Timeline for Genus of Y^3 = X^4 - 1.
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2013 at 16:49 | comment | added | Ian Agol | Once you know it is non singular, the genus is the number of lattice points in the interior of the Newton polygon. | |
| Feb 26, 2013 at 14:36 | answer | added | Alex Suciu | timeline score: 1 | |
| Feb 26, 2013 at 14:03 | comment | added | François Brunault | Indeed, the computation of singular points should be done by hand with this kind of equation. For "generic" curves, Magma or equivalent free programs might be the best way to go. | |
| Feb 26, 2013 at 13:34 | answer | added | Alex Fok | timeline score: 3 | |
| Feb 26, 2013 at 12:35 | comment | added | Joe Silverman | Possibly "teach me" means "explain how to compute", rather than "tell me the answer". One can check that a curve of this sort is nonsingular (as a projective curve) in a minute or two by hand, one really doesn't need Magma. Indeed, it's a nice exercise in a first-year algebraic geometry course to compute the genus of X^n + Y^m = 1 (by hand!). The sequence of blow-ups needed to resolve the singularity mimics the Euclidean algorithm used to compute gcd(m,n). | |
| Feb 26, 2013 at 11:05 | history | edited | Gerry Myerson |
re-tagged
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| Feb 26, 2013 at 10:53 | comment | added | François Brunault | Magma tells me that the projective closure of this curve has no singular points, so it is a smooth quartic curve and has genus 3. | |
| Feb 26, 2013 at 10:10 | comment | added | Adrien Hardy | "the genus" of a planar curve ? | |
| Feb 26, 2013 at 9:54 | comment | added | Francesco Polizzi | "math-philosophy?" :-) | |
| Feb 26, 2013 at 9:49 | history | asked | Pierre MATSUMI | CC BY-SA 3.0 |