Timeline for Is each rationally chain connected surface rational?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2014 at 6:10 | vote | accept | Mikhail Skopenkov | ||
| Sep 4, 2014 at 6:06 | comment | added | Mikhail Skopenkov | @DanielLitt: Yes, exactly, this proves that a ruled rationally connected surface must be rational. The point is that for the proof of assertion 1 we also need assertion 3 now proved by Roberto Pignatelli | |
| Sep 4, 2014 at 1:12 | comment | added | Daniel Litt | @MikhailSkopenkov: Can't you just choose two points in different fibers and find a rational curve connecting them; then that rational curve dominates $X$, hence $X$ is rational? | |
| Sep 3, 2014 at 17:32 | history | edited | Mikhail Skopenkov | CC BY-SA 3.0 |
added 141 characters in body
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| Sep 3, 2014 at 13:07 | answer | added | Roberto Pignatelli | timeline score: 2 | |
| Sep 3, 2014 at 10:33 | comment | added | Mikhail Skopenkov | Anyway you helped much by giving a new approach to the problem, thank you! | |
| Sep 3, 2014 at 10:27 | comment | added | abx | You are right, you need a more sophisticated argument. Let me think about it. | |
| Sep 3, 2014 at 9:58 | comment | added | Mikhail Skopenkov | Thank you! I am sorry I cannot get the details of the argument. Yes, a ruled rationally connected surface must be rational. But we have proved that the initial surface S is uniruled. How to conclude that it is ruled? It seems we have come back to the original question, namely, to assertion 3. | |
| Sep 3, 2014 at 9:11 | comment | added | abx | Actually you don't need Castelnuovo, sorry. Once you know your surface is ruled (and rationally connected), you get plenty of rational curves which map onto your base curve $X$. | |
| Sep 3, 2014 at 8:16 | comment | added | Mikhail Skopenkov | Thanx! Then, by Noether-Enriques theorem, the initial surface is dominated by $X\times \mathbb{P}^1$ (not $\mathbb{P}^1\times \mathbb{P}^1$). In other words, the initial surface is uniruled, not yet unirational and not yet ruled, cf. assertion 3 in the question. How can I apply the Castelnuovo theorem? | |
| Sep 3, 2014 at 7:33 | comment | added | abx | No. Once you know you have a family of rational curves through each point, just take a general curve in the parameter space of the family and you have a surface fibered with rational curves dominating your surface. | |
| Sep 3, 2014 at 7:31 | comment | added | Mikhail Skopenkov | Thank you very much. It seems that the resulting excercise requires further nontrivial results (like the Enriques-Kodaira classification of surfaces) to construct the fibration with rational fibers and relate it to the initial surface. That is why a reference is preferrable. | |
| Sep 3, 2014 at 6:48 | comment | added | abx | You need three nontrivial results : rationally chain connected $\ \Rightarrow\ $ rationally connected (see Koll\'ar's book), the Noether-Enriques theorem which says that any surface fibered over a curve with rational general fibers is ruled, and the Castelnuovo theorem which says that a surface dominated by $\mathbb{P}^1\times \mathbb{P}^1$ is rational. The last two can be found in any book on surfaces. The rest is an exercise. | |
| Sep 3, 2014 at 6:31 | history | asked | Mikhail Skopenkov | CC BY-SA 3.0 |