Timeline for Explicit formula for the j-invariant of binary quartic form
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 3 at 13:43 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title, while this is on the front page
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| May 3 at 7:55 | answer | added | user165013 | timeline score: 2 | |
| Jul 8, 2010 at 13:55 | vote | accept | David Marín | ||
| Jul 7, 2010 at 22:32 | answer | added | Abdelmalek Abdesselam | timeline score: 13 | |
| Jul 7, 2010 at 14:23 | comment | added | Robin Chapman | If you have access to a computer algebra system, you can do the following. Let $\xi$ denote a solution to $f(1,\xi)=0$ where $f$ is your quartic. Then $f(X,Y+\xi X)=b'X^3Y+\cdots+Y^4$. The elliptic curve is now isomorphic to $y^2=b'x^3+c'x^2+d'x+1$. Transform it to the usual Weierstrass form and take the $j$-invariant. Note that $b'$ etc. will have $\xi$s in them, but they should all cancel out via the equation $f(1,\xi)=0$ in the final result. | |
| Jul 7, 2010 at 14:00 | comment | added | David Marín | Yes, that is exactly what I am looking for. | |
| Jul 7, 2010 at 13:54 | comment | added | Robin Chapman | Do you mean the j-invariant of the elliptic curve $y^2=ax^4+bx^3+cx^2+dx+1$? | |
| Jul 7, 2010 at 13:51 | history | asked | David Marín | CC BY-SA 2.5 |