Timeline for A K3 over $P^1$ with six singular $A_1$- fibers?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2011 at 19:40 | comment | added | Noam D. Elkies | You're welcome. Of course I in turned committed a typo: the last factor should be $x_0^3 + 8 x_1^3$, not $x_0^3 + 8 x_0^3$... | |
| Sep 25, 2011 at 17:37 | history | edited | JME | CC BY-SA 3.0 |
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| Sep 25, 2011 at 17:35 | comment | added | Noam D. Elkies | Wait, $x_0^8 + x_1^8 = 0$ gives a regular octagon, not an octahedron (which has six vertices, not eight). If you want 8 points with octahedral symmetry, you're looking for vertices of a cube, e.g. the roots of $x_0 x_1 (x_0^3-x_1^3) (x_0^3 + 8 x_0^3)$. | |
| Sep 25, 2011 at 17:03 | history | edited | JME | CC BY-SA 3.0 |
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| Sep 25, 2011 at 16:34 | history | edited | JME | CC BY-SA 3.0 |
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| Sep 25, 2011 at 16:06 | history | answered | JME | CC BY-SA 3.0 |