Timeline for Algebraic varieties and UFD
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Dec 7, 2011 at 19:56 | vote | accept | HYL | ||
Dec 7, 2011 at 19:55 | comment | added | HYL | Many thanks to all of you. I think this question has been sufficiently discussed and answered. | |
Dec 7, 2011 at 14:38 | comment | added | Will Sawin | Indeed, which is why the distinction between Picard/Cartier and Weil is important. | |
Dec 7, 2011 at 9:12 | comment | added | François Brunault | Note also that the fact that $k[V]$ is a UFD does not necessarily imply that $V$ is smooth. This is true if $V$ is a curve, but in general something like $k[X,Y,Z,T]/(XY-Z^2-T^3)$ should give a counterexample. | |
Dec 7, 2011 at 6:43 | comment | added | Will Sawin | The Cartier group is geometric because it classifies line bundles, which are certainly geometric. While Weil divisors are geometric, the Weil class group is less so, being a quotient by an entirely algebraically-defined subgroup (though perhaps still more geometric than the Picard group? The distinctions seem silly at this point.) | |
Dec 7, 2011 at 6:27 | comment | added | Sándor Kovács | I completely agree with Angelo. Weil divisors are totally geometric: they are defined as (linear combinations of) subvarieties. Cartier divisors are on the other hand more algebraic as they are defined by their defining equations. Their geometric nature lies in their associated Weil divisor. | |
Dec 7, 2011 at 6:07 | comment | added | Angelo | To Will: I would have said that the Weil class group is more geometric that the Picard group. Look at what happens for a nodal curve, or a curve with a cusp. | |
Dec 7, 2011 at 2:44 | history | edited | Will Sawin | CC BY-SA 3.0 |
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Dec 7, 2011 at 0:21 | comment | added | François Brunault | Maybe one needs to replace the Picard group by the Weil divisor class group, see this previous question mathoverflow.net/questions/25758/… | |
Dec 6, 2011 at 23:30 | history | answered | Will Sawin | CC BY-SA 3.0 |