Timeline for Wikipedia's definition of 'locally free sheaf'
Current License: CC BY-SA 2.5
5 events
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| Nov 6, 2010 at 3:21 | comment | added | BCnrd | Dear roger123: The example in my comment, viewed as an $R$-algebra, is a counterexample to the sufficiency aspect in EGA IV$_4$, 18.4.12(ii) (the mistake is assuming locally finite type rather than locally finite presentation, and the error in the proof occurs when they say "donc $j$ est une immersion ouverte"). This bogus sufficiency claim is invoked in the proof of 18.4.14, but that proof can be easily made OK by verifying the locally finite presentation condition holds in the relevant cases for that argument. | |
| Nov 4, 2010 at 18:36 | comment | added | roger123 | Thank you for the comment. In my case $R$ is noetherian and $M$ is unfortunately not finitely generated. | |
| Nov 4, 2010 at 17:55 | comment | added | BCnrd | Dear Michael: your defn of "locally free" for finitely generated modules is wrong in the sense that it isn't too useful when $R$ is not noetherian. The right defn for f.gentd modules is local freeness for Zariski topology (i.e., over Zariski-open covering, acquires a basis), and this is equivalent to the stalk condition in the noetherian case. But not otherwise. A counterexample is $M=R/I$ for $R=\prod_{n \ge 0} \mathbf{F}_2$ and $I$ an ideal that isn't finitely generated. This $M$ is not loc. free but each $R_P$ is $\mathbf{F}_2$ (needs some thought) and $M_P$ is 0 or 1-diml. | |
| Nov 4, 2010 at 17:28 | history | edited | Harry Gindi | CC BY-SA 2.5 |
added 4 characters in body
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| Nov 4, 2010 at 17:25 | history | answered | Michael | CC BY-SA 2.5 |