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Jan 4, 2021 at 7:01 comment added abx @R. van Dobben de Bruyn: Oops, you are right of course. I forgot the $\otimes \mathbb{Q}$.
Jan 3, 2021 at 21:19 comment added R. van Dobben de Bruyn @abx over finite fields and when tensored with $\mathbf Q$, it should be injective for curves, as $\operatorname{Pic}^0(C)$ is torsion and smooth projective curves have Picard rank $1$.
Jan 3, 2021 at 8:35 review Close votes
Jan 10, 2021 at 21:47
Jan 3, 2021 at 8:16 comment added abx Not true either for a curve in $\Bbb{P}^3$, for the same reason.
Jan 3, 2021 at 5:35 comment added R. van Dobben de Bruyn This is not true for surfaces $S \subseteq \mathbf P^3$ (embed them linearly in $\mathbf P^4$ if you insist on codimension $2$ complete intersections). The Picard rank can be more than $1$.
Jan 3, 2021 at 3:06 history asked user127776 CC BY-SA 4.0