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Oct 18, 2011 at 11:46 comment added Donu Arapura For example, if $X$ is projective space then Hilbert's syzygy theorem would allow you to resolve the ideal of $Z$ by graded free modules. This means that the ideal sheaf could be resolved by sums of line bundles. You could compute cohomology of $H^*(Z,\mathcal{O}_Z)$ using this.
Oct 18, 2011 at 3:22 comment added Mikhail Bondarko Could you say more about resolutions? For which $S$ the computation of $H^\ast(Z,i^*S)$ is easy?! Thank you!
Oct 17, 2011 at 22:12 comment added Donu Arapura Take $Z=x$ to be a point and $S=\mathcal{O}_X$, then $H^0(i^*S)=k$ and $H^0(i^{-1}S)=\mathcal{O}_x$. So they are certainly different. Computing $H^∗(Z,i^∗S)$ is not so easy in general, and usually it is more useful to resolve the ideal sheaf than to play with inf. nbhds I'm not quite sure what else to say.
Oct 17, 2011 at 19:52 history asked Mikhail Bondarko CC BY-SA 3.0