Timeline for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Mar 23, 2010 at 22:02 | comment | added | Peter Tingley | I don't think this counter example is correct. The w be a primitive third root of unity. Then $(x,y)= (-1,1), (-w^2, w), (-w,w^2)$ are all sent to $(0,1)$ by your map. So it is not bijective. | |
| Mar 21, 2010 at 17:56 | comment | added | Qfwfq | And in this case $X$ is even smooth (though not projective). | |
| Mar 21, 2010 at 17:55 | comment | added | Qfwfq | Something like: $(x,y)\mapsto(y^2+x,y^3)$ | |
| Mar 21, 2010 at 17:51 | comment | added | Qfwfq | I don't think it's true: $\mathbb{A}^2$ is irreducible, but consider the map from $\mathbb{A}^2$ to itself that sends each vertical line into a "translated" cuspidal cubic... | |
| Mar 2, 2010 at 18:56 | answer | added | damiano | timeline score: 8 | |
| Mar 2, 2010 at 16:03 | vote | accept | Peter Tingley | ||
| Mar 2, 2010 at 0:03 | answer | added | Frank | timeline score: 7 | |
| Mar 1, 2010 at 23:44 | comment | added | Andrea Ferretti | Basically the smooth case follows from a computation of differential and Zariski main theorem. | |
| Mar 1, 2010 at 23:43 | comment | added | Andrea Ferretti | So, are you OK with the smooth case? That is much easier and I can sketch it if you want. | |
| Mar 1, 2010 at 23:27 | answer | added | Pavel Etingof | timeline score: 7 | |
| Mar 1, 2010 at 21:41 | comment | added | Mariano Suárez-Álvarez | A bijective regular morphism between irreducible varieties in characteristic zero is biregular. | |
| Mar 1, 2010 at 21:35 | history | asked | Peter Tingley | CC BY-SA 2.5 |