Timeline for Are generically trivial finite unramified morphisms trivial
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 9, 2014 at 18:40 | vote | accept | Unit section | ||
| Nov 7, 2014 at 13:47 | answer | added | Ariyan Javanpeykar | timeline score: 2 | |
| Nov 7, 2014 at 12:54 | comment | added | Jason Starr | @Unitsection: Okay, let me be more clear. Look up the original formulation of "Zariski's Main Theorem", not necessarily the formulation given in some textbooks. There is a nice comparison of the different formulations in Mumford's "Red Book". Also the Wikipedia entry is pretty good. | |
| Nov 7, 2014 at 12:38 | history | edited | Unit section | CC BY-SA 3.0 |
Added a positive answer under the assumption that $\dim S = 1$.
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| Nov 7, 2014 at 11:39 | comment | added | Unit section | @abx I see your point. I meant to assume that the generic fibre $X_{K(S)}$ of $X\to S$ has a $K(S)$-rational point. This is stronger than the set $X(K(S))$ being non-empty (unless we consider only elements of $X(K(S))$ that correspond to the generic point Spec $K(S) \to S$. Then $X(K(S)) = X_{K(S)}(K(S))$ which can be certainly be empty (take a curve with no rational points over $\mathbb C(t)$). | |
| Nov 7, 2014 at 11:37 | comment | added | Unit section | @JasonStarr How is ZMT relevant here? Am I missing something obvious? | |
| Nov 7, 2014 at 11:34 | comment | added | abx | $X(K(S))$ is always non-empty : just take some (strange) embedding of $K(S)$ into $\mathbb{C}$ ... | |
| Nov 7, 2014 at 10:55 | comment | added | Jason Starr | You should look up "Zariski's Main Theorem". | |
| Nov 7, 2014 at 9:36 | review | First posts | |||
| Nov 7, 2014 at 10:39 | |||||
| Nov 7, 2014 at 9:33 | history | asked | Unit section | CC BY-SA 3.0 |