Timeline for What is the right definition of the Picard group of a commutative ring?
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9 events
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| Feb 2, 2010 at 12:23 | comment | added | Pete L. Clark | The "of course" doesn't make sense to me, because that's not what was meant in the standard text that I referenced (Eisenbud). The point is that it's subtle whether the weaker sense of locally implies the stronger sense (yes for Noetherian rings or with local constancy of the rank; no in general). | |
| Feb 2, 2010 at 12:05 | comment | added | Martin Brandenburg | Yes of course locally refers to the Zariski topology ;-). I didn't read the proof of Bourbaki, but the argument above is the most slick proof I can imagine. | |
| Feb 2, 2010 at 11:25 | comment | added | Pete L. Clark | I took the weaker use of "locally free" from Eisenbud, loc. cit. | |
| Feb 2, 2010 at 11:19 | comment | added | Pete L. Clark | I suppose it also depends on what you mean by locally free: I meant that the localization at each prime ideal is free. If instead you mean "locally in the Zariski topology" -- your condition about the f_i's above -- then that implies the local constancy of the rank (same theorem in Bourbaki). | |
| Feb 2, 2010 at 8:40 | comment | added | Pete L. Clark | Bourbaki, Section II.5.2: For an A-module P, TFAE: (a) P is finitely generated projective. (c) P is finitely generated, for each p in Spec(A), P_{p} is free, and the rank function p |-> rank(P_{p}) is locally constant on Spec(A). | |
| Feb 2, 2010 at 8:25 | comment | added | Martin Brandenburg | hm? I don't need any constancy. | |
| Feb 2, 2010 at 3:31 | comment | added | Pete L. Clark | I believe this is not quite correct, depending upon what you mean by "finite rank". It is true though if the rank is finite and constant, which is the situation I asked about. | |
| Feb 2, 2010 at 2:52 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
added 104 characters in body
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| Feb 2, 2010 at 2:44 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |