Timeline for Schemes with no finite morphisms onto themselves
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2019 at 17:15 | vote | accept | CommunityBot | ||
| Apr 24, 2019 at 20:28 | answer | added | Piotr Achinger | timeline score: 1 | |
| Apr 24, 2019 at 14:58 | comment | added | Bort | Sort of similarly to the Frobenius, toric varieties have "toric Frobenius" endomorphisms $F_k$ for each $k$, obtained by extended the map $(x_1,\ldots,x_n) \mapsto (x_1^k, \ldots, x_n^k)$ from the torus to the whole variety. | |
| Apr 24, 2019 at 14:49 | comment | added | Daniel Litt | Just a remark -- Will's comment is for smooth projective schemes over a field of characteristic zero; in characteristic p every scheme has the Frobenius endomorphism, which will often be an example of what you're looking for. | |
| Apr 24, 2019 at 14:42 | answer | added | Francesco Polizzi | timeline score: 2 | |
| Apr 24, 2019 at 14:34 | comment | added | Will Sawin | Well for instance if the canonical bundle is very ample, because any finite map has to be injective on sections of the canonical bundle, the map would be isomorphic on sections of the canonical bundle, which makes it injective because the canonical bundle is very ample, hence an isomorphism, hence the identity by assumption. | |
| Apr 24, 2019 at 14:13 | comment | added | user138661 | @WillSawin maybe you could give some necessary conditions then? | |
| Apr 24, 2019 at 13:58 | comment | added | Will Sawin | I think the answer is pretty clearly "no". The existence of such a morphism is a rare event. It's possible to give many obstructions to its existence, but the only way to guarantee its existence is to exhibit one. | |
| Apr 24, 2019 at 13:41 | history | asked | user138661 | CC BY-SA 4.0 |