Timeline for Extending birational isomorphisms between planar curves to the P^2
Current License: CC BY-SA 3.0
11 events
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| May 11, 2012 at 21:15 | comment | added | Maarten Derickx | Now $E(\Gamma(3))/\mathbb{P}^1$ doesn't have a section of infinite order. But by choosing a suitable line $L \in \P^2$ to parameterize the elliptic elliptic curves in the Hesse pensil by one can find a different elliptic curve $E \to \P^1$ wich has a section of infite order, and this $E$ will also be birational equivalent to $\P^2$. Now the translation by this point of infinite order is an isomorphism from $E$ to itself and hence gives a birational map from $\P^2$ to itself. This birational automorphism will be the translation by a point of infinite order in most of its fibers over $L$. | |
| May 11, 2012 at 21:02 | comment | added | Maarten Derickx | Yesterday during lunch at Univeriteit Leiden I also heard another argument (by bas Edixhoven) why the "negative result for elliptic curves" argument cannot work in the birational case. The argument is by considering the Hesse pencil $u(x3+y3+z3)+vxyz$ . Now by blowing up in the 9 points where rational map $\mathbb{P}^2 \to \mathbb{P}^1$ given by $(x:y:z)\mapsto(x^3+y^3+z^3:xyz)$ is not defined we get $E(\Gamma(3))\mapsto \mathbb{P}^2$. $E(\Gamma(3))$ together with the map to $\mathbb{P}^1$ (via $\mathbb{P}^2$) will be the universal elliptic curve with full level 3 structure. | |
| May 7, 2012 at 15:15 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 7, 2012 at 9:38 | comment | added | Maarten Derickx | Thanks for the answer. Altough it was not the answer to my question it stil was instructive. | |
| May 7, 2012 at 6:11 | comment | added | François Brunault | Isn't it the case that any point on a cubic curve can be made into an inflection point using a birational transformation ? There is an algorithm due to Nagell to transform any cubic curve into Weierstrass form, but I don't know whether this applies here. | |
| May 7, 2012 at 1:34 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 7, 2012 at 1:02 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 7, 2012 at 0:41 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 6, 2012 at 23:52 | comment | added | R.P. | Re your counter-example: Maarten's $F$ was allowed to be a birational automorphism. Hence $L$ is not necessarily mapped to a line, I guess. | |
| May 6, 2012 at 23:19 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 6, 2012 at 23:08 | history | answered | Karl Schwede | CC BY-SA 3.0 |