Timeline for A morphism from proper to affine is constant?
Current License: CC BY-SA 2.5
10 events
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| Aug 15, 2010 at 9:34 | history | edited | Behrang Noohi | CC BY-SA 2.5 |
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| Aug 11, 2010 at 17:57 | history | edited | Behrang Noohi | CC BY-SA 2.5 |
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| Aug 10, 2010 at 11:40 | comment | added | Behrang Noohi | That was very helpful, thanks (also to Angelo). I am trying to come up with a general statement which is valid for ANY base scheme (but I am a bit flexible with $f$). Although flatness of $f$ may be the wrong kind of assumption here, the non-normal counterexamples I can think of turn out to be non flat. | |
| Aug 10, 2010 at 11:23 | history | edited | Behrang Noohi | CC BY-SA 2.5 |
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| Aug 7, 2010 at 16:11 | answer | added | Angelo | timeline score: 2 | |
| Aug 7, 2010 at 14:50 | history | edited | Bruce Westbury | CC BY-SA 2.5 |
Fixed typo in title
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| Aug 7, 2010 at 13:28 | comment | added | BCnrd | By the way, surely despite the weakening of properness, you still assume $f$ is separated, right? Please edit the question so it is accurate as stated. | |
| Aug 7, 2010 at 13:26 | comment | added | BCnrd | Problem local on base, so $S =$ Spec($A$), $Y =$ Spec($B$), $g$ is $A$-alg. map $B \rightarrow O(X)$, & ask if image = $A$ ($A \rightarrow O(X)$ is inj.). If OK for all $B$ f. pres. over $A$ then OK for all $A$-alg. $B$ by limits. Then $B = O(X)$ forces $O(X) = A$. Conversely if $O(X) = A$ all OK. So $Y$ irrelevant; question is if $O_S \rightarrow f_ {\ast}O_X$ is =. Surely $U$ dense in sense preserved by localization, so localize @ generic pt of complement of max. open where $O_S = f_ {\ast}O_X$, so $S$ local and "$O_S = f_ {\ast}O_X$" away from closed pt. Try Nagata or non-normal counterex. | |
| Aug 7, 2010 at 13:09 | answer | added | Ricky | timeline score: 2 | |
| Aug 7, 2010 at 12:03 | history | asked | Behrang Noohi | CC BY-SA 2.5 |