Timeline for Surjective implies local affine surjective?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2011 at 17:48 | comment | added | Qing Liu | Sorry, you don't need openess of $f$. Yes for local rings, flat=faithfully flat. | |
| Feb 14, 2011 at 12:33 | vote | accept | evgeniamerkulova | ||
| Feb 14, 2011 at 12:33 | comment | added | evgeniamerkulova | Professor Liu thank you for answer, but I am not sure I understand it correctly because a word lacks between "don't" and "openess"! For local morphism of local rings flat= faithfully flat, right? | |
| Feb 14, 2011 at 10:15 | comment | added | Qing Liu | Your conclusion is correct, but you don't openess of $f$ : if $y\in Y$ and $x\in f^{-1}(y)$, then $O_{Y,y}\to O_{X,x}$ is faithfully flat. | |
| Feb 13, 2011 at 22:43 | comment | added | evgeniamerkulova | So now we deduce thanks to your wonderfull answer following result, because we can reduce to affine case. If $f:X \to Y$ is faithfully flat morphism of integral schemes and $X$ is normal, then also $Y$ is normal.( Maybe should add $f$ locally of finite presentation to be sure $f$ is open?). Is that true, Professor? | |
| Feb 13, 2011 at 22:29 | comment | added | evgeniamerkulova | Wonderfull, Professor! This is very amusing because source of question is your book! In example 3.5 of 4.3.1 you say that normalization of integral scheme is flat only if scheme was already normal but you do not give proof and say to do exercise 1.2.10 (which is affine case) . I can do exercise just with faithfully flat assumption: if $A\subset B$ is faithfully flat ring extension and $B$ is integrally closed domain, then $A$ is also integrally closed ( no finiteness or noether assumption and I do not assume that $B$ is integral closure of $A$). Is that correct? | |
| Feb 13, 2011 at 21:46 | history | edited | Qing Liu | CC BY-SA 2.5 |
Explanations
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| Feb 13, 2011 at 21:34 | history | answered | Qing Liu | CC BY-SA 2.5 |