : Y \to X$ exists? Say I take $X$ to be an elliptic curve over $k$. | |
| Oct 1, 2017 at 21:04 | comment | added | Jason Starr | The simplest such example is where $X$ is a smooth quasi-projective curve over an algebraically closed field $k$ such that the projective model of $X$ has genus $g>0$. | |
| Oct 1, 2017 at 20:35 | history | asked | user2831784 | CC BY-SA 3.0 |