Timeline for Examples of injective morphisms which are not universally injective
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2012 at 16:47 | comment | added | Jérémy Blanc | Ok, thanks anyway. I will have a look. | |
| Nov 26, 2012 at 12:32 | comment | added | Angelo | Dear Jérémy, sorry, I don't read anything, therefore I never know what is in the literature. The result is certainly well know; it is the sort of thing I would use without mention. | |
| Nov 25, 2012 at 22:16 | comment | added | Jérémy Blanc | @Angelo: Thanks for your answers. Do you think that your result could be found somewhere in the literature? For examples, this says that for maps between varieties over an algebraically closed field, open immersion= injective + étale, which seems nice to be observed | |
| Nov 25, 2012 at 21:29 | vote | accept | Jérémy Blanc | ||
| Nov 25, 2012 at 20:23 | comment | added | anon | "Universally injective" is equivalent to "injective and all the maps on the residue fields are radicial" EGA I, 3.7.1. | |
| Nov 25, 2012 at 16:45 | comment | added | Angelo | To Jérémy: the map $x \mapsto x^p$ is certainly not étale in characteristic $p$. A finite flat map is étale if and only if it non-ramificated, that is, if the fibers are reduced. About your other remark, you are absolutely right, I edited the post. | |
| Nov 25, 2012 at 16:43 | history | edited | Angelo | CC BY-SA 3.0 |
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| Nov 25, 2012 at 16:06 | comment | added | Jérémy Blanc | @ Damian: I had already read EGA before asking the question (especially 3.5.5 and 3.5.11) and I do not find that it answer the question (but maybe I missed something). For me, it just says that radicial is equal to universally injective and that condition is only necessary to be checked on ALL extensions K where K is algebraically closed. It seems implicit from what is said in EGA that one algebraically closed field is not sufficient. (This is indeed why I asked the question). But again, I could have missed something. | |
| Nov 25, 2012 at 16:05 | comment | added | Jérémy Blanc | @Angelo Thanks for the answer. I do not get all the points. 1) For your first case, you say that finite, étale and injective implies isomorphism. What about x-> x^p in characteristic p? It seems finite, étale and injective but is not an isomorphism. What did I miss? 2) You say "we need to show that k(p) is not a purely inseparable extension of k(f(p))". For me, we need to show the converse. | |
| Nov 25, 2012 at 10:02 | history | answered | Angelo | CC BY-SA 3.0 |