Timeline for smooth connected affine scheme over Z has good reduction almost everywhere
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 4, 2014 at 12:16 | answer | added | Hugo Chapdelaine | timeline score: 0 | |
| Oct 4, 2014 at 1:59 | vote | accept | Hugo Chapdelaine | ||
| Oct 4, 2014 at 1:54 | vote | accept | Hugo Chapdelaine | ||
| Oct 4, 2014 at 1:59 | |||||
| Oct 3, 2014 at 21:14 | answer | added | user59003 | timeline score: 8 | |
| Oct 3, 2014 at 20:43 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 2, 2014 at 14:14 | comment | added | Felipe Voloch | It's basically what Ari wrote. | |
| Oct 2, 2014 at 12:26 | comment | added | Hugo Chapdelaine | Thanks @Felipe. You said in your answer (in the reference link you gave) "...with schemes it is completely obvious...", so may be you could write down or give me a sketch? Thanks in advance | |
| Oct 2, 2014 at 11:55 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 2, 2014 at 1:03 | comment | added | Felipe Voloch | mathoverflow.net/questions/59071/… | |
| Oct 1, 2014 at 23:44 | comment | added | Ariyan Javanpeykar | Possibly. But what about working with the sheaf of differentials. This is a coherent sheaf on the arithmetic scheme defined by your polynomial. It is generically free by your assumption (of smoothness over the field of rational numbers). Do you see that it is locally free outside a finite set of primes? | |
| Oct 1, 2014 at 23:42 | answer | added | Ariyan Javanpeykar | timeline score: 2 | |
| Oct 1, 2014 at 23:21 | comment | added | Hugo Chapdelaine | @Ari, is it possible to prove this result using the notion of resultant combined with an decreasing induction on the dimension? | |
| Oct 1, 2014 at 23:19 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 1, 2014 at 23:19 | comment | added | Hugo Chapdelaine | sorry sorry, I meant smooth connected (so irreducible), I'll reedit it! | |
| Oct 1, 2014 at 22:39 | comment | added | Ariyan Javanpeykar | Not true. Take $f = y^2- x^3$. This is not smooth over the generic fibre. You need more than an irreducible polynomial. The type of statements you are looking for are sometimes coined "spreading out". See Bjorn Poonen's notes on rational points. For example, any nice morphism of schemes $X \to S$ which is generically smooth is smooth over an open of $S$. That's the statement you are looking for. (It's a consequence of the fact that the locus of smoothness is open. Thus, if your polynomial $f$ defines a smooth variety modulo some $p$, then it defines a smooth variety over $\mathbb Q$.) | |
| Oct 1, 2014 at 22:20 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |