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Mar 1, 2016 at 17:43 comment added Sándor Kovács Yes. It's the same as $H^0$ of $R^1$. You should be able to extract this from the above proof.
Mar 1, 2016 at 17:39 history edited Sándor Kovács CC BY-SA 3.0
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Mar 1, 2016 at 17:13 comment added Ron Thank you very much for the very detailed and helpful answer. I also wanted to ask, does this also mean $H^1(\mathcal{O}_{X\backslash S}) \not-0$?
Mar 1, 2016 at 17:11 vote accept Ron
Feb 29, 2016 at 22:48 history answered Sándor Kovács CC BY-SA 3.0