Timeline for Proving that any two points on a variety can be joined by a curve; why does Bertini apply?
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 11, 2014 at 7:48 | vote | accept | Jesko Hüttenhain | ||
| Nov 8, 2014 at 1:07 | answer | added | Bjorn Poonen | timeline score: 13 | |
| Apr 9, 2014 at 12:43 | vote | accept | Jesko Hüttenhain | ||
| Nov 11, 2014 at 7:48 | |||||
| Apr 7, 2014 at 18:16 | answer | added | Puzzled | timeline score: 4 | |
| Apr 7, 2014 at 13:04 | comment | added | Jesko Hüttenhain | @user76758: Thank you! This looks promising. | |
| Apr 7, 2014 at 12:41 | comment | added | user76758 | @JeskoHüttenhain: Are you reading the proof of the theorem of the cube in Mumford's book on abelian varieties? An excellent characteristic-free reference on Bertini theorems (of many flavors!) is the book by Jouanolou (just Google that name and "Bertini"). And Chow's Lemma is valid with separatedness rather than completeness (even though Hartshorne's textbook exercise imposes completeness); e.g., look in EGA II, 5.6. | |
| Apr 7, 2014 at 12:26 | comment | added | Jesko Hüttenhain | That's true, but as you said, it only works in characteristic zero. | |
| Apr 7, 2014 at 12:21 | comment | added | Alex Degtyarev | If you are in characteristic $0$, you can resolve the singularities, apply the statement to any pair of points in the respective exceptional divisors, and then map the curve back to $X$. | |
| Apr 7, 2014 at 11:07 | history | edited | Jesko Hüttenhain | CC BY-SA 3.0 |
added 860 characters in body
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| Apr 7, 2014 at 7:54 | comment | added | abx | There is no smoothness assumption in the second Bertini theorem, see for instance this paper, Theorem 5.3. | |
| Apr 7, 2014 at 7:46 | comment | added | Jesko Hüttenhain | But here we are back to the original problem: Why can I apply Bertini when $X$ is not smooth? | |
| Apr 7, 2014 at 7:28 | comment | added | abx | If you accept to have your $X$ quasi-projective, you can just apply directly Bertini's theorem to the linear system of hyperplanes passing through $x$ and $y$. If $X$ contains the line $\langle x,y\rangle$ you are done, otherwise Bertini tells you that a general divisor in the system is irreducible, and you get the result by induction on the dimension. | |
| Apr 7, 2014 at 6:51 | history | asked | Jesko Hüttenhain | CC BY-SA 3.0 |