Timeline for Cubic surface in $\mathbb{P}^3$ singular along a line
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2020 at 5:04 | review | Close votes | |||
| Nov 29, 2020 at 13:08 | |||||
| Nov 27, 2020 at 0:08 | comment | added | gigi | @dhy yes thank you! I've started reasoning like this: by the useful suggestion of JoeSilverman I have that the curve is a twisted cubic and so, up to a linear transformation (which does not affect the problem) we have that the curve is locally of the form (t,t^2,t^3) and so the second coordinate is forced to be a perfect square. Sorry to bother you but can you please explain why there is no nontrivial section of the projection into the second factor $\mathbb{P}^1$? | |
| Nov 26, 2020 at 23:52 | comment | added | dhy | I think just thinking about rationality of the cubic surface (or in @JoeSilverman's formulation the curve) isn't particularly useful: they are rational, for instance in your case via the parametrization $n=\frac{p^2}{q^2}$. The more relevant fact is that there is no nontrivial section of the map to $\mathbb{P}^1$, given by forgetting $p$ and $q$. | |
| Nov 26, 2020 at 23:23 | comment | added | Joe Silverman | Might it be easier to consider your bihomogeneous equation $q^2n=p^2z$ as defining a (non-singular) curve of type $(2,1)$ in $\mathbb P^1\times\mathbb P^1$, where the first $\mathbb P^1$ has coordinates $[p,q]$ and the second has coordinates $[n,z]$. | |
| Nov 26, 2020 at 22:36 | history | asked | gigi | CC BY-SA 4.0 |