Timeline for Why is the inverse of a bijective rational map rational?
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22 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Oct 31, 2014 at 8:58 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Oct 30, 2014 at 20:17 | comment | added | Mikhail Skopenkov | Francesco, Thank you for your irrefragable answer and great references! Now I got it. You helped me very much! | |
| Oct 30, 2014 at 20:11 | vote | accept | Mikhail Skopenkov | ||
| Oct 30, 2014 at 9:09 | comment | added | Francesco Polizzi | Yes, the two definitions of open immersion are equivalent. A reference is Grothendieck EGA I, Chapitre I Proposition 4.2.2 a), page 122. See also this MSE question and the beautiful answer of Georges Elencwajg: math.stackexchange.com/questions/70293/… | |
| Oct 29, 2014 at 19:55 | comment | added | Mikhail Skopenkov | thank you very much! Point (2) is clear, thanks. Point (1) turns us back to the initial question: what is the exact REFERENCE to the proof that the inverse map is rational? Theorem 12.83 in page 355 from [1] does NOT tell this because Definition 3.40 from the same book [1] does not say that an open immersion is an isomorphism onto its image. The authors of [1] use the term immersion in their own sense (they GIVE a formal definition DIFFERENT from the one from Glossary_of_scheme_theory). If these defs are equivalent then a reference to the proof of equivalence is very much wanted. | |
| Oct 29, 2014 at 19:43 | comment | added | Francesco Polizzi | (2) The map $z \to z^2$ does not satisfy the assumption that $\mathcal{O}_Y \to f_* \mathcal{O}_X$ is an isomorphism. The point is that its fibres are not integral, since it is a (ramified) double cover. In fact, $f_* \mathcal{O}_X$ is a rank $2$ vector bundle in this case, containing $\mathcal{O}_X$ as a direct summand. | |
| Oct 29, 2014 at 19:40 | comment | added | Francesco Polizzi | (1) By definition, "immersions" are maps which factors through isomorphisms with subschemes, i.e. they give an isomorphism onto their image. en.wikipedia.org/wiki/Glossary_of_scheme_theory | |
| Oct 29, 2014 at 19:15 | comment | added | Mikhail Skopenkov | I have looked carefully through the provided references and did not find any indication to the proof that the inverse map is rational. Theorem 12.83 in page 355 from [1] does not contain the part of Proposition 2 starting from 'hence an isomorphism onto its image'. Definition 3.40 from [1] does not tell that an open immersion is an isomorphism onto its image. Finally, I cannot see immediately which assumptions of Proposition 2 are not satisfied by the map $C\to C$, $z\mapsto z^2$ with no rational inverse. Could you explain what am I missing? [1] Görz-Wedhorn, Algebraic Geometry I. | |
| Oct 2, 2014 at 6:40 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 18:09 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 15:46 | comment | added | Mikhail Skopenkov | Thank you for your very detailed answer! Now I see. You helped me very much! | |
| Sep 30, 2014 at 15:45 | vote | accept | Mikhail Skopenkov | ||
| Oct 29, 2014 at 19:33 | |||||
| Sep 30, 2014 at 15:03 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 14:48 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 14:42 | comment | added | Francesco Polizzi | You are welcome. Actually, it is a consequence of (a version of) Zariski's Main Theorem. I have edited the answer, adding references. | |
| Sep 30, 2014 at 14:41 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 11:35 | comment | added | Mikhail Skopenkov | Thanks you very much! My question is what is a REFERENCE to this proposition or to the initial question being a particular case. In the discussion of the question Isomorphism between varieties of char 0 I could not find any indication how it is proved that the inverse map is rational. The version of the Zarisky main theorem is not sufficient because of the assumption that the initial map is birational. This answer does not prove anything on the rationality of $f^{-1}$. | |
| Sep 29, 2014 at 9:30 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 29, 2014 at 9:25 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Sep 29, 2014 at 9:17 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |