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Mar 13, 2018 at 5:36 comment added naf For a projective example, simply take any cycle which is homologically equivalent to zero but not algebraically equivalent to zero. (On a surface, any 1-cycle which is homologically trivial is algebraically trivial.)
Mar 12, 2018 at 15:49 comment added Jason Starr I suspect that this fails already for Hironaka's example, as described in Appendix B of Hartshorne's "Algebraic geometry".
Mar 12, 2018 at 14:55 history edited YCor
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Mar 12, 2018 at 14:39 comment added pi_1 Indeed, my question is whether one can choose a class on $\widetilde S$ which is homologically trivial.
Mar 12, 2018 at 14:19 comment added byu If you take any surface containing the curves in the support of $\gamma$, and take a resolution of it, you get a class in $Pic(\widetilde S)$ that pushes forward to $\gamma$ like you say. Is your question whether we can take this to be homologically trivial?
Mar 12, 2018 at 12:38 comment added pi_1 Thank you. No, I do mean "curves".
Mar 12, 2018 at 12:26 comment added Qfwfq You mean $\gamma$ in $\mathrm{CH}^1(X)$ (codimension $1$)?
Mar 12, 2018 at 11:53 history asked pi_1 CC BY-SA 3.0