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Feb 17, 2012 at 17:45 history edited gio CC BY-SA 3.0
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Jun 17, 2011 at 13:50 answer added Remke Kloosterman timeline score: 6
Jun 17, 2011 at 13:26 comment added Francesco Polizzi Sorry yiou are right, I misread your comment. Now I understand what you say. Well, usually "factorial" means that every codimension 1 subvariety is cut out by an hypersurface in the ambient space, and my answer shows that it is not always the case. But, as you remark, if C is not smooth this is not equivalent to $\textrm{Pic}(C)=Z$ .
Jun 17, 2011 at 13:04 comment added Parsa Why do you need smoothness? In the case $D$ is reduced, you have the Lefschetz hyperplane theorem, and if $D$ is reduced means its exponential sequence is exact, so you use the Kodaira vanishings on $X$ to get the vanishings you need on $D$, and the 5-lemma on the long exact sequence associated to the exponential sequences of $X$ and $D$ gives you the isomorphism $H^1(X,\mathcal {O_X}^*) \cong H^1(D,\mathcal {O_D}^*)$.
Jun 17, 2011 at 12:55 answer added Francesco Polizzi timeline score: 4
Jun 17, 2011 at 12:39 comment added Francesco Polizzi Parsa, this is true when $C$ is smooth.
Jun 17, 2011 at 12:36 comment added Parsa By the Grothendieck-Lefschetz theorem on Picard groups, the restriction map $Pic(X) \rightarrow Pic(D)$ is an isomorphism when $D$ is an ample effective divisor on $X$ and the dimension of $X$ is at least $4$. So the second part of your question is always the case. See Ample Subvarieties of Alg Var's by Hartshorne (Corollary IV.3.3) and Positivity in Alg. Geom. I Remark (3.1.26.)
Jun 17, 2011 at 9:47 history asked gio CC BY-SA 3.0