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May 24, 2010 at 2:20 history edited Emerton CC BY-SA 2.5
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Mar 2, 2010 at 15:01 comment added Emerton Dear Bjorn, I think you're right.
Mar 2, 2010 at 5:56 comment added Bjorn Poonen All Andrea is really using is that E.f^* C = 0, which is easy because C can be replaced by a linearly equivalent divisor disjoint from P. Aren't you implicitly using the same kind of argument to know that D_1.D_2 = C_1.C_2 ?
Mar 2, 2010 at 3:33 comment added Emerton Dear Bjorn, I saw that, but thought it might be helpful to have a computation that didn't use anything (counting the push-pull formula as something).
Mar 2, 2010 at 2:50 comment added Bjorn Poonen @David Speyer and Emerton: See the last two paragraphs of Andrea Ferretti's version of the computation: it's simpler to move one curve instead of two. Maybe somebody should consolidate all these answers into one community wiki answer?
Mar 2, 2010 at 2:11 history edited Emerton CC BY-SA 2.5
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Mar 2, 2010 at 0:55 history edited Emerton CC BY-SA 2.5
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Mar 1, 2010 at 22:03 history answered Emerton CC BY-SA 2.5