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Sep 28, 2010 at 15:46 answer added Hailong Dao timeline score: 1
Sep 25, 2010 at 21:53 comment added Karl Schwede In fact, BCnrd's argument implies aniket's as well (at least in sufficiently geometric settings). This is because $H^{i+1}_{X\setminus U}(X, \widetilde{M})$ is non-zero at the generic points of $X \setminus U$ for appropriate $i$ (depending on the dimension of those generic points).
Sep 25, 2010 at 10:12 comment added pinaki A necessary condition: $I$ has to be of pure codimension one (see e.g. jstor.org/pss/1970814)
Sep 25, 2010 at 9:11 comment added BCnrd Assume $U$ is q-compact; equivalently, choose $I$ finitely generated. Inclusion $j:U \rightarrow X := {\rm{Spec}}(A)$ is a q-compact open immersion, so $j_{\ast}(F)$ is q-coh. on $X$ for all q-coh. $F$ on $U$. Restricting back to $U$ gives $F$, so $F = \widetilde{M}|_U$ for an $A$-module $M$. Then by excision ${\rm{H}}^i(U,F) = {\rm{H}}^i(U,\widetilde{M}) = {\rm{H}}^{i+1}_{X-U}(X,\widetilde{M}) = \injlim {\rm{Ext}}^{i+1}(A/I^n,M)$ (limit over $n \rightarrow \infty$), final equality since $I$ f. gen'td (univ. $\delta$-functor argument). Vanishing for $i > 0$ and all $M$ seems "impractical"...
Sep 25, 2010 at 9:04 comment added Martin Brandenburg This was discussed here: mathoverflow.net/questions/20782/…
Sep 25, 2010 at 8:53 history asked unknown CC BY-SA 2.5