Timeline for Why is it a non-basepoint?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Jul 12, 2011 at 23:49 | comment | added | Emerton | By the very definition of linear equivalence, $n$ times a hyperplane and $S$ are linearly equivalent in $\mathbb P^3$, and so their intersections with $C$ are linearly equivalent too. Subtracting $P_1,\ldots,P_{i−1}$ from each of these intersections then also gives linearly equivalent divisors. | |
| Jul 12, 2011 at 22:49 | comment | added | phil | Thank you for your answer. You mean that the points cut on $C$ by $S$ and different from $P_1,..., P_{i-1}$ form a divisor linearly equivalent to $nD-P_1-...-P_{i-1}$. But can you explain me, why these two are linearly equivalent. I do not get it immediately. | |
| Jul 12, 2011 at 21:56 | history | answered | Jack Huizenga | CC BY-SA 3.0 |