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Timeline for Algebraic varieties and UFD

Current License: CC BY-SA 3.0

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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