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Aug 7, 2014 at 14:41 vote accept Hugo Chapdelaine
Aug 6, 2014 at 22:40 answer added Will Sawin timeline score: 2
Aug 6, 2014 at 19:49 comment added Hugo Chapdelaine I thought a bit a bout my problem and now I realized that the way I set it up is probably not ideal. It is probably better to work with homology since then one can make pictures. The group $H_2(E,\mathbf{Z})$ has $3$ natural $\mathbf{Z}$-linearly independant elements, namely $E_1=E\times\{0\}$, $E_2=\{0\}\times E$ and $\Delta$ (the diagonal). Intuitively we should have $E_2\cdot \Gamma=1$ and $E_1\cdot\Gamma=D$. Though the intersection $\Delta\cdot \Gamma$ seems to be more complicated to compute.
Aug 6, 2014 at 15:20 history edited Hugo Chapdelaine CC BY-SA 3.0
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Aug 6, 2014 at 15:09 history edited GH from MO
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S Aug 6, 2014 at 14:13 history suggested Jesper Petersen
added elliptic curves tag
Aug 6, 2014 at 14:10 review Suggested edits
S Aug 6, 2014 at 14:13
Aug 6, 2014 at 13:49 history edited Hugo Chapdelaine CC BY-SA 3.0
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Aug 6, 2014 at 13:24 history asked Hugo Chapdelaine CC BY-SA 3.0