Timeline for Homomorphism of Legendre curve
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2010 at 10:51 | history | edited | Robin Chapman | CC BY-SA 2.5 |
spelling correction
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| Apr 11, 2010 at 8:51 | answer | added | Bjorn Poonen | timeline score: 6 | |
| Apr 11, 2010 at 8:14 | answer | added | Robin Chapman | timeline score: 2 | |
| Apr 11, 2010 at 5:18 | comment | added | Pete L. Clark | That's a kind of strange question, honestly. The j-invariant of any elliptic curve in Weierstrass form is given by $j(E) = c_4^3/\Delta$, where $c_4$ and $\Delta$ are explicit polynomials in the coefficients: see e.g. en.wikipedia.org/wiki/J-invariant So you just multiply out $(x-a)*(x-b)*(x-c)$ and apply the formula. Or you get a computer algebra package to do this for you... | |
| Apr 11, 2010 at 5:16 | comment | added | S. Carnahan♦ | Take the cross-ratio of $a,b,c,\infty$ to get $\lambda$, and use the formula for $j$. | |
| Apr 11, 2010 at 5:09 | answer | added | Pete L. Clark | timeline score: 3 | |
| Apr 11, 2010 at 4:30 | comment | added | Josh | How do we calculate the j-invariant of y^2 = (x-a)(x-b)(x-c)? | |
| Apr 11, 2010 at 4:28 | history | asked | Josh | CC BY-SA 2.5 |