Is a (Cartier) divisor on a variety uniquely determined by its restriction to curves inside the variety? If so, how do we see this?
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$\begingroup$ at least you have to assume that the variety is projective, so that it is covered by algebriac curves. Then I do not know the answer to your question. $\endgroup$– GiulioCommented Jun 30, 2016 at 16:39
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If you assume projective and smooth, this is not hard. Induct on dimension, $\dim X=1$ being the hypothesis. If $\dim X=n\geq 2$ and result proved for smaller dimensions, if $Y\in \mathcal{O}(mH)$, $m>>0$, $H$ a hyperplane section, then we may assume $Y$ is smooth and by induction, for the line bundle $L$ in question, we may assume $L_{|Y}$ is trivial. Since we may assume $H^1(L(-Y))=0$ by choosing $m$ large, we see that the section $1\in H^0(L_{|Y})$ lifts and thus $L$ is effective, but disjoint from a hypersurface $Y$. This is impossible unless $L$ is trivial, since $n\geq 2$.
